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0475
Determine a equação geral da circunferência λ \lambda λ , que se encontra no 2° quadrante, tangencia os eixos coordenados e tem raio( R ) (R) ( R ) igual a 3.
0475 - Resposta
λ : x 2 + y 2 + 6 x − 6 y + 9 = 0 \lambda: x^2+y^2+6x-6y+9=0 λ : x 2 + y 2 + 6 x − 6 y + 9 = 0
0475 - Solução
Pelas informações, temos o centro em C ( − 3 ; 3 ) C(-3;\,3) C ( − 3 ; 3 ) e, além disso, R = 3 R=3 R = 3
Assim, a equação reduzida será: λ : ( x + 3 ) 2 + ( y − 3 ) 2 = 9 \lambda: (x+3)^2+(y-3)^2=9 λ : ( x + 3 ) 2 + ( y − 3 ) 2 = 9
Para encontrarmos a equação geral, basta desenvolver a equação reduzida, agrupando os termos semelhantes e igualando todas as parcelas a zero; assim:
λ : x 2 + 6 x + 9 + y 2 − 6 y + 9 − 9 = 0 → \lambda: x^2+6x+9+y^2-6y+9-9=0\to λ : x 2 + 6 x + 9 + y 2 − 6 y + 9 − 9 = 0 →
λ : x 2 + y 2 + 6 x − 6 y + 9 = 0 \boxed{\lambda: x^2+y^2+6x-6y+9=0} λ : x 2 + y 2 + 6 x − 6 y + 9 = 0 (resposta final)
0474
Calcule:
a) 0 , 3 − 1 4 − 1 5 + 0 , 036 ÷ 0 , 04 \dfrac{0,3-\frac{1}{4}}{\sqrt[5]{-1}}+0,036\div 0,04 5 − 1 0 , 3 − 4 1 + 0 , 036 ÷ 0 , 04
b) 2 28 + 2 30 10 3 \sqrt[3]{\dfrac{2^{28}+2^{30}}{10}} 3 10 2 28 + 2 30
c) 3 + 1 3 − 1 + 3 − 1 3 + 1 \dfrac{\sqrt{3}+1}{\sqrt{3}-1}+\dfrac{\sqrt{3}-1}{\sqrt{3}+1} 3 − 1 3 + 1 + 3 + 1 3 − 1
0474 - Respostas
a) 0 , 85 0,85\quad 0 , 85 b) 2 9 2^9\quad 2 9 c) 4 4 4
0474 - Soluções
a) 0 , 05 − 1 + 0 , 9 → − 0 , 05 + 0 , 9 = 0 , 85 \dfrac{0,05}{-1}+0,9\to -0,05+0,9=0,85 − 1 0 , 05 + 0 , 9 → − 0 , 05 + 0 , 9 = 0 , 85
2 28 . ( 1 + 4 ) 10 3 → ( 2 9 ) 3 . 2 . 5 10 2 3 = 2 9 \sqrt[3]{\dfrac{2^{28}\,.\,(1+4)}{10}}\to\sqrt[\cancel{3}]{\dfrac{(2^9)^{\cancel{3}}\,.\,\cancel{2}\,.\,\cancel{5}}{\cancel{10}^{\,\,\cancel{2}}}}=2^9 3 10 2 28 . ( 1 + 4 ) → 3 10 2 ( 2 9 ) 3 . 2 . 5 = 2 9
c) ( 3 + 1 ) 2 + ( 3 − 1 ) 2 2 = 2 ( 2 + 3 + 2 − 3 ) 2 = 4 \dfrac{(\sqrt{3}+1)^2+(\sqrt{3}-1)^2}{2}=\dfrac{\cancel{2}(2+\cancel{\sqrt{3}}+2-\cancel{\sqrt{3}})}{\cancel{2}}=4 2 ( 3 + 1 ) 2 + ( 3 − 1 ) 2 = 2 2 ( 2 + 3 + 2 − 3 ) = 4
0473
Em cada cado, expresse 1 u \dfrac{1}{u} u 1 , em função de "v v v "
a) u = v − 1 u=v^{-1} u = v − 1
b) u = 1 v + 1 2 u=\dfrac{1}{\frac{v+1}{2}} u = 2 v + 1 1
c) u = 2 + 1 v u=2+\dfrac{1}{v} u = 2 + v 1
d) u = 1 v + 1 2 u=\dfrac{1}{v}+\dfrac{1}{2} u = v 1 + 2 1
0473 - Respostas
a) v v\quad v b) v + 1 2 \dfrac{v+1}{2}\quad 2 v + 1 c) v 2 v + 1 \dfrac{v}{2v+1}\quad 2 v + 1 v d) 2 v v + 2 \dfrac{2v}{v+2} v + 2 2 v
0473 - Soluções
a) u = v − 1 1 v → 1 u = 1 v − 1 = v u=v^{-1}\quad\underrightarrow{\frac{1}{v}}\quad\dfrac{1}{u}=\dfrac{1}{v^{-1}}=v u = v − 1 v 1 u 1 = v − 1 1 = v
b) u = 1 v + 1 2 1 v → 1 u = v + 1 2 u=\dfrac{1}{\frac{v+1}{2}}\quad\underrightarrow{\frac{1}{v}}\quad\dfrac{1}{u}=\dfrac{v+1}{2} u = 2 v + 1 1 v 1 u 1 = 2 v + 1
c) u = 2 + 1 v 1 v → 1 u = v 2 v + 1 u=2+\dfrac{1}{v}\quad\underrightarrow{\frac{1}{v}}\quad\dfrac{1}{u}=\dfrac{v}{2v+1} u = 2 + v 1 v 1 u 1 = 2 v + 1 v
d) u = 1 v + 1 2 1 v → 1 u = 2 v v + 2 u=\dfrac{1}{v}+\dfrac{1}{2}\quad\underrightarrow{\frac{1}{v}}\quad\dfrac{1}{u}=\dfrac{2v}{v+2} u = v 1 + 2 1 v 1 u 1 = v + 2 2 v
0472
Desenvolva
a) ( 3 a + 2 b ) 2 (3a + 2b)^2 ( 3 a + 2 b ) 2
b) ( 3 a + 2 b ) 3 (3a + 2b)^3 ( 3 a + 2 b ) 3
c) ( 3 a − 2 b ) 3 (3a − 2b)^3 ( 3 a − 2 b ) 3
d) ( x 2 − 1 ) ⋅ ( x 2 + 1 ) (x^2 − 1)\cdot(x^2 + 1) ( x 2 − 1 ) ⋅ ( x 2 + 1 )
e) [ ( x − y ) + 1 ] ⋅ [ ( x − y ) − 1 ] [(x − y) + 1]\cdot[(x − y) − 1] [( x − y ) + 1 ] ⋅ [( x − y ) − 1 ]
f) ( a + b + c ) 2 (a + b + c)^2 ( a + b + c ) 2
0472 - Soluções
a) ( 3 a + 2 b ) 2 = 9 a 2 + 12 a b + 4 b 2 (3a + 2b)^2=9a^2+12ab+4b^2 ( 3 a + 2 b ) 2 = 9 a 2 + 12 ab + 4 b 2
b) ( 3 a + 2 b ) 3 = 27 a 3 + 54 a 2 b + 36 a b 2 + 8 b 3 (3a + 2b)^3=27a^3+54a^2b+36ab^2+8b^3 ( 3 a + 2 b ) 3 = 27 a 3 + 54 a 2 b + 36 a b 2 + 8 b 3
c) ( 3 a − 2 b ) 3 = 27 a 3 − 54 a 2 b + 36 a b 2 − 8 b 3 (3a − 2b)^3=27a^3-54a^2b+36ab^2-8b^3 ( 3 a − 2 b ) 3 = 27 a 3 − 54 a 2 b + 36 a b 2 − 8 b 3
d) ( x 2 − 1 ) ⋅ ( x 2 + 1 ) = x 4 − 1 (x^2 − 1)\cdot(x^2 + 1)=x^4-1 ( x 2 − 1 ) ⋅ ( x 2 + 1 ) = x 4 − 1
e) [ ( x − y ) + 1 ] ⋅ [ ( x − y ) − 1 ] = ( x − y ) 2 − 1 = x 2 − 2 x y + y 2 − 1 [(x − y) + 1]\cdot[(x − y) − 1]=(x-y)^2-1=x^2-2xy+y^2-1 [( x − y ) + 1 ] ⋅ [( x − y ) − 1 ] = ( x − y ) 2 − 1 = x 2 − 2 x y + y 2 − 1
f) ( a + b + c ) 2 = [ ( a + b ) + c ] 2 = (a + b + c)^2=[(a+b)+c]^2= ( a + b + c ) 2 = [( a + b ) + c ] 2 =
a 2 + 2 a b + b 2 + 2 ⋅ ( a + b ) ⋅ c + c 2 = \quad a^2+2ab+b^2+2\cdot(a+b)\cdot c+c^2= a 2 + 2 ab + b 2 + 2 ⋅ ( a + b ) ⋅ c + c 2 =
a 2 + b 2 + c 2 + 2 a b + 2 a c + 2 b c \quad a^2+ b^2+ c^2+2ab+2ac+2bc a 2 + b 2 + c 2 + 2 ab + 2 a c + 2 b c
0471
Mostre que
a) ( x + y ) 2 = x 2 + y 2 (x+y)^2=x^2+ y^2 ( x + y ) 2 = x 2 + y 2 se e somente se x = 0 x=0\,\, x = 0 ou y = 0 \,\,y=0 y = 0
b) ( x + y ) 3 = x 3 + y 3 (x+y)^3=x^3+y^3 ( x + y ) 3 = x 3 + y 3 se e somente se x = 0 x=0\,\, x = 0 ou y = 0 \,\, y=0\,\, y = 0 ou x = − y \,\, x=-y x = − y
0471 - Soluções
a) ( x + y ) 2 − x 2 − y 2 = 0 → \,\,(x+y)^2-x^2-y^2=0\to ( x + y ) 2 − x 2 − y 2 = 0 →
x 2 − x 2 + y 2 − y 2 + 2 x y = 0 ⏟ → \quad\cancel{x^2}-\cancel{x^2}+\cancel{y^2}-\cancel{y^2}+\underbrace{2xy=0}\to x 2 − x 2 + y 2 − y 2 + 2 x y = 0 →
2 x y = 0 → x = 0 ou y = 0 \quad 2xy=0\to\boxed{x=0}\,\,\text{ou}\,\,\boxed{y=0} 2 x y = 0 → x = 0 ou y = 0
b) ( x + y ) 3 − x 3 − y 3 = 0 → (x+y)^3-x^3-y^3=0\to ( x + y ) 3 − x 3 − y 3 = 0 →
x 3 − x 3 + y 3 − y 3 + 3 x 2 y + 3 x y 2 = 0 ⏟ → \quad \cancel{x^3}-\cancel{x^3}+\cancel{y^3}-\cancel{y^3}+\underbrace{3x^2y+3xy^2=0}\to x 3 − x 3 + y 3 − y 3 + 3 x 2 y + 3 x y 2 = 0 →
3 x y ( x + y ) = 0 → x = 0 \quad 3xy(x+y)=0\to\boxed{x=0}\,\, 3 x y ( x + y ) = 0 → x = 0 ou y = 0 \,\,\boxed{y=0}\,\, y = 0 ou x = − y \,\,\boxed{x=-y} x = − y
0470
Fatore
a) 4 y 2 − 16 4y^2-16 4 y 2 − 16
b) ( x + b ) 2 − a 2 (x+b)^2-a^2 ( x + b ) 2 − a 2
c) a 2 x + b 2 y + a 2 y + b 2 x a^2x + b^2y + a^2y + b^2x a 2 x + b 2 y + a 2 y + b 2 x
d) 2 x 2 − x + 4 x y − 2 y 2x^2-x+4xy-2y 2 x 2 − x + 4 x y − 2 y
e) x 2 − a 2 − 2 a b − b 2 x^2-a^2-2ab-b^2 x 2 − a 2 − 2 ab − b 2
f) x 2 − 6 x + 9 − y 2 x^2-6x+9-y^2 x 2 − 6 x + 9 − y 2
g) x 3 + 1 x 3 x^3+\dfrac{1}{x^3} x 3 + x 3 1
h) x 6 + 1 x^6+1 x 6 + 1
0470 - Soluções
a) 4 y 2 − 16 = 4 ( y 2 − 4 ) = 4 ( y − 2 ) ( 2 y + 2 ) 4y^2-16=4(y^2-4)=4(y-2)(2y+2) 4 y 2 − 16 = 4 ( y 2 − 4 ) = 4 ( y − 2 ) ( 2 y + 2 )
b) ( x + b ) 2 − a 2 = ( x + b − a ) ( x + b + a ) (x+b)^2-a^2=(x+b-a)(x+b+a) ( x + b ) 2 − a 2 = ( x + b − a ) ( x + b + a )
c) a 2 x + b 2 y + a 2 y + b 2 x = a^2x + b^2y + a^2y + b^2x= a 2 x + b 2 y + a 2 y + b 2 x =
a 2 ( x + y ) + b 2 ( x + y ) = \quad a^2(x+y)+b^2(x+y)= a 2 ( x + y ) + b 2 ( x + y ) =
( a 2 + b 2 ) ( x + y ) \quad (a^2+b^2)(x+y) ( a 2 + b 2 ) ( x + y )
d) 2 x 2 − x + 4 x y − 2 y = 2x^2-x+4xy-2y= 2 x 2 − x + 4 x y − 2 y =
2 x ( x + 2 y ) − ( x + 2 y ) = \quad 2x(x+2y)-(x+2y)= 2 x ( x + 2 y ) − ( x + 2 y ) =
( x + 2 y ) ( 2 x − 1 ) \quad (x+2y)(2x-1) ( x + 2 y ) ( 2 x − 1 )
e) x 2 − a 2 − 2 a b − b 2 = x 2 − ( a + b ) 2 = x^2-a^2-2ab-b^2=x^2-(a+b)^2= x 2 − a 2 − 2 ab − b 2 = x 2 − ( a + b ) 2 =
( x − a − b ) ( x + a + b ) \quad (x-a-b)(x+a+b) ( x − a − b ) ( x + a + b )
f) x 2 − 6 x + 9 − y 2 = ( x − 3 ) 2 − y 2 = x^2-6x+9-y^2=(x-3)^2-y^2= x 2 − 6 x + 9 − y 2 = ( x − 3 ) 2 − y 2 =
( x − y − 3 ) ( x + y − 3 ) \quad (x-y-3)(x+y-3) ( x − y − 3 ) ( x + y − 3 )
g) x 3 + 1 x 3 = ( x + 1 x ) ( x 2 − 1 + 1 x 2 ) x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)\left(x^2-1+\dfrac{1}{x^2}\right) x 3 + x 3 1 = ( x + x 1 ) ( x 2 − 1 + x 2 1 )
h) x 6 + 1 = ( x 2 ) 3 + 1 1 3 = x^6+1=(x^2)^3+\dfrac{1}{1^3}= x 6 + 1 = ( x 2 ) 3 + 1 3 1 =
( x 2 + 1 ) ( x 4 − x 2 + 1 ) \quad(x^2+1)(x^4-x^2+1) ( x 2 + 1 ) ( x 4 − x 2 + 1 )
0469
Simplifique as expressões
a) 2 ( x − 2 ) ( x − 3 ) 3 − 3 ( x − 2 ) 2 ( x − 3 ) 2 ( x − 3 ) 6 \dfrac{2(x-2)(x-3)^3-3(x-2)^2(x-3)^2}{(x-3)^6} ( x − 3 ) 6 2 ( x − 2 ) ( x − 3 ) 3 − 3 ( x − 2 ) 2 ( x − 3 ) 2
b) x 2 + 1 x 2 − 1 − 1 x − 1 \dfrac{x^2+1}{x^2-1}-\dfrac{1}{x-1} x 2 − 1 x 2 + 1 − x − 1 1
0469 - Respostas
a) x ( 2 − x ) ( x − 3 ) 4 \dfrac{x(2-x)}{(x-3)^4}\quad ( x − 3 ) 4 x ( 2 − x ) b) x x + 1 \dfrac{x}{x+1} x + 1 x
0469 - Soluções
a) 2 ( x − 2 ) ( x − 3 ) 3 − 3 ( x − 2 ) 2 ( x − 3 ) 2 ( x − 3 ) 6 = \dfrac{2(x-2)(x-3)^3-3(x-2)^2(x-3)^2}{(x-3)^6}= ( x − 3 ) 6 2 ( x − 2 ) ( x − 3 ) 3 − 3 ( x − 2 ) 2 ( x − 3 ) 2 =
[ ( x − 2 ) ( x − 3 ) 2 ] ⋅ [ 2 ( x − 3 ) − 3 ( x − 2 ) ] ( x − 3 ) 6 = \quad \dfrac{[(x-2)(x-3)^2]\cdot[2(x-3)-3(x-2)]}{(x-3)^6}= ( x − 3 ) 6 [( x − 2 ) ( x − 3 ) 2 ] ⋅ [ 2 ( x − 3 ) − 3 ( x − 2 )] =
[ ( x − 2 ) ( x − 3 ) 2 ] ⋅ ( 2 x − 6 − 3 x − 6 ) ⏞ − x ( x − 3 ) 6 = \quad\dfrac{[(x-2)(x-3)^2]\cdot(\overbrace{2x-6-3x-6)}^{-x}}{(x-3)^6}= ( x − 3 ) 6 [( x − 2 ) ( x − 3 ) 2 ] ⋅ ( 2 x − 6 − 3 x − 6 ) − x =
− x ( x − 2 ) ( x − 3 ) 2 ( x − 3 ) 6 = x ( 2 − x ) ( x − 3 ) 4 \quad\dfrac{-x(x-2)\cancel{(x-3)^2}}{(x-3)^{\cancel{6}}}=\dfrac{x(2-x)}{(x-3)^4} ( x − 3 ) 6 − x ( x − 2 ) ( x − 3 ) 2 = ( x − 3 ) 4 x ( 2 − x )
b) x 2 + 1 x 2 − 1 − 1 x − 1 = x 2 + 1 − ( x + 1 ) ( x − 1 ) ( x + 1 ) = \dfrac{x^2+1}{x^2-1}-\dfrac{1}{x-1}=\dfrac{x^2+1-(x+1)}{(x-1)(x+1)}= x 2 − 1 x 2 + 1 − x − 1 1 = ( x − 1 ) ( x + 1 ) x 2 + 1 − ( x + 1 ) =
x 2 − x x 2 − 1 = x ( x − 1 ) ( x + 1 ) ( x − 1 ) = x x + 1 \quad\dfrac{x^2-x}{x^2-1}=\dfrac{x\cancel{(x-1)}}{(x+1)\cancel{(x-1)}}=\dfrac{x}{x+1} x 2 − 1 x 2 − x = ( x + 1 ) ( x − 1 ) x ( x − 1 ) = x + 1 x
0468
Sabendo que a + 1 a = b a+\dfrac{1}{a}=b a + a 1 = b , determine, em função de "b b b "
a) a 2 + 1 a 2 a^2+\dfrac{1}{a^2} a 2 + a 2 1
b) a 3 + 1 a 3 a^3+\dfrac{1}{a^3} a 3 + a 3 1
c) a 4 + 1 a 4 a^4+\dfrac{1}{a^4} a 4 + a 4 1
0468 - Respostas
a) b 2 − 2 b^2-2\quad b 2 − 2 b) b 3 − 3 b b^3-3b\quad b 3 − 3 b c) b 4 − 4 b 2 + 2 b^4-4b^2+2 b 4 − 4 b 2 + 2
0468 - Soluções
a) Façamos ( a + 1 a ) 2 = b 2 → \left(a+\dfrac{1}{a}\right)^2=b^2\to ( a + a 1 ) 2 = b 2 →
a 2 + 1 a 2 + 2 = b 2 → \quad a^2+\dfrac{1}{a^2}+2=b^2\to a 2 + a 2 1 + 2 = b 2 →
a 2 + 1 a 2 = b 2 − 2 \quad a^2+\dfrac{1}{a^2}=b^2-2 a 2 + a 2 1 = b 2 − 2
b) Façamos ( a + 1 a ) 3 = b 3 → \left(a+\dfrac{1}{a}\right)^3=b^3\to ( a + a 1 ) 3 = b 3 →
( a 3 + 3 a + 3 a + 1 a 3 ) = b 3 → \quad \left(a^3+3a+\dfrac{3}{a}+\dfrac{1}{a^3}\right)=b^3\to ( a 3 + 3 a + a 3 + a 3 1 ) = b 3 →
a 3 + 1 a 3 + 3 ( a + 1 a ) ⏟ 3 b = b 3 → \quad a^3+\dfrac{1}{a^3}+\underbrace{3\left(a+\dfrac{1}{a}\right)}_{3b}=b^3\to a 3 + a 3 1 + 3 b 3 ( a + a 1 ) = b 3 →
a 3 + 1 a 3 = b 3 − 3 b \quad a^3+\dfrac{1}{a^3}=b^3-3b a 3 + a 3 1 = b 3 − 3 b
c) Utilizando Binômio de Newton, vamos desenvolver:
( a + 1 a ) 4 = b 4 → \quad\left(a+\dfrac{1}{a}\right)^4=b^4\to ( a + a 1 ) 4 = b 4 →
a 4 + 1 a 4 + 4 a 2 + 4 a 2 + 6 = b 4 → a^4+\dfrac{1}{a^4}+4a^2+\dfrac{4}{a^2}+6=b^4\to a 4 + a 4 1 + 4 a 2 + a 2 4 + 6 = b 4 →
a 4 + 1 a 4 + 4 ( a 2 + 1 a 2 ) ⏟ b 2 − 2 = b 4 − 6 → \quad a^4+\dfrac{1}{a^4}+4\underbrace{\left(a^2+\dfrac{1}{a^2}\right)}_{b^2-2}=b^4-6\to a 4 + a 4 1 + 4 b 2 − 2 ( a 2 + a 2 1 ) = b 4 − 6 →
a 4 + 1 a 4 = b 4 − 6 − 4 ( b 2 − 2 ) → \quad a^4+\dfrac{1}{a^4}=b^4-6-4(b^2-2)\to a 4 + a 4 1 = b 4 − 6 − 4 ( b 2 − 2 ) →
a 4 + 1 a 4 = b 4 − 4 b 2 + 2 \quad a^4+\dfrac{1}{a^4}=b^4-4b^2+2 a 4 + a 4 1 = b 4 − 4 b 2 + 2
0467
Escreva cada expressão usando apenas um radical e simplifique
a) x \sqrt{\sqrt{x}} x
b) x \sqrt{\sqrt{\sqrt{x}}} x
c) 25 x 3 \sqrt{\sqrt[3]{25x}} 3 25 x
d) x ⋅ x 3 \sqrt{x}\cdot\sqrt[3]{x} x ⋅ 3 x
e) x y 5 x y 3 \dfrac{\sqrt[5]{xy}}{\sqrt[3]{xy}} 3 x y 5 x y
f) x y 5 x 3 ⋅ y \dfrac{\sqrt[5]{xy}}{\sqrt[3]{x}\cdot\sqrt{y}} 3 x ⋅ y 5 x y
0467 - Soluções
a) x = x 2 × 2 = x 4 \sqrt{\sqrt{x}}=\sqrt[2\times 2]{x}=\sqrt[4]{x} x = 2 × 2 x = 4 x
b) x = x 8 \sqrt{\sqrt{\sqrt{x}}}=\sqrt[8]{x} x = 8 x
c) 25 x 3 = 5 x 3 = \sqrt{\sqrt[3]{25x}}=\sqrt[3]{5\sqrt{x}}= 3 25 x = 3 5 x =
5 3 ⋅ x 6 \quad\sqrt[3]{5}\cdot\sqrt[6]{x} 3 5 ⋅ 6 x
d) x ⋅ x 3 = x 1 2 ⋅ x 1 3 = \sqrt{x}\cdot\sqrt[3]{x}=x^{\frac{1}{2}}\cdot x^{\frac{1}{3}}= x ⋅ 3 x = x 2 1 ⋅ x 3 1 =
x 1 2 + 1 3 = x 5 6 = x 5 6 \quad x^{\frac{1}{2}+\frac{1}{3}}=x^{\frac{5}{6}}=\sqrt[6]{x^5} x 2 1 + 3 1 = x 6 5 = 6 x 5
e) x y 5 x y 3 = x 1 5 ⋅ y 1 5 x 1 3 ⋅ y 1 3 = \dfrac{\sqrt[5]{xy}}{\sqrt[3]{xy}}=\dfrac{x^{\frac{1}{5}}\cdot y^{\frac{1}{5}}}{x^{\frac{1}{3}}\cdot y^{\frac{1}{3}}}= 3 x y 5 x y = x 3 1 ⋅ y 3 1 x 5 1 ⋅ y 5 1 =
x 1 5 − 1 3 ⋅ y 1 5 − 1 3 = x − 2 15 ⋅ y − 2 15 = \quad x^{\frac{1}{5}-\frac{1}{3}}\cdot y^{\frac{1}{5}-\frac{1}{3}}=x^{-\frac{2}{15}}\cdot y^{-\frac{2}{15}}= x 5 1 − 3 1 ⋅ y 5 1 − 3 1 = x − 15 2 ⋅ y − 15 2 =
1 x 2 15 ⋅ y 2 15 = 1 x 2 y 2 15 \quad \dfrac{1}{x^{\frac{2}{15}}\cdot y^{\frac{2}{15}}}=\dfrac{1}{\sqrt[15]{x^2y^2}} x 15 2 ⋅ y 15 2 1 = 15 x 2 y 2 1
f) x y 5 x 3 ⋅ y = x 1 5 ⋅ y 1 5 x 1 3 ⋅ y 1 2 = \dfrac{\sqrt[5]{xy}}{\sqrt[3]{x}\cdot\sqrt{y}}=\dfrac{x^{\frac{1}{5}}\cdot y^{\frac{1}{5}}}{x^{\frac{1}{3}}\cdot y^{\frac{1}{2}}}= 3 x ⋅ y 5 x y = x 3 1 ⋅ y 2 1 x 5 1 ⋅ y 5 1 =
x 1 5 − 1 3 ⋅ y 1 5 − 1 2 = x − 2 15 ⋅ y − 3 10 = \quad x^{\frac{1}{5}-\frac{1}{3}}\cdot y^{\frac{1}{5}-\frac{1}{2}}=x^{-\frac{2}{15}}\cdot y^{-\frac{3}{10}}= x 5 1 − 3 1 ⋅ y 5 1 − 2 1 = x − 15 2 ⋅ y − 10 3 =
1 x 2 15 ⋅ y 3 10 = 1 x 2 15 ⋅ y 3 10 \quad \dfrac{1}{x^{\frac{2}{15}}\cdot y^{\frac{3}{10}}}=\dfrac{1}{\sqrt[15]{x^2}\cdot\sqrt[10]{y^3}} x 15 2 ⋅ y 10 3 1 = 15 x 2 ⋅ 10 y 3 1
0466
Simplifique as expressões ( ( ( em que a , b > 0 ) a,\,b\,>\,0) a , b > 0 )
a) a 3 5 ⋅ a 2 7 a 1 3 \dfrac{a^{\frac{3}{5}}\cdot a^{\frac{2}{7}}}{a^{\frac{1}{3}}} a 3 1 a 5 3 ⋅ a 7 2
b) a 2 5 ⋅ b 3 4 ⋅ ( 3 a ) 2 b 3 5 ⋅ a 1 3 \dfrac{a^{\frac{2}{5}}\cdot b^{\frac{3}{4}}\cdot(3a)^2}{b^{\frac{3}{5}}\cdot a^{\frac{1}{3}}} b 5 3 ⋅ a 3 1 a 5 2 ⋅ b 4 3 ⋅ ( 3 a ) 2
c) ( a 9 ⋅ b 6 ) − 1 3 ( a 6 ⋅ b 4 ) − 1 2 \dfrac{\left(a^9\cdot b^6\right)^{-\frac{1}{3}}}{\left(a^6\cdot b^4\right)^{-\frac{1}{2}}} ( a 6 ⋅ b 4 ) − 2 1 ( a 9 ⋅ b 6 ) − 3 1
d) ( a 2 ⋅ b 4 ) 1 2 ( 81 a 6 ⋅ b 9 ) 1 3 \dfrac{\left(a^2\cdot b^4\right)^{\frac{1}{2}}}{\left(81a^6\cdot b^9\right)^{\frac{1}{3}}} ( 81 a 6 ⋅ b 9 ) 3 1 ( a 2 ⋅ b 4 ) 2 1
0466 - Respostas
a) a 58 105 a^{\frac{58}{105}}\quad a 105 58 b) 9 ⋅ a 31 15 ⋅ b 3 20 9\cdot a^{\frac{31}{15}}\cdot b^{\frac{3}{20}}\quad 9 ⋅ a 15 31 ⋅ b 20 3 c) 1 1\quad 1 c) 1 3 3 3 ⋅ a ⋅ b \dfrac{1}{3\sqrt[3]{3}\cdot a\cdot b} 3 3 3 ⋅ a ⋅ b 1
0466 - Soluções
a) a 3 5 ⋅ a 2 7 a 1 3 = a 3 5 ⋅ a 2 7 ⋅ a − 1 3 = \dfrac{a^{\frac{3}{5}}\cdot a^{\frac{2}{7}}}{a^{\frac{1}{3}}}=a^{\frac{3}{5}}\cdot a^{\frac{2}{7}}\cdot a^{-\frac{1}{3}}= a 3 1 a 5 3 ⋅ a 7 2 = a 5 3 ⋅ a 7 2 ⋅ a − 3 1 =
a 3 5 + 2 7 − 1 3 = a 63 + 30 − 35 105 = a 58 105 \quad a^{\frac{3}{5}+\frac{2}{7}-\frac{1}{3}}=a^{\frac{63+30-35}{105}}=a^{\frac{58}{105}} a 5 3 + 7 2 − 3 1 = a 105 63 + 30 − 35 = a 105 58
b) a 2 5 ⋅ b 3 4 ⋅ ( 3 a ) 2 b 3 5 ⋅ a 1 3 = a 2 5 ⋅ b 3 4 ⋅ 9 a 2 ⋅ b − 3 5 ⋅ a − 1 3 = \dfrac{a^{\frac{2}{5}}\cdot b^{\frac{3}{4}}\cdot(3a)^2}{b^{\frac{3}{5}}\cdot a^{\frac{1}{3}}}=a^{\frac{2}{5}}\cdot b^{\frac{3}{4}}\cdot 9a^2\cdot b^{-\frac{3}{5}}\cdot a^{-\frac{1}{3}}= b 5 3 ⋅ a 3 1 a 5 2 ⋅ b 4 3 ⋅ ( 3 a ) 2 = a 5 2 ⋅ b 4 3 ⋅ 9 a 2 ⋅ b − 5 3 ⋅ a − 3 1 =
9 ⋅ a 2 5 + 2 − 1 3 ⋅ b 3 4 − 3 5 = 9 ⋅ a 31 15 ⋅ b 3 20 \quad 9\cdot a^{\frac{2}{5}+2-\frac{1}{3}}\cdot b^{\frac{3}{4}-\frac{3}{5}}=9\cdot a^{\frac{31}{15}}\cdot b^{\frac{3}{20}} 9 ⋅ a 5 2 + 2 − 3 1 ⋅ b 4 3 − 5 3 = 9 ⋅ a 15 31 ⋅ b 20 3
c) ( a 9 ⋅ b 6 ) − 1 3 ( a 6 ⋅ b 4 ) − 1 2 = a − 3 ⋅ b − 2 a − 3 ⋅ b − 2 = 1 \dfrac{\left(a^9\cdot b^6\right)^{-\frac{1}{3}}}{\left(a^6\cdot b^4\right)^{-\frac{1}{2}}}=\dfrac{a^{-3}\cdot b^{-2}}{a^{-3}\cdot b^{-2}}=1 ( a 6 ⋅ b 4 ) − 2 1 ( a 9 ⋅ b 6 ) − 3 1 = a − 3 ⋅ b − 2 a − 3 ⋅ b − 2 = 1
d) ( a 2 ⋅ b 4 ) 1 2 ( 81 a 6 ⋅ b 9 ) 1 3 = a ⋅ b 2 81 3 ⋅ a 2 ⋅ b 3 = 1 3 3 3 ⋅ a ⋅ b \dfrac{\left(a^2\cdot b^4\right)^{\frac{1}{2}}}{\left(81a^6\cdot b^9\right)^{\frac{1}{3}}}=\dfrac{a\cdot b^2}{\sqrt[3]{81}\cdot a^2\cdot b^3}=\dfrac{1}{3\sqrt[3]{3}\cdot a\cdot b} ( 81 a 6 ⋅ b 9 ) 3 1 ( a 2 ⋅ b 4 ) 2 1 = 3 81 ⋅ a 2 ⋅ b 3 a ⋅ b 2 = 3 3 3 ⋅ a ⋅ b 1
0465
Simplifique
a) 4 x 3 y 2 ( x − 2 ) 4 6 x 2 y ( x − 2 ) 3 / 2 \dfrac{\dfrac{4x^3y^2}{(x-2)^4}}{\dfrac{6x^2y}{(x-2)^{3/2}}} ( x − 2 ) 3/2 6 x 2 y ( x − 2 ) 4 4 x 3 y 2
b) x 2 − y 2 3 x 2 y 5 y + x x y \dfrac{\dfrac{x^2-y^2}{3x^2y^5}}{\dfrac{y+x}{xy}} x y y + x 3 x 2 y 5 x 2 − y 2
c) 1 ( x + h ) 2 − 1 x 2 h \dfrac{\dfrac{1}{(x+h)^2}-\dfrac{1}{x^2}}{h} h ( x + h ) 2 1 − x 2 1
d) 1 a + 1 b b a − a b \dfrac{\dfrac{1}{a}+\dfrac{1}{b}}{\dfrac{b}{a}-\dfrac{a}{b}} a b − b a a 1 + b 1
e) ( z + w ) − 1 ( z − w ) − 1 \dfrac{(z+w)^{-1}}{(z-w)^{-1}} ( z − w ) − 1 ( z + w ) − 1
f) ( p − 1 + q − 1 ) − 1 \left(p^{-1}+q^{-1}\right)^{-1} ( p − 1 + q − 1 ) − 1
0465 - Respostas
0465 - Soluções
a) 4 x 3 y 2 ( x − 2 ) 4 6 x 2 y ( x − 2 ) 3 / 2 = 4 x 3 y 2 ⋅ ( x − 2 ) 3 / 2 6 x 2 y ⋅ ( x − 2 ) 4 = \dfrac{\dfrac{4x^3y^2}{(x-2)^4}}{\dfrac{6x^2y}{(x-2)^{3/2}}}=\dfrac{4x^3y^2\cdot(x-2)^{3/2}}{6x^2y\cdot(x-2)^4}= ( x − 2 ) 3/2 6 x 2 y ( x − 2 ) 4 4 x 3 y 2 = 6 x 2 y ⋅ ( x − 2 ) 4 4 x 3 y 2 ⋅ ( x − 2 ) 3/2 =
2 x y 3 ⋅ ( x − 2 ) 5 / 2 \quad\dfrac{2xy}{3\cdot(x-2)^{5/2}} 3 ⋅ ( x − 2 ) 5/2 2 x y ou 2 x y 3 ⋅ ( x − 2 ) 2 ⋅ x − 2 \dfrac{2xy}{3\cdot(x-2)^{2}\cdot\sqrt{x-2}} 3 ⋅ ( x − 2 ) 2 ⋅ x − 2 2 x y
b) x 2 − y 2 3 x 2 y 5 y + x x y = ( x + y ) ⋅ ( x − y ) ⋅ x y 3 x 2 y 5 ⋅ ( x + y ) = x − y 3 x y 4 \dfrac{\dfrac{x^2-y^2}{3x^2y^5}}{\dfrac{y+x}{xy}}=\dfrac{(\cancel{x+y})\cdot(x-y)\cdot xy}{3x^2y^5\cdot(\cancel{x+y})}=\dfrac{x-y}{3xy^4} x y y + x 3 x 2 y 5 x 2 − y 2 = 3 x 2 y 5 ⋅ ( x + y ) ( x + y ) ⋅ ( x − y ) ⋅ x y = 3 x y 4 x − y
c) 1 ( x + h ) 2 − 1 x 2 h = x 2 − ( x + h ) 2 h ⋅ ( x + h ) 2 ⋅ x 2 = \dfrac{\dfrac{1}{(x+h)^2}-\dfrac{1}{x^2}}{h}=\dfrac{x^2-(x+h)^2}{h\cdot(x+h)^2\cdot x^2}= h ( x + h ) 2 1 − x 2 1 = h ⋅ ( x + h ) 2 ⋅ x 2 x 2 − ( x + h ) 2 =
− h ( 2 x + h ) h ⋅ ( x + h ) 2 ⋅ x 2 = − ( 2 x + h ) ( x + h ) 2 ⋅ x 2 \quad \dfrac{-\cancel{h}(2x+h)}{\cancel{h}\cdot(x+h)^2\cdot x^2}=\dfrac{-(2x+h)}{(x+h)^2\cdot x^2} h ⋅ ( x + h ) 2 ⋅ x 2 − h ( 2 x + h ) = ( x + h ) 2 ⋅ x 2 − ( 2 x + h )
d) 1 a + 1 b b a − a b = a + b a b b 2 − a 2 a b = \dfrac{\dfrac{1}{a}+\dfrac{1}{b}}{\dfrac{b}{a}-\dfrac{a}{b}}=\dfrac{\dfrac{a+b}{\cancel{ab}}}{\dfrac{b^2-a^2}{\cancel{ab}}}= a b − b a a 1 + b 1 = ab b 2 − a 2 ab a + b =
( a + b ) ( a + b ) ⋅ ( b − a ) = 1 b − a \quad\dfrac{(\cancel{a+b})}{(\cancel{a+b})\cdot(b-a)}=\dfrac{1}{b-a} ( a + b ) ⋅ ( b − a ) ( a + b ) = b − a 1
e) ( z + w ) − 1 ( z − w ) − 1 = z − w z + w \dfrac{(z+w)^{-1}}{(z-w)^{-1}}=\dfrac{z-w}{z+w} ( z − w ) − 1 ( z + w ) − 1 = z + w z − w
f) ( p − 1 + q − 1 ) − 1 = 1 p + q p q = p q p + q \left(p^{-1}+q^{-1}\right)^{-1}=\dfrac{1}{\dfrac{p+q}{pq}}=\dfrac{pq}{p+q} ( p − 1 + q − 1 ) − 1 = pq p + q 1 = p + q pq
0464
Efetue as seguintes divisões de polinômios
a) ( 5 x 2 + 4 x + 2 ) ÷ ( 6 x + 2 ) (5x^2 + 4x + 2) \div (6x + 2) ( 5 x 2 + 4 x + 2 ) ÷ ( 6 x + 2 )
b) ( x 2 + x − 2 ) ÷ ( x − 1 ) (x^2 + x − 2) \div (x − 1) ( x 2 + x − 2 ) ÷ ( x − 1 )
c) ( x 2 − a 2 ) ÷ ( x − a ) (x^2 − a^2) \div (x − a) ( x 2 − a 2 ) ÷ ( x − a )
d) ( x 4 − 256 ) ÷ ( x − 4 ) (x^4 − 256) \div (x − 4) ( x 4 − 256 ) ÷ ( x − 4 )
e) ( x 4 − a 4 ) ÷ ( x 3 + x 2 a + x a 2 + a 3 ) (x^4 − a^4) \div (x^3 + x^2 a + xa^2 + a^3) ( x 4 − a 4 ) ÷ ( x 3 + x 2 a + x a 2 + a 3 )
f) ( x 5 + x 3 − 2 ) ÷ ( x − 1 ) (x^5 + x^3 − 2) \div (x − 1) ( x 5 + x 3 − 2 ) ÷ ( x − 1 )
g) ( 4 x 3 + 2 x + 1 ) ÷ ( x + 1 ) (4x^3 + 2x + 1) \div (x + 1) ( 4 x 3 + 2 x + 1 ) ÷ ( x + 1 )
h) x 3 ÷ ( x − a ) x^3\div(x-a) x 3 ÷ ( x − a )
0464 - Respostas
a) 5 x 2 + 4 x + 2 = ( 6 x + 2 ) ⋅ ( 5 x 6 + 7 18 ) + 11 9 5x^2+4x+2=(6x+2)\cdot\left(\frac{5x}{6}+\frac{7}{18}\right)+\frac{11}{9} 5 x 2 + 4 x + 2 = ( 6 x + 2 ) ⋅ ( 6 5 x + 18 7 ) + 9 11
b) x 2 + x − 2 = ( x − 1 ) ⋅ ( x + 2 ) + 0 x^2+x-2=(x-1)\cdot(x+2)+0 x 2 + x − 2 = ( x − 1 ) ⋅ ( x + 2 ) + 0
c) x 2 − a 2 = ( x − a ) ⋅ ( x + a ) + 0 x^2-a^2=(x-a)\cdot(x+a)+0 x 2 − a 2 = ( x − a ) ⋅ ( x + a ) + 0
d) x 4 − 256 = ( x − 4 ) ⋅ ( x + 4 ) ⋅ ( x 2 + 16 ) + 0 x^4-256=(x-4)\cdot(x+4)\cdot(x^2+16)+0 x 4 − 256 = ( x − 4 ) ⋅ ( x + 4 ) ⋅ ( x 2 + 16 ) + 0
e) x 4 − a 4 = ( x 3 + x 2 a + x a 2 + a 3 ) ⋅ ( x − a ) + 0 x^4-a^4=(x^3 + x^2 a + xa^2 + a^3)\cdot(x-a)+0 x 4 − a 4 = ( x 3 + x 2 a + x a 2 + a 3 ) ⋅ ( x − a ) + 0
f) x 5 + x 3 − 2 = ( x − 1 ) ⋅ ( x 4 + x 3 + 2 x 2 + 2 x + 2 ) + 0 x^5+x^3-2=(x-1)\cdot(x^4+x^3+ 2x^2+2x+2)+0 x 5 + x 3 − 2 = ( x − 1 ) ⋅ ( x 4 + x 3 + 2 x 2 + 2 x + 2 ) + 0
g) 4 x 3 + 2 x + 1 = ( x + 1 ) ⋅ ( 4 x 2 − 4 x + 6 ) − 5 4x^3+2x+1=(x+1)\cdot(4x^2-4x+6)-5 4 x 3 + 2 x + 1 = ( x + 1 ) ⋅ ( 4 x 2 − 4 x + 6 ) − 5
h) x 3 = ( x − a ) ⋅ ( x 2 + a x + a 2 ) + a 3 x^3=(x-a)\cdot(x^2+ax+a^2)+a^3 x 3 = ( x − a ) ⋅ ( x 2 + a x + a 2 ) + a 3
0464 - Soluções
a) Divisão por Chave para a divisão( 5 x 2 + 4 x + 2 ) ÷ ( 6 x + 2 ) (5x^2 + 4x + 2) \div (6x + 2) ( 5 x 2 + 4 x + 2 ) ÷ ( 6 x + 2 ) :
5 x 2 + 4 x + 2 ⏞ dividendo ∣ 6 x + 2 ‾ ⏞ divisor \,\,\,\,\overbrace{\cancel{5x^2}+4x+2}^{\text{dividendo}}\quad\quad |\overbrace{\underline{6x+2}}^{\text{divisor}} 5 x 2 + 4 x + 2 dividendo ∣ 6 x + 2 divisor
− 5 x 2 − 5 x 3 ‾ 5 x 6 + 7 18 } ← quociente \underline{-\cancel{5x^2}-\frac{5x}{3}}\quad\quad\quad\quad \frac{5x}{6}+\frac{7}{18}\Big\}\leftarrow\text{quociente} − 5 x 2 − 3 5 x 6 5 x + 18 7 } ← quociente
7 x 3 + 2 \quad\quad\quad\,\, \cancel{\frac{7x}{3}}+2 3 7 x + 2
− 7 x 3 − 7 9 ‾ \quad\quad\,\,\,\underline{-\cancel{\frac{7x}{3}}-\frac{7}{9}} − 3 7 x − 9 7
11 9 } ← resto \quad\quad\quad\quad\quad\frac{11}{9}\big\}\leftarrow\text{resto} 9 11 } ← resto
Portanto: 5 x 2 + 4 x + 2 = ( 6 x + 2 ) ⋅ ( 5 x 6 + 7 18 ) + 11 9 5x^2+4x+2=(6x+2)\cdot\left(\frac{5x}{6}+\frac{7}{18}\right)+\frac{11}{9} 5 x 2 + 4 x + 2 = ( 6 x + 2 ) ⋅ ( 6 5 x + 18 7 ) + 9 11
b) Algoritmo de Briot-Ruffini para a divisão ( x 2 + x − 2 ) ÷ ( x − 1 ) (x^2 + x − 2) \div (x − 1) ( x 2 + x − 2 ) ÷ ( x − 1 ) :
1 ‾ ∣ 1 1 − 2 ‾ \quad\underline{1} |\underline{\,\,1\quad\,\, 1\quad -2} 1 ∣ 1 1 − 2
∣ 1 2 0 \quad\,\,\,|\,\,1\quad\,\,2\quad\quad 0 ∣ 1 2 0
Portanto: x 2 + x − 2 = ( x − 1 ) ⋅ ( x + 2 ) + 0 x^2+x-2=(x-1)\cdot(x+2)+0 x 2 + x − 2 = ( x − 1 ) ⋅ ( x + 2 ) + 0
c) Farotação para a divisão ( x 2 − a 2 ) ÷ ( x − a ) (x^2 − a^2) \div (x − a) ( x 2 − a 2 ) ÷ ( x − a ) :
( x + a ) ( x − a ) ( x − a ) = x − a \dfrac{(\cancel{x+a})(x-a)}{(\cancel{x-a})}=x-a ( x − a ) ( x + a ) ( x − a ) = x − a
Portanto: x 2 − a 2 = ( x − a ) ⋅ ( x + a ) + 0 x^2-a^2=(x-a)\cdot(x+a)+0 x 2 − a 2 = ( x − a ) ⋅ ( x + a ) + 0
d) Fatoração para a divisão ( x 4 − 256 ) ÷ ( x − 4 ) (x^4 − 256) \div (x − 4) ( x 4 − 256 ) ÷ ( x − 4 ) :
( x − 4 ) ⋅ ( x + 4 ) ⋅ ( x 2 + 16 ) ( x − 4 ) = ( x + 4 ) ⋅ ( x 2 + 16 ) \dfrac{(\cancel{x-4})\cdot(x+4)\cdot(x^2+16)}{(\cancel{x-4})}=(x+4)\cdot(x^2+16) ( x − 4 ) ( x − 4 ) ⋅ ( x + 4 ) ⋅ ( x 2 + 16 ) = ( x + 4 ) ⋅ ( x 2 + 16 )
Portanto: x 4 − 256 = ( x − 4 ) ⋅ ( x + 4 ) ⋅ ( x 2 + 16 ) + 0 x^4-256=(x-4)\cdot(x+4)\cdot(x^2+16)+0 x 4 − 256 = ( x − 4 ) ⋅ ( x + 4 ) ⋅ ( x 2 + 16 ) + 0
e) Fatoração para a divisão( x 4 − a 4 ) ÷ ( x 3 + x 2 a + x a 2 + a 3 ) (x^4 − a^4) \div (x^3 + x^2 a + xa^2 + a^3) ( x 4 − a 4 ) ÷ ( x 3 + x 2 a + x a 2 + a 3 ) :
( x 2 + a 2 ) ⋅ ( x + a ) ⋅ ( x − a ) x 2 ( x + a ) + a 2 ( x + a ) = \dfrac{(x^2+a^2)\cdot(x+a)\cdot(x-a)}{x^2(x+a)+a^2(x+a)}= x 2 ( x + a ) + a 2 ( x + a ) ( x 2 + a 2 ) ⋅ ( x + a ) ⋅ ( x − a ) =
( x 2 + a 2 ) ⋅ ( x + a ) ⋅ ( x − a ) ( x 2 + a 2 ) ⋅ ( x + a ) = x − a \dfrac{(\cancel{x^2+a^2})\cdot(\cancel{x+a})\cdot(x-a)}{(\cancel{x^2+a^2})\cdot(\cancel{x+a})}=x-a ( x 2 + a 2 ) ⋅ ( x + a ) ( x 2 + a 2 ) ⋅ ( x + a ) ⋅ ( x − a ) = x − a
Portanto: x 4 − a 4 = ( x 3 + x 2 a + x a 2 + a 3 ) ⋅ ( x − a ) + 0 x^4-a^4=(x^3 + x^2 a + xa^2 + a^3)\cdot(x-a)+0 x 4 − a 4 = ( x 3 + x 2 a + x a 2 + a 3 ) ⋅ ( x − a ) + 0
f) Algoritmo de Briot-Ruffini para a divisão ( x 5 + x 3 − 2 ) ÷ ( x − 1 ) (x^5 + x^3 − 2) \div (x − 1) ( x 5 + x 3 − 2 ) ÷ ( x − 1 ) :
1 ‾ ∣ 1 0 1 0 0 − 2 ‾ \quad\underline{1} |\underline{\,\,1\quad\,\, 0\quad 1\quad 0\quad 0\quad -2} 1 ∣ 1 0 1 0 0 − 2
∣ 1 1 2 2 2 0 \quad\,\,\,|\,\,1\quad\,\,1\quad 2\quad 2\quad 2\quad\quad\,\, 0 ∣ 1 1 2 2 2 0
Portanto: x 5 + x 3 − 2 = ( x − 1 ) ⋅ ( x 4 + x 3 + 2 x 2 + 2 x + 2 ) + 0 x^5+x^3-2=(x-1)\cdot(x^4+x^3+ 2x^2+2x+2)+0 x 5 + x 3 − 2 = ( x − 1 ) ⋅ ( x 4 + x 3 + 2 x 2 + 2 x + 2 ) + 0
g) Algoritmo de Briot-Ruffini para a divisão ( 4 x 3 + 2 x + 1 ) ÷ ( x + 1 ) (4x^3 + 2x + 1) \div (x + 1) ( 4 x 3 + 2 x + 1 ) ÷ ( x + 1 ) :
− 1 ‾ ∣ 4 0 2 1 ‾ \quad\underline{-1} |\underline{\,\,4\quad\,\,\,\,0\quad 2\quad\,\,\,\,\,1} − 1 ∣ 4 0 2 1
∣ 4 − 4 6 − 5 \quad\quad\,\,|\,\,4\,\,\,-4\quad 6\,\,\, -5 ∣ 4 − 4 6 − 5
Portanto: 4 x 3 + 2 x + 1 = ( x + 1 ) ⋅ ( 4 x 2 − 4 x + 6 ) − 5 4x^3+2x+1=(x+1)\cdot(4x^2-4x+6)-5 4 x 3 + 2 x + 1 = ( x + 1 ) ⋅ ( 4 x 2 − 4 x + 6 ) − 5
h) Fatoração para a divisão x 3 ÷ ( x − a ) x^3\div(x-a) x 3 ÷ ( x − a ) :
Se x 3 − a 3 = ( x − a ) ⋅ ( x 2 + a x + a 2 ) x^3-a^3=(x-a)\cdot (x^2+ax+a^2) x 3 − a 3 = ( x − a ) ⋅ ( x 2 + a x + a 2 )
Então: x 3 = ( x − a ) ⋅ ( x 2 + a x + a 2 ) + a 3 x^3=(x-a)\cdot(x^2+ax+a^2)+a^3 x 3 = ( x − a ) ⋅ ( x 2 + a x + a 2 ) + a 3
0463
De um recipiente cheio de água tiram-se 2 3 \frac{2}{3} 3 2 do seu conteúdo. Colocando 30 litros de água o conteúdo passa a ocupar a metade do volume inicial. Determine a capacidade do recipiente.
0463 - Resposta
180 180 180 litros
0463 - Solução
Vamos chamar de "x x x " o valor do conteúdo total desse recipiente. Equacionando o texto dado, encontraremos a capacidade pedida; assim:
x − 2 x 3 + 30 = x 2 → 6 x − 4 x + 180 = 3 x 6 → x-\dfrac{2x}{3}+30=\dfrac{x}{2}\to\dfrac{6x-4x+180=3x}{\cancel{6}}\to x − 3 2 x + 30 = 2 x → 6 6 x − 4 x + 180 = 3 x →
6 x − 4 x − 3 x = − 180 → − x = − 180 → x = 180 6x-4x-3x=-180\to -x=-180\to\boxed{x=180} 6 x − 4 x − 3 x = − 180 → − x = − 180 → x = 180
Portanto, a capacidade do recipiente é de 180 180 180 litros.
0462
Determine o domínio e resolva as seguintes equações em R \mathbb{R} R
a) x x − 2 + 4 x − 1 = 5 \dfrac{x}{x-2}+\dfrac{4}{x-1}=5 x − 2 x + x − 1 4 = 5
b) 2 x 2 − 1 − x x − 1 = 1 \dfrac{2}{x^2-1}-\dfrac{x}{x-1}=1 x 2 − 1 2 − x − 1 x = 1
0462 - Respostas
0462 - Soluções
O domínio( D ) (D) ( D ) de uma função são os valores de "x x x " que proporcionam a existência(ou não) dessas equações.
a) D = { x ∈ R ∣ x ≠ 1 e x ≠ 2 } D=\{x\in\mathbb{R}\,|\,x\neq 1\,\,\text{e}\,\,x\neq 2\} D = { x ∈ R ∣ x = 1 e x = 2 }
Solução da equação: x x − 2 + 4 x − 1 = 5 \dfrac{x}{x-2}+\dfrac{4}{x-1}=5 x − 2 x + x − 1 4 = 5
x ( x − 1 ) + 4 ( x − 2 ) = 5 ( x − 1 ) ( x − 2 ) ( x − 2 ) ( x − 1 ) → \dfrac{x(x-1)+4(x-2)=5(x-1)(x-2)}{\cancel{(x-2)(x-1)}}\to ( x − 2 ) ( x − 1 ) x ( x − 1 ) + 4 ( x − 2 ) = 5 ( x − 1 ) ( x − 2 ) →
x 2 − x + 4 x − 8 = 5 ( x 2 − 3 x + 2 ) → x^2-x+4x-8=5(x^2-3x+2)\to x 2 − x + 4 x − 8 = 5 ( x 2 − 3 x + 2 ) →
x 2 − 5 x 2 − x + 4 x + 15 x − 8 − 10 = 0 → x^2-5x^2-x+4x+15x-8-10=0\to x 2 − 5 x 2 − x + 4 x + 15 x − 8 − 10 = 0 →
− 4 x 2 + 18 x − 18 = 0 [ ÷ ( − 2 ) ] → -4x^2+18x-18=0\,\,[\div(-2)]\to − 4 x 2 + 18 x − 18 = 0 [ ÷ ( − 2 )] →
2 x 2 − 9 x + 9 = 0 → 2x^2-9x+9=0\to 2 x 2 − 9 x + 9 = 0 → Fórmula Quadrática
x = − b ± b 2 − 4 × a × c 2 × a → x=\dfrac{-b\pm\sqrt{b^2-4\times a\times c}}{2\times a}\to x = 2 × a − b ± b 2 − 4 × a × c →
x = − ( − 9 ) ± ( − 9 ) 2 − 4 × 2 × 9 2 × 2 → x=\dfrac{-(-9)\pm\sqrt{(-9)^2-4\times 2\times 9}}{2\times 2}\to x = 2 × 2 − ( − 9 ) ± ( − 9 ) 2 − 4 × 2 × 9 →
x = 9 ± 9 4 → x = 9 ± 3 4 → x=\dfrac{9\pm\sqrt{9}}{4}\to x=\dfrac{9\pm 3}{4}\to x = 4 9 ± 9 → x = 4 9 ± 3 →
x 1 = 9 − 3 4 → x 1 = 3 2 x_{1}=\dfrac{9-3}{4}\to\boxed{x_{1}=\dfrac{3}{2}} x 1 = 4 9 − 3 → x 1 = 2 3
x 2 = 9 + 3 4 → x 2 = 3 x_{2}=\dfrac{9+3}{4}\to\boxed{x_{2}=3} x 2 = 4 9 + 3 → x 2 = 3
Portanto, a solução( S ) (S) ( S ) será: S = { 3 2 ; 3 } S=\left\{\dfrac{3}{2};\,\,3\right\} S = { 2 3 ; 3 }
b) D = { x ∈ R ∣ x ≠ − 1 e x ≠ 1 } D=\{x\in\mathbb{R}\,|\, x\neq -1\,\,\text{e}\,\,x\neq 1\} D = { x ∈ R ∣ x = − 1 e x = 1 }
Solução da equação: 2 x 2 − 1 − x x − 1 = 1 \dfrac{2}{x^2-1}-\dfrac{x}{x-1}=1 x 2 − 1 2 − x − 1 x = 1
2 − x ( x + 1 ) = 1 ( x 2 − 1 ) x 2 − 1 → \dfrac{2-x(x+1)=1(x^2-1)}{\cancel{x^2-1}}\to x 2 − 1 2 − x ( x + 1 ) = 1 ( x 2 − 1 ) →
2 − x 2 − x − x 2 + 1 = 0 → − 2 x 2 − x + 3 = 0 [ ÷ ( − 1 ) ] → 2-x^2-x-x^2+1=0\to -2x^2-x+3=0\,\,[\div(-1)]\to 2 − x 2 − x − x 2 + 1 = 0 → − 2 x 2 − x + 3 = 0 [ ÷ ( − 1 )] →
2 x 2 + x − 3 = 0 → 2x^2+x-3=0\to 2 x 2 + x − 3 = 0 → Fórmula Quadrática
x = − b ± b 2 − 4 × a × c 2 × a → x=\dfrac{-b\pm\sqrt{b^2-4\times a\times c}}{2\times a}\to x = 2 × a − b ± b 2 − 4 × a × c →
x = − 1 ± 1 2 − 4 × 2 × 3 2 × 2 → x=\dfrac{-1\pm\sqrt{1^2-4\times 2\times 3}}{2\times 2}\to x = 2 × 2 − 1 ± 1 2 − 4 × 2 × 3 →
x = − 1 ± 25 4 → x = − 1 ± 5 4 → x=\dfrac{-1\pm \sqrt{25}}{4}\to x=\dfrac{-1\pm 5}{4}\to x = 4 − 1 ± 25 → x = 4 − 1 ± 5 →
x 1 = − 1 + 5 4 → x 1 = − 1 x_{1}=\dfrac{-1+5}{4}\to \cancel{x_{1}=-1}\quad x 1 = 4 − 1 + 5 → x 1 = − 1 Não serve, pois − 1 ∉ D -1\not\in\,\,D − 1 ∈ D
x 2 = − 1 − 5 4 → x 2 = − 3 2 x_{2}=\dfrac{-1-5}{4}\to\boxed{x_{2}=-\dfrac{3}{2}} x 2 = 4 − 1 − 5 → x 2 = − 2 3
Portanto, a solução( S ) (S) ( S ) será: S = { − 3 2 } S=\left\{-\dfrac{3}{2}\right\} S = { − 2 3 }
0461
Determine o domínio e resolva as seguintes equações em Z \mathbb{Z} Z
a) x + 1 x − 1 + 4 − x x = 4 \dfrac{x+1}{x-1}+\dfrac{4-x}{x}=4 x − 1 x + 1 + x 4 − x = 4
b) 2 x ( x − 2 ) = 1 + x − 1 x − 2 \dfrac{2}{x(x-2)}=1+\dfrac{x-1}{x-2} x ( x − 2 ) 2 = 1 + x − 2 x − 1
0461 - Respostas
0461 - Soluções
O domínio( D ) (D) ( D ) de uma função são os valores de "x x x " que proporcionam a existência(ou não) dessas equações.
a) D = { x ∈ Z / x ≠ 0 e x ≠ 1 } D=\{x\in\mathbb{Z}\,/\,x\neq 0\,\,\text{e}\,\,x\neq 1\} D = { x ∈ Z / x = 0 e x = 1 }
Solução da equação: x + 1 x − 1 + 4 − x x = 4 \dfrac{x+1}{x-1}+\dfrac{4-x}{x}=4 x − 1 x + 1 + x 4 − x = 4
x ( x + 1 ) + ( 4 − x ) ( x − 1 ) = 4 x ( x − 1 ) x ( x − 1 ) → \dfrac{x(x+1)+(4-x)(x-1)=4x(x-1)}{\cancel{x(x-1)}}\to x ( x − 1 ) x ( x + 1 ) + ( 4 − x ) ( x − 1 ) = 4 x ( x − 1 ) →
x 2 + x + 4 x − 4 − x 2 + x − 4 x 2 + 4 x = 0 → \cancel{x^2}+x+4x-4-\cancel{x^2}+x-4x^2+4x=0\to x 2 + x + 4 x − 4 − x 2 + x − 4 x 2 + 4 x = 0 →
− 4 x 2 + 10 x − 4 = 0 [ ÷ ( − 2 ) ] → -4x^2+10x-4=0\,\,[\div(-2)]\to − 4 x 2 + 10 x − 4 = 0 [ ÷ ( − 2 )] →
2 x 2 − 5 x + 2 = 0 → 2x^2-5x+2=0\to 2 x 2 − 5 x + 2 = 0 → Fórmula Quadrática
x = − b ± b 2 − 4 × a × c 2 × a → x=\dfrac{-b\pm\sqrt{b^2-4\times a\times c}}{2\times a}\to x = 2 × a − b ± b 2 − 4 × a × c →
x = − ( − 5 ) ± ( − 5 ) 2 − 4 × 2 × 2 2 × 2 → x=\dfrac{-(-5)\pm\sqrt{(-5)^2-4\times 2\times 2}}{2\times 2}\to x = 2 × 2 − ( − 5 ) ± ( − 5 ) 2 − 4 × 2 × 2 →
x = 5 ± 9 4 → x = 5 ± 3 4 → x=\dfrac{5\pm\sqrt{9}}{4}\to x=\dfrac{5\pm3}{4}\to x = 4 5 ± 9 → x = 4 5 ± 3 →
x 1 = 5 − 3 4 → x 1 = 1 2 x_{1}=\dfrac{5-3}{4}\to \cancel{x_{1}=\dfrac{1}{2}}\quad x 1 = 4 5 − 3 → x 1 = 2 1 Não serve, pois 1 2 ∉ Z \dfrac{1}{2}\not\in\mathbb{Z} 2 1 ∈ Z
x 2 = 5 + 3 4 → x 2 = 2 x_{2}=\dfrac{5+3}{4}\to\boxed{x_{2}=2} x 2 = 4 5 + 3 → x 2 = 2
Portanto, a solução( S ) (S) ( S ) será: S = { 2 } S=\{2\} S = { 2 }
b) D = { x ∈ Z / x ≠ 0 e x ≠ 2 } D=\{x\in\mathbb{Z}\,/\,x\neq 0\,\,\text{e}\,\,x\neq 2\} D = { x ∈ Z / x = 0 e x = 2 }
Solução da equação: 2 x ( x − 2 ) = 1 + x − 1 x − 2 \dfrac{2}{x(x-2)}=1+\dfrac{x-1}{x-2} x ( x − 2 ) 2 = 1 + x − 2 x − 1
2 = x ( x − 2 ) + x ( x − 1 ) x ( x − 2 ) → \dfrac{2=x(x-2)+x(x-1)}{\cancel{x(x-2)}}\to x ( x − 2 ) 2 = x ( x − 2 ) + x ( x − 1 ) →
2 = x 2 − 2 x + x 2 − x → 2=x^2-2x+x^2-x\to 2 = x 2 − 2 x + x 2 − x →
2 x 2 − 3 x − 2 = 0 → 2x^2-3x-2=0\to 2 x 2 − 3 x − 2 = 0 → Fórmula Quadrática
x = − b ± b 2 − 4 × a × c 2 × a → x=\dfrac{-b\pm\sqrt{b^2-4\times a\times c}}{2\times a}\to x = 2 × a − b ± b 2 − 4 × a × c →
x = − ( − 3 ) ± ( − 3 ) 2 − 4 × 2 × ( − 2 ) 2 × 2 → x=\dfrac{-(-3)\pm\sqrt{(-3)^2-4\times 2\times (-2)}}{2\times 2}\to x = 2 × 2 − ( − 3 ) ± ( − 3 ) 2 − 4 × 2 × ( − 2 ) →
x = 3 ± 25 4 → x = 3 ± 5 4 → x=\dfrac{3\pm\sqrt{25}}{4}\to x=\dfrac{3\pm 5}{4}\to x = 4 3 ± 25 → x = 4 3 ± 5 →
x 1 = 3 − 5 4 → x = − 1 2 x_{1}=\dfrac{3-5}{4}\to \cancel{x=-\dfrac{1}{2}}\quad x 1 = 4 3 − 5 → x = − 2 1 Não serve, pois − 1 2 ∉ Z -\dfrac{1}{2}\not\in\mathbb{Z} − 2 1 ∈ Z
x 2 = 3 + 5 4 → x = 2 x_{2}=\dfrac{3+5}{4}\to \cancel{x=2}\quad x 2 = 4 3 + 5 → x = 2 Não serve, pois 2 ∉ D 2\not\in\,\,D 2 ∈ D
Portanto, a solução( S ) (S) ( S ) será: S = ∅ S=\varnothing S = ∅
0460
Resolva, em R \mathbb{R} R , a equação:
2 x 2 − 1 + 1 x + 1 = − 1 \dfrac{2}{x^2-1}+\dfrac{1}{x+1}=-1 x 2 − 1 2 + x + 1 1 = − 1
0460 - Resposta
S = { 0 } S=\{0\} S = { 0 }
0460 - Solução
Observe que o domínio( D ) (D) ( D ) dessa equação é: D = { x ∈ R / x ≠ − 1 e x ≠ 1 } D=\{x\in\mathbb{R}\,/\,x\neq -1\,\,\text{e}\,\,x\neq 1\} D = { x ∈ R / x = − 1 e x = 1 }
2 + x − 1 = − ( x 2 − 1 ) x 2 − 1 → \dfrac{2+x-1=-(x^2-1)}{\cancel{x^2-1}}\to x 2 − 1 2 + x − 1 = − ( x 2 − 1 ) →
x 2 + x = 0 x^2+x=0 x 2 + x = 0 ou x ( x + 1 ) = 0 x(x+1)=0\quad x ( x + 1 ) = 0 Duas possibilidades:
x = 0 \boxed{x=0} x = 0 ou
x = − 1 \cancel{x=-1}\quad x = − 1 Não serve, pois − 1 ∉ D -1\not\in\,\,D − 1 ∈ D
Portanto, a solução( S ) (S) ( S ) , será: S = { 0 } S=\{0\} S = { 0 }
0459
Determine o domínio e resolva as seguintes equações em R \mathbb{R} R
a) x = 2 x \sqrt{x}=2x x = 2 x
b) x = − 2 x \sqrt{x}=-2x x = − 2 x
c) 3 − x = x − 3 \sqrt{3-x}=x-3 3 − x = x − 3
d) x + 1 = 8 − 3 x + 1 \sqrt{x+1}=8-\sqrt{3x+1} x + 1 = 8 − 3 x + 1
e) 1 + 3 x + 5 = x 1+\sqrt{3x+5}=x 1 + 3 x + 5 = x
f) 4 x − 3 + 5 x − 1 = 15 x + 4 \sqrt{4x-3}+\sqrt{5x-1}=\sqrt{15x+4} 4 x − 3 + 5 x − 1 = 15 x + 4
g) x + 34 3 − x − 3 3 = 1 \sqrt[3]{x+34}-\sqrt[3]{x-3}=1 3 x + 34 − 3 x − 3 = 1
0459 - Soluções
O domínio( D ) (D) ( D ) são os valores de "x x x " que proporcionam a existência(ou não) dessas equações irracionais. Importante lembrar, também, que as soluções desses tipos de equações devem ser testadas (o que chamamos de teste de verificação ), para termos uma maior segurança ao definirmos o conjunto solução. Posto isto, vamos às soluções:
a) D = { x ∈ R / x ⩾ 0 } D=\{x\in\mathbb{R}\,\,/\,\,x\geqslant 0\} D = { x ∈ R / x ⩾ 0 }
Solução da equação: x = 2 x \sqrt{x}=2x x = 2 x
( x ) 2 = ( 2 x ) 2 → x = 4 x 2 → \left(\sqrt{x}\right)^2=(2x)^2\to x=4x^2\to ( x ) 2 = ( 2 x ) 2 → x = 4 x 2 →
4 x 2 − x = 0 4x^2-x=0 4 x 2 − x = 0 ou x ( 4 x − 1 ) = 0 → x(4x-1)=0\to x ( 4 x − 1 ) = 0 →
Duas possibilidades:
x = 0 \boxed{x=0} x = 0 ou
4 x − 1 = 0 → x = 1 4 4x-1=0\to \boxed{x=\dfrac{1}{4}} 4 x − 1 = 0 → x = 4 1
Portanto, a solução( S ) (S) ( S ) , será: S = { 0 ; 1 4 } S=\left\{0;\,\,\dfrac{1}{4}\right\} S = { 0 ; 4 1 }
b) D = { x ∈ R / x ⩾ 0 } D=\{x\in\mathbb{R}\,/\,x\geqslant 0\} D = { x ∈ R / x ⩾ 0 }
Solução da equação: x = − 2 x \sqrt{x}=-2x x = − 2 x
( x ) 2 = ( − 2 x ) 2 \left(\sqrt{x}\right)^2=(-2x)^2 ( x ) 2 = ( − 2 x ) 2
4 x 2 − x = 0 4x^2-x=0 4 x 2 − x = 0 ou x ( 4 x − 1 ) = 0 x(4x-1)=0 x ( 4 x − 1 ) = 0
Duas possibilidades:
x = 0 \boxed{x=0} x = 0 ou
4 x − 1 = 0 → x = 1 4 4x-1=0\to \cancel{x=\dfrac{1}{4}}\quad 4 x − 1 = 0 → x = 4 1 Não serve, pois este valor não é válido no teste de verificação .
Portanto, a solução( S ) (S) ( S ) , será: S = { 0 } S=\{0\} S = { 0 }
c) D = { x ∈ R / x ⩽ 3 } D=\{x\in\mathbb{R}\,/\,x\leqslant 3\} D = { x ∈ R / x ⩽ 3 }
Solução da equação: 3 − x = x − 3 \sqrt{3-x}=x-3 3 − x = x − 3
( 3 − x ) 2 = ( x − 3 ) 2 → \left(\sqrt{3-x}\right)^2=(x-3)^2\to ( 3 − x ) 2 = ( x − 3 ) 2 →
3 − x = x 2 − 6 x + 9 → 3-x=x^2-6x+9\to 3 − x = x 2 − 6 x + 9 →
x 2 − 5 x + 6 = 0 → x^2-5x+6=0\to\quad x 2 − 5 x + 6 = 0 → Fórmula Quadrática
x = − b ± b 2 − 4 × a × c 2 × a → x=\dfrac{-b\pm\sqrt{b^2-4\times a\times c}}{2\times a}\to x = 2 × a − b ± b 2 − 4 × a × c →
x = − ( − 5 ) ± ( − 5 ) 2 − 4 × 1 × 6 2 × 1 → x=\dfrac{-(-5)\pm\sqrt{(-5)^2-4\times 1\times 6}}{2\times 1}\to x = 2 × 1 − ( − 5 ) ± ( − 5 ) 2 − 4 × 1 × 6 →
x = 5 ± 1 2 → x = 5 ± 1 2 x=\dfrac{5\pm\sqrt{1}}{2}\to x=\dfrac{5\pm 1}{2} x = 2 5 ± 1 → x = 2 5 ± 1
x 1 = 5 + 1 2 → x = 3 x_{1}=\dfrac{5+1}{2}\to\boxed{x=3} x 1 = 2 5 + 1 → x = 3
x 2 = 5 − 1 2 → x = 2 x_{2}=\dfrac{5-1}{2}\to \cancel{x=2}\quad x 2 = 2 5 − 1 → x = 2 Não serve, pois 2 2 2 não é válido no teste de verificação .
Portanto, a solução( S ) (S) ( S ) , será: S = { 3 } S=\{3\} S = { 3 }
d) D = { x ∈ R / x ⩾ − 1 3 } D=\left\{x\in\mathbb{R}\,/\,x\geqslant-\dfrac{1}{3}\right\} D = { x ∈ R / x ⩾ − 3 1 }
Solução da equação: x + 1 = 8 − 3 x + 1 \sqrt{x+1}=8-\sqrt{3x+1} x + 1 = 8 − 3 x + 1
( x + 1 + 3 x + 1 ) = 8 2 → \left(\sqrt{x+1}+\sqrt{3x+1}\right)=8^2\to ( x + 1 + 3 x + 1 ) = 8 2 →
x + 1 + 2 ( x + 1 ) ⋅ ( 3 x + 1 ) + 3 x + 1 = 64 → x+1+2\sqrt{(x+1)\cdot(3x+1)}+3x+1=64\to x + 1 + 2 ( x + 1 ) ⋅ ( 3 x + 1 ) + 3 x + 1 = 64 →
( ( x + 1 ) ⋅ ( 3 x + 1 ) ) 2 = ( 31 − 2 x ) 2 → \left(\sqrt{(x+1)\cdot(3x+1)}\right)^2=(31-2x)^2\to ( ( x + 1 ) ⋅ ( 3 x + 1 ) ) 2 = ( 31 − 2 x ) 2 →
( x + 1 ) ⋅ ( 3 x + 1 ) = 961 − 124 x + 4 x 2 → (x+1)\cdot(3x+1)=961-124x+4x^2\to ( x + 1 ) ⋅ ( 3 x + 1 ) = 961 − 124 x + 4 x 2 →
3 x 2 + 4 x + 1 = 961 − 124 x + 4 x 2 → 3x^2+4x+1=961-124x+4x^2\to 3 x 2 + 4 x + 1 = 961 − 124 x + 4 x 2 →
x 2 − 128 x + 960 = 0 x^2-128x+960=0\quad x 2 − 128 x + 960 = 0 Fórmula Quadrática
x = − b ± b 2 − 4 × a × c 2 × a → x=\dfrac{-b\pm\sqrt{b^2-4\times a\times c}}{2\times a}\to x = 2 × a − b ± b 2 − 4 × a × c →
x = 128 ± 16384 − 3840 2 → x=\dfrac{128\pm\sqrt{16384-3840}}{2}\to x = 2 128 ± 16384 − 3840 →
x = 128 ± 12544 2 → x = 128 ± 112 2 → x=\dfrac{128\pm\sqrt{12544}}{2}\to x=\dfrac{128\pm 112}{2}\to x = 2 128 ± 12544 → x = 2 128 ± 112 →
x 1 = 128 − 112 2 → x = 8 x_{1}=\dfrac{128-112}{2}\to\boxed{x=8} x 1 = 2 128 − 112 → x = 8
x 2 = 128 + 112 2 → x = 120 x_{2}=\dfrac{128+112}{2}\to \cancel{x=120}\quad x 2 = 2 128 + 112 → x = 120 Não serve, pois 120 120 120 não é válido no teste de verificação .
Portanto, a solução( S ) (S) ( S ) , será: S = { 8 } S=\{8\} S = { 8 }
e) D = { x ∈ R / x ⩾ − 5 3 } D=\left\{x\in\mathbb{R}\,/\,x\geqslant -\dfrac{5}{3}\right\} D = { x ∈ R / x ⩾ − 3 5 }
Solução da equação: 1 + 3 x + 5 = x 1+\sqrt{3x+5}=x 1 + 3 x + 5 = x
( 3 x + 5 ) 2 = ( x − 1 ) 2 → \left(\sqrt{3x+5}\right)^2=(x-1)^2\to ( 3 x + 5 ) 2 = ( x − 1 ) 2 →
3 x + 5 = x 2 − 2 x + 1 → 3x+5=x^2-2x+1\to 3 x + 5 = x 2 − 2 x + 1 →
x 2 − 5 x − 4 = 0 x^2-5x-4=0\quad x 2 − 5 x − 4 = 0 Fórmula Quadrática
x = − b ± b 2 − 4 × a × c 2 × a → x=\dfrac{-b\pm\sqrt{b^2-4\times a\times c}}{2\times a}\to x = 2 × a − b ± b 2 − 4 × a × c →
x = 5 ± 25 + 16 2 → x = 5 ± 41 2 → x=\dfrac{5\pm\sqrt{25+16}}{2}\to x=\dfrac{5\pm\sqrt{41}}{2}\to x = 2 5 ± 25 + 16 → x = 2 5 ± 41 →
x 1 ≈ 5 − 6 , 4 2 → x ≈ − 0 , 7 x_{1}\approx \dfrac{5-6,4}{2}\to \cancel{x\approx -0,7}\quad x 1 ≈ 2 5 − 6 , 4 → x ≈ − 0 , 7 Não serve, pois − 0 , 7 -0,7 − 0 , 7 não é válido no teste de verificação .
x 2 = 5 + 41 2 \boxed{x_{2}=\dfrac{5+\sqrt{41}}{2}} x 2 = 2 5 + 41
Portanto, a solução( S ) (S) ( S ) será: S = { 5 + 41 2 } S=\left\{\dfrac{5+\sqrt{41}}{2}\right\} S = { 2 5 + 41 }
f) O domínio( D ) (D) ( D ) será a intersecção de: x ⩾ 3 4 , x\geqslant\dfrac{3}{4},\quad x ⩾ 4 3 , x ⩾ 1 5 x\geqslant \dfrac{1}{5}\quad x ⩾ 5 1 e x ⩾ − 4 15 x\geqslant-\dfrac{4}{15} x ⩾ − 15 4 , portanto
D = { x ∈ R / x ⩾ 3 4 } D=\left\{x\in\mathbb{R}\,/\,x\geqslant\dfrac{3}{4}\right\} D = { x ∈ R / x ⩾ 4 3 }
Solução da equação: 4 x − 3 + 5 x − 1 = 15 x + 4 \sqrt{4x-3}+\sqrt{5x-1}=\sqrt{15x+4} 4 x − 3 + 5 x − 1 = 15 x + 4
( 4 x − 3 + 5 x − 1 ) 2 = ( 15 x + 4 ) 2 → \left(\sqrt{4x-3}+\sqrt{5x-1}\right)^2=\left(\sqrt{15x+4}\right)^2\to ( 4 x − 3 + 5 x − 1 ) 2 = ( 15 x + 4 ) 2 →
4 x − 3 + 2 ⋅ ( 4 x − 3 ) + ( 5 x − 1 ) + 5 x − 1 = 15 x + 4 → 4x-3+2\cdot\sqrt{(4x-3)+(5x-1)}+5x-1=15x+4\to 4 x − 3 + 2 ⋅ ( 4 x − 3 ) + ( 5 x − 1 ) + 5 x − 1 = 15 x + 4 →
( ( 4 x − 3 ) + ( 5 x − 1 ) ) 2 = ( 3 x + 4 ) 2 → \left(\sqrt{(4x-3)+(5x-1)}\right)^2=(3x+4)^2\to ( ( 4 x − 3 ) + ( 5 x − 1 ) ) 2 = ( 3 x + 4 ) 2 →
20 x 2 − 19 x + 3 − 9 x 2 − 24 x − 16 = 0 → 20x^2-19x+3-9x^2-24x-16=0\to 20 x 2 − 19 x + 3 − 9 x 2 − 24 x − 16 = 0 →
11 x 2 − 43 x − 13 = 0 11x^2-43x-13=0\quad 11 x 2 − 43 x − 13 = 0 Fórmula Quadrática
x = − b ± b 2 − 4 × a × c 2 × a → x=\dfrac{-b\pm\sqrt{b^2-4\times a\times c}}{2\times a}\to x = 2 × a − b ± b 2 − 4 × a × c →
x = 43 ± 1849 + 572 22 → x=\dfrac{43\pm\sqrt{1849+572}}{22}\to x = 22 43 ± 1849 + 572 →
x = 43 ± 2421 22 → x = 43 ± 3 269 22 → x=\dfrac{43\pm\sqrt{2421}}{22}\to x=\dfrac{43\pm3\sqrt{269}}{22}\to x = 22 43 ± 2421 → x = 22 43 ± 3 269 →
x 1 ≈ 43 − 49 , 2 22 → x 1 ≈ − 0 , 2 81 ‾ x_{1}\approx \dfrac{43-49,2}{22}\to \cancel{x_{1}\approx -0,2\overline{81}} x 1 ≈ 22 43 − 49 , 2 → x 1 ≈ − 0 , 2 81 , pois este valor não pertence ao conjunto domínio.
x 2 = 43 + 3 269 22 \boxed{x_{2}=\dfrac{43+3\sqrt{269}}{22}} x 2 = 22 43 + 3 269
Portanto, a solução( S ) (S) ( S ) , será: S = { 43 + 3 269 22 } S=\left\{\dfrac{43+3\sqrt{269}}{22}\right\} S = { 22 43 + 3 269 }
g) D = R D=R D = R
Solução da equação: x + 34 3 − x − 3 3 = 1 \sqrt[3]{x+34}-\sqrt[3]{x-3}=1 3 x + 34 − 3 x − 3 = 1
( x + 34 3 ) 3 = ( x − 3 3 + 1 ) 3 → \left(\sqrt[3]{x+34}\right)^3=\left(\sqrt[3]{x-3}+1\right)^3\to ( 3 x + 34 ) 3 = ( 3 x − 3 + 1 ) 3 →
x + 34 = x − 3 + 3 ⋅ ( x − 3 3 ) 2 ⋅ 1 + 3 ⋅ ( x − 3 3 ) ⋅ 1 2 + 1 3 → \cancel{x}+34=\cancel{x}-3+3\cdot\left(\sqrt[3]{x-3}\right)^2\cdot 1+3\cdot\left(\sqrt[3]{x-3}\right)\cdot 1^2+1^3\to x + 34 = x − 3 + 3 ⋅ ( 3 x − 3 ) 2 ⋅ 1 + 3 ⋅ ( 3 x − 3 ) ⋅ 1 2 + 1 3 →
36 = 3 ⋅ [ ( x − 3 3 ) 2 + ( x − 3 3 ) ] → 36=3\cdot\left[\left(\sqrt[3]{x-3}\right)^2+\left(\sqrt[3]{x-3}\right)\right]\to 36 = 3 ⋅ [ ( 3 x − 3 ) 2 + ( 3 x − 3 ) ] →
12 = [ ( x − 3 3 ) 2 + ( x − 3 3 ) ] → 12=\left[\left(\sqrt[3]{x-3}\right)^2+\left(\sqrt[3]{x-3}\right)\right]\to 12 = [ ( 3 x − 3 ) 2 + ( 3 x − 3 ) ] →
Incógnita auxiliar : a = ( x − 3 3 ) a=\left(\sqrt[3]{x-3}\right) a = ( 3 x − 3 )
12 = a 2 + a → a 2 + a − 12 = 0 12=a^2+a\to a^2+a-12=0\quad 12 = a 2 + a → a 2 + a − 12 = 0 Fórmula Quadrática
a = − b ± b 2 − 4 × a × c 2 × a → a=\dfrac{-b\pm\sqrt{b^2-4\times a\times c}}{2\times a}\to a = 2 × a − b ± b 2 − 4 × a × c →
a = − 1 ± 1 + 48 2 → a = − 1 ± 7 2 a=\dfrac{-1\pm\sqrt{1+48}}{2}\to a=\dfrac{-1\pm 7}{2} a = 2 − 1 ± 1 + 48 → a = 2 − 1 ± 7
a 1 = − 1 − 7 2 → a 1 = − 4 a_{1}=\dfrac{-1-7}{2}\to a_{1}=-4 a 1 = 2 − 1 − 7 → a 1 = − 4
a 2 = − 1 + 7 2 → a 2 = 3 a_{2}=\dfrac{-1+7}{2}\to a_{2}=3 a 2 = 2 − 1 + 7 → a 2 = 3
Voltando à incógnita principal, onde a = ( x − 3 3 ) a=\left(\sqrt[3]{x-3}\right) a = ( 3 x − 3 )
Para a = − 4 a=-4\,\, a = − 4 :
( − 4 ) 3 = ( x − 3 3 ) 3 → − 64 = x − 3 → x = − 61 (-4)^3=\left(\sqrt[3]{x-3}\right)^3\to -64=x-3\to\boxed{x=-61} ( − 4 ) 3 = ( 3 x − 3 ) 3 → − 64 = x − 3 → x = − 61
Para a = 3 a=3\,\, a = 3 :
3 3 = ( x − 3 3 ) 3 → 27 = x − 3 → x = 30 3^3=\left(\sqrt[3]{x-3}\right)^3\to 27=x-3\to\boxed{x=30} 3 3 = ( 3 x − 3 ) 3 → 27 = x − 3 → x = 30
Portanto, a solução( S ) (S) ( S ) , será: S = { − 61 ; 30 } S=\{-61;\,\,30\} S = { − 61 ; 30 }
0458
Obtenha a soma das raízes da equação 3 x − 2 = x + 2 \sqrt{3x-2}=\sqrt{x}+2 3 x − 2 = x + 2
0458 - Resposta
S = 7 S=7 S = 7
0458 - Solução
Antes de resolver, vamos observar o conjunto domínio( D ) (D) ( D ) para essa equação, que será dado pela intersecção de x ⩾ 2 3 x\geqslant\dfrac{2}{3}\quad x ⩾ 3 2 e x ⩾ 0 \quad x\geqslant 0 x ⩾ 0 , dentro do conjunto dos números reais. Assim,
D = { x ∈ R / x ⩾ 2 3 } D=\left\{x\in\mathbb{R}\,/\,x\geqslant \dfrac{2}{3}\right\} D = { x ∈ R / x ⩾ 3 2 }
Vamos à solução:
( 3 x − 2 ) 2 = ( x + 2 ) 2 → \left(\sqrt{3x-2}\right)^2=\left(\sqrt{x}+2\right)^2\to ( 3 x − 2 ) 2 = ( x + 2 ) 2 →
3 x − 2 = x + 2 ⋅ x + 4 → 3x-2=x+2\cdot\sqrt{x}+4\to 3 x − 2 = x + 2 ⋅ x + 4 →
( x − 3 ) 2 = ( x ) 2 → (x-3)^2=\left(\sqrt{x}\right)^2\to ( x − 3 ) 2 = ( x ) 2 →
x 2 − 6 x + 9 = x → x 2 − 7 x + 9 = 0 x^2-6x+9=x\to x^2-7x+9=0\quad x 2 − 6 x + 9 = x → x 2 − 7 x + 9 = 0 Fórmula Quadrática
x = − b ± b 2 − 4 × a × c 2 × a x=\dfrac{-b\pm\sqrt{b^2-4\times a\times c}}{2\times a} x = 2 × a − b ± b 2 − 4 × a × c
x = 7 ± 49 − 36 2 → x = 7 ± 13 2 x=\dfrac{7\pm\sqrt{49-36}}{2}\to x=\dfrac{7\pm\sqrt{13}}{2} x = 2 7 ± 49 − 36 → x = 2 7 ± 13
Dessa forma, a soma( S ) (S) ( S ) das raízes será:
S = 7 2 − 13 2 + 7 2 + 13 2 → S = 7 S= \dfrac{7}{2}-\cancel{\dfrac{\sqrt{13}}{2}}+\dfrac{7}{2}+\cancel{\dfrac{\sqrt{13}}{2}}\to\boxed{S=7} S = 2 7 − 2 13 + 2 7 + 2 13 → S = 7
Observação :
A soma( S ) (S) ( S ) das raízes de uma equação quadrática genérica a x 2 + b x + c = 0 ax^2+bx+c=0 a x 2 + b x + c = 0 é dada pela relação:
S = − b a \boxed{S=-\dfrac{b}{a}} S = − a b Portanto, poderíamos utilizá-la, assim:
S = − ( − 7 ) 1 → S = 7 S=-\dfrac{(-7)}{1}\to\boxed{S=7} S = − 1 ( − 7 ) → S = 7
0457
Resolva, em R \mathbb{R} R , as equações abaixo, usando as substituições indicadas. Observe o domínio de cada variável e teste as soluções encontradas.
0457 - Soluções
a) u 2 − 10 u + 21 = 0 u^2-10u+21=0\quad u 2 − 10 u + 21 = 0 Fórmula Quadrática
u = − b ± b 2 − 4 × a × c 2 × a u=\dfrac{-b\pm\sqrt{b^2-4\times a\times c}}{2\times a} u = 2 × a − b ± b 2 − 4 × a × c
u = 10 ± 100 − 84 2 → u=\dfrac{10\pm\sqrt{100-84}}{2}\to u = 2 10 ± 100 − 84 →
u = 10 ± 16 2 → u = 10 ± 4 2 → u=\dfrac{10\pm\sqrt{16}}{2}\to u=\dfrac{10\pm4}{2}\to u = 2 10 ± 16 → u = 2 10 ± 4 →
u 1 = 10 − 4 2 → u 1 = 3 u_{1}=\dfrac{10-4}{2}\to u_{1}=3 u 1 = 2 10 − 4 → u 1 = 3
u 2 = 10 + 4 2 → u 2 = 7 u_{2}=\dfrac{10+4}{2}\to u_{2}=7 u 2 = 2 10 + 4 → u 2 = 7
Voltando à incógnita principal onde u = x 2 u=x^2 u = x 2
Para u = 3 u=3 u = 3
x 2 = 3 → x = − 3 x^2=3\to x=-\sqrt{3} x 2 = 3 → x = − 3 ou x = 3 x=\sqrt{3} x = 3
Para u = 7 u=7 u = 7
x 2 = 7 → x = − 7 x^2=7\to x=-\sqrt{7} x 2 = 7 → x = − 7 ou x = 7 x=\sqrt{7} x = 7
Portanto, a solução( S ) (S) ( S ) , será: S = { − 7 ; − 7 ; 3 ; 7 } S=\{-\sqrt{7};\,-\sqrt{7};\,\sqrt{3};\,\sqrt{7}\} S = { − 7 ; − 7 ; 3 ; 7 }
b) u 2 − 4 u − 21 = 0 u^2-4u-21=0\quad u 2 − 4 u − 21 = 0 Fórmula Quadrática
u = − b ± b 2 − 4 × a × c 2 × a u=\dfrac{-b\pm\sqrt{b^2-4\times a\times c}}{2\times a} u = 2 × a − b ± b 2 − 4 × a × c
u = 4 ± 16 + 84 2 → u=\dfrac{4\pm\sqrt{16+84}}{2}\to u = 2 4 ± 16 + 84 →
u = 4 ± 100 2 → u = 4 ± 10 2 → u=\dfrac{4\pm\sqrt{100}}{2}\to u=\dfrac{4\pm 10}{2}\to u = 2 4 ± 100 → u = 2 4 ± 10 →
u 1 = 4 − 10 2 → u 1 = − 3 u_{1}=\dfrac{4-10}{2}\to u_{1}=-3 u 1 = 2 4 − 10 → u 1 = − 3
u 2 = 4 + 10 2 → u 2 = 7 u_{2}=\dfrac{4+10}{2}\to u_{2}=7 u 2 = 2 4 + 10 → u 2 = 7
Voltando à incógnita principal, onde u = x 2 u=x^2 u = x 2
Para u = − 3 → x = ± − 3 ∉ R u=-3\to x=\pm\sqrt{-3}\,\not\in\mathbb{R} u = − 3 → x = ± − 3 ∈ R
Para u = 7 u=7 u = 7
x 2 = 7 → x = − 7 x^2=7\to x=-\sqrt{7} x 2 = 7 → x = − 7 ou x = 7 x=\sqrt{7} x = 7
Portanto, a solução( S ) (S) ( S ) , será: S = { − 7 ; 7 } S=\{-\sqrt{7};\,\,\sqrt{7}\} S = { − 7 ; 7 }
c) u 2 − 4 u + 3 = 0 u^2-4u+3=0\quad u 2 − 4 u + 3 = 0 Fórmula Quadrática
u = − b ± b 2 − 4 × a × c 2 × a u=\dfrac{-b\pm\sqrt{b^2-4\times a\times c}}{2\times a} u = 2 × a − b ± b 2 − 4 × a × c
u = 4 ± 16 − 12 2 → u = 4 ± 2 2 u=\dfrac{4\pm\sqrt{16-12}}{2}\to u=\dfrac{4\pm 2}{2} u = 2 4 ± 16 − 12 → u = 2 4 ± 2
u 1 = 1 , u 2 = 3 u_{1}=1,\quad u_{2}=3 u 1 = 1 , u 2 = 3
Voltando à incógnita principal, onde u = x u=\sqrt{x} u = x
Para u = 1 u=1 u = 1
x = 1 → x = 1 \sqrt{x}=1\to\boxed{x=1} x = 1 → x = 1
Para u = 3 u=3 u = 3
x = 3 → x = 9 \sqrt{x}=3\to\boxed{x=9} x = 3 → x = 9
Portanto, a solução( S ) (S) ( S ) , será: S = { 1 ; 9 } S=\{1;\,\,9\} S = { 1 ; 9 }
d) 2 u 2 − 7 u + 3 = 0 2u^2-7u+3=0\quad 2 u 2 − 7 u + 3 = 0 Fórmula Quadrática
u = − b ± b 2 − 4 × a × c 2 × a u=\dfrac{-b\pm\sqrt{b^2-4\times a\times c}}{2\times a} u = 2 × a − b ± b 2 − 4 × a × c
u = 7 ± 49 − 24 4 → u = 7 ± 5 4 → u=\dfrac{7\pm\sqrt{49-24}}{4}\to u=\dfrac{7\pm 5}{4}\to u = 4 7 ± 49 − 24 → u = 4 7 ± 5 →
u 1 = 7 − 5 4 → u 1 = 1 2 u_{1}=\dfrac{7-5}{4}\to u_{1}=\dfrac{1}{2} u 1 = 4 7 − 5 → u 1 = 2 1
u 1 = 7 + 5 4 → u 2 = 3 u_{1}=\dfrac{7+5}{4}\to u_{2}=3 u 1 = 4 7 + 5 → u 2 = 3
Voltando à incógnita principal, onde u = x u=\sqrt{x} u = x
Para u = 1 2 u=\dfrac{1}{2} u = 2 1
x = 1 2 → x = 1 4 \sqrt{x}=\dfrac{1}{2}\to\boxed{x=\dfrac{1}{4}} x = 2 1 → x = 4 1
Para u = 3 u=3 u = 3
x = 3 → x = 9 \sqrt{x}=3\to\boxed{x=9} x = 3 → x = 9
Portanto, a solução( S ) (S) ( S ) , será: S = { 1 4 ; 9 } S=\left\{\dfrac{1}{4};\,\,9\right\} S = { 4 1 ; 9 }
e) u 2 + u − 6 = 0 u^2+u-6=0\quad u 2 + u − 6 = 0 Fórmula Quadrática
u = − b ± b 2 − 4 × a × c 2 × a u=\dfrac{-b\pm\sqrt{b^2-4\times a\times c}}{2\times a} u = 2 × a − b ± b 2 − 4 × a × c
u = − 1 ± 1 + 24 2 → u = − 1 ± 5 2 u=\dfrac{-1\pm\sqrt{1+24}}{2}\to u=\dfrac{-1\pm 5}{2} u = 2 − 1 ± 1 + 24 → u = 2 − 1 ± 5
u 1 = − 3 , u 2 = 2 u_{1}=-3,\quad u_{2}=2 u 1 = − 3 , u 2 = 2
Voltando à incógnita principal, onde u = x u=\sqrt{x} u = x
Para u = − 3 u=-3 u = − 3
x = − 3 → x = 9 \sqrt{x}=-3\to \cancel{x=9}\quad x = − 3 → x = 9 Não serve, pois 9 9 9 não é válido no teste de verificação .
Para u = 2 u=2 u = 2
2 = 2 → x = 4 \sqrt{2}=2\to\boxed{x=4} 2 = 2 → x = 4
Portanto, a solução( S ) (S) ( S ) , será: S = { 4 } S=\{4\} S = { 4 }
f) 2 u 2 − 5 u − 3 = 0 2u^2-5u-3=0\quad 2 u 2 − 5 u − 3 = 0 Fórmula Quadrática
u = − b ± b 2 − 4 × a × c 2 × a u=\dfrac{-b\pm\sqrt{b^2-4\times a\times c}}{2\times a} u = 2 × a − b ± b 2 − 4 × a × c
u = 5 ± 25 + 24 4 → u = 5 ± 7 4 → u=\dfrac{5\pm\sqrt{25+24}}{4}\to u=\dfrac{5\pm 7}{4}\to u = 4 5 ± 25 + 24 → u = 4 5 ± 7 →
u 1 = − 1 2 , u 2 = 3 u_{1}=-\dfrac{1}{2},\quad u_{2}=3 u 1 = − 2 1 , u 2 = 3
Voltando à incógnita principal, onde u = x 1 / 3 u=x^{1/3} u = x 1/3
Para u = − 1 2 u=-\dfrac{1}{2} u = − 2 1
( x 1 3 ) 3 = ( − 1 2 ) 3 → x = − 1 8 \left(x^{\frac 13}\right)^3=\left(-\dfrac{1}{2}\right)^3\to \boxed{x=-\dfrac{1}{8}} ( x 3 1 ) 3 = ( − 2 1 ) 3 → x = − 8 1
Para u = 3 u=3 u = 3
( x 1 3 ) 3 = 3 3 → x = 27 \left(x^{\frac 13}\right)^3=3^3\to\boxed{x=27} ( x 3 1 ) 3 = 3 3 → x = 27
Portanto, a solução( S ) (S) ( S ) , será: S = ( − 1 8 ; 27 ) S=\left(-\dfrac{1}{8};\,\,27\right) S = ( − 8 1 ; 27 )
0456
Resolva, em R \mathbb{R} R , as seguintes...
0456 - Respostas das ...
0456 - Soluções
a) 20 x − 4 = 5 x → 20x-4=5x\to 20 x − 4 = 5 x →
20 x − 5 x = 4 → x = 4 15 20x-5x=4\to\boxed{x=\dfrac{4}{15}} 20 x − 5 x = 4 → x = 15 4
b) 5 ( 1 − x ) − 2 x + 1 = − 3 ( 2 + x ) → 5(1-x)-2x+1=-3(2+x)\to 5 ( 1 − x ) − 2 x + 1 = − 3 ( 2 + x ) →
5 − 5 x − 2 x + 1 = − 6 − 3 x → 5-5x-2x+1=-6-3x\to 5 − 5 x − 2 x + 1 = − 6 − 3 x →
− 4 x = − 12 → x = 3 -4x=-12\to\boxed{x=3} − 4 x = − 12 → x = 3
c) 4 x = − 8 x + 36 → 4x=-8x+36\to 4 x = − 8 x + 36 →
4 x + 8 x = 36 → x = 3 4x+8x=36\to\boxed{x=3} 4 x + 8 x = 36 → x = 3
d) 2 + 3 [ x − ( 3 x + 1 ) ] = 5 [ x − ( 2 x − 1 ) ] → 2+3[x-(3x+1)]=5[x-(2x-1)]\to 2 + 3 [ x − ( 3 x + 1 )] = 5 [ x − ( 2 x − 1 )] →
2 + 3 ( − 2 x − 1 ) = 5 ( − x + 1 ) → 2+3(-2x-1)=5(-x+1)\to 2 + 3 ( − 2 x − 1 ) = 5 ( − x + 1 ) →
2 − 6 x − 3 = − 5 x + 5 → − 6 x + 5 x = 5 + 3 − 2 → 2-6x-3=-5x+5\to -6x+5x=5+3-2\to 2 − 6 x − 3 = − 5 x + 5 → − 6 x + 5 x = 5 + 3 − 2 →
− x = 6 → x = − 6 -x=6\to\boxed{x=-6} − x = 6 → x = − 6
e) 4 ( x − 3 ) = 2 x − 5 → 4(x-3)=2x-5\to 4 ( x − 3 ) = 2 x − 5 →
4 x − 12 = 2 x − 5 → 4 x − 2 x = − 5 + 12 → 4x-12=2x-5\to 4x-2x=-5+12\to 4 x − 12 = 2 x − 5 → 4 x − 2 x = − 5 + 12 →
2 x = 7 → x = 7 2 2x=7\to\boxed{x=\dfrac{7}{2}} 2 x = 7 → x = 2 7
f) 1 − 2 x = x 3 − x 2 → 1-2x=\dfrac{x}{3}-\dfrac{x}{2}\to 1 − 2 x = 3 x − 2 x →
6 − 12 x = 2 x − 3 x 6 → \dfrac{6-12x=2x-3x}{\cancel{6}}\to 6 6 − 12 x = 2 x − 3 x →
− 11 x = − 6 → x = 6 11 -11x=-6\to\boxed{x=\dfrac{6}{11}} − 11 x = − 6 → x = 11 6
g) 3 ( x − 1 ) − 2 x 5 = 5 ( x − 3 ) 6 → \dfrac{3(x-1)-2x}{5}=\dfrac{5(x-3)}{6}\to 5 3 ( x − 1 ) − 2 x = 6 5 ( x − 3 ) →
6 x − 18 = 25 x − 75 30 → \dfrac{6x-18=25x-75}{\cancel{30}}\to 30 6 x − 18 = 25 x − 75 →
− 19 x = − 57 → x = 3 -19x=-57\to\boxed{x=3} − 19 x = − 57 → x = 3
h) 2 x + 5 3 x = 1 4 → \dfrac{2x+5}{3x}=\dfrac{1}{4}\to 3 x 2 x + 5 = 4 1 →
4 ( 2 x + 5 ) = 3 x 12 x ; ∀ x ∈ R − { 0 } → \dfrac{4(2x+5)=3x}{\cancel{12x}};\,\,\forall x\in\mathbb{R}-\{0\}\to 12 x 4 ( 2 x + 5 ) = 3 x ; ∀ x ∈ R − { 0 } →
5 x = − 20 → x = − 4 5x=-20\to\boxed{x=-4} 5 x = − 20 → x = − 4
0455
Resolva, em R \mathbb{R} R , as equações:
a) 7 x − 3 = x + 5 − 2 x 7x-3=x+5-2x 7 x − 3 = x + 5 − 2 x
b) 2 x 2 − 8 = 10 2x^2-8=10 2 x 2 − 8 = 10
c) x 3 + 1 = − 7 x^3+1=-7 x 3 + 1 = − 7
0455 - Respostas
a) x = 1 x=1\quad x = 1
b) x = − 3 x=-3\,\, x = − 3 ou x = 3 \,\,x=3 x = 3
c) x = − 2 x=-2 x = − 2
0455 - Soluções
a) 7 x − 3 = x + 5 − 2 x → 8 x = 8 → x = 1 7x-3=x+5-2x\to 8x=8\to\boxed{x=1} 7 x − 3 = x + 5 − 2 x → 8 x = 8 → x = 1
b) 2 x 2 − 8 = 10 → x 2 = 9 → x = − 3 ou x = 3 2x^2-8=10\to x^2=9\to \boxed{x=-3\,\text{ou}\,x=3} 2 x 2 − 8 = 10 → x 2 = 9 → x = − 3 ou x = 3
c) x 3 + 1 = − 7 → x 3 = − 8 → x = − 8 3 → x = − 2 x^3+1=-7\to x^3=-8\to x=\sqrt[3]{-8}\to\boxed{x=-2} x 3 + 1 = − 7 → x 3 = − 8 → x = 3 − 8 → x = − 2
0454
Resolva, em R \mathbb{R} R , a seguinte inequação:
4 − x < 3 − 2 x 4-x<3-2x 4 − x < 3 − 2 x
0454 - Resposta
S = { x ∈ R / x < − 1 } S=\left\{x\in\mathbb{R}\,/\,x<-1\right\} S = { x ∈ R / x < − 1 }
0454 - Solução
Vamos resolver a inequação:
− x + 2 x < 3 − 4 → x + 1 < 0 → x < − 1 -x+2x<3-4\to x+1<0\to x<-1 − x + 2 x < 3 − 4 → x + 1 < 0 → x < − 1
Observe o gráfico da função y = x + 1 y=x+1 y = x + 1
Portanto, a solução( S ) (S) ( S ) , será: S = { x ∈ R / x < − 1 } \boxed{S=\left\{x\in\mathbb{R}\,/\,x<-1\right\}} S = { x ∈ R / x < − 1 }
0453
Resolva, em R \mathbb{R} R , a seguinte inequação:
5 − x 2 < 8 5-x^2<8 5 − x 2 < 8 .
0453 - Resposta
S = { x ∈ R } S=\left\{x\in\mathbb{R}\right\} S = { x ∈ R }
0453 - Solução
Vamos desenvolver a inequação:
− x 2 − 3 < 0 -x^2-3<0 − x 2 − 3 < 0 ou x 2 + 3 > 0 x^2+3>0 x 2 + 3 > 0 (Não há raízes reais)
Entretanto, observe o gráfico da função y = x 2 + 3 y=x^2+3 y = x 2 + 3 :
Portanto, a solução( S ) (S) ( S ) , será: S = { x ∈ R } \boxed{S=\left\{x\in\mathbb{R}\right\}} S = { x ∈ R }
0452
Resolva, em R \mathbb{R} R , a seguinte inequação:
5 − x 2 < − 2 5-x^2<-2 5 − x 2 < − 2
0452 - Resposta
S = { x ∈ R / x < − 7 ou x > 7 } S=\left\{x\in\mathbb{R}\,/\, x<-\sqrt{7}\,\,\text{ou}\,\,x>\sqrt{7}\right\} S = { x ∈ R / x < − 7 ou x > 7 }
0452 - Solução
Vamos desenvolver a inequação:
5 − x 2 < − 2 → − x 2 + 7 < 0 5-x^2<-2\to -x^2+7<0 5 − x 2 < − 2 → − x 2 + 7 < 0 .
Resolvendo a equação − x 2 + 7 = 0 -x^2+7=0 − x 2 + 7 = 0 , encontraremos:
x 1 = − 7 x_{1}=-\sqrt{7} x 1 = − 7 e x 2 = 7 x_{2}=\sqrt{7} x 2 = 7
Observe o gráfico da função y = − x 2 + 7 y=-x^2+7 y = − x 2 + 7 :
Portanto, a solução( S ) (S) ( S ) , será: S = { x ∈ R / x < − 7 ou x > 7 } \boxed{S=\left\{x\in\mathbb{R}\,/\, x<-\sqrt{7}\,\,\text{ou}\,\,x>\sqrt{7}\right\}} S = { x ∈ R / x < − 7 ou x > 7 }
0451
Resolva, em R \mathbb{R} R , a seguinte inequação:
( x − 1 ) ( x − 3 ) ⩾ 0 (x-1)(x-3)\geqslant 0 ( x − 1 ) ( x − 3 ) ⩾ 0
0451 - Resposta
S = { x ∈ R / x ⩽ 1 ou x ⩾ 3 } S=\left\{x\in\mathbb{R}\,/\,x\leqslant 1\,\,\text{ou}\,\,x\geqslant 3\right\} S = { x ∈ R / x ⩽ 1 ou x ⩾ 3 }
0451 - Solução
Observe o quadro de sinais:
Portanto, a solução( S ) (S) ( S ) , será: S = { x ∈ R / x ⩽ 1 ou x ⩾ 3 } \boxed{S=\left\{x\in\mathbb{R}\,/\,x\leqslant 1\,\,\text{ou}\,\,x\geqslant 3\right\}} S = { x ∈ R / x ⩽ 1 ou x ⩾ 3 }