Página06
0150
Duas questões:
a) Provar que ( 1 + i ) 2 = 2 i (1+i)^2=2i ( 1 + i ) 2 = 2 i
b) Colocar na forma algébrica o número: z = ( 1 + i ) 80 − ( 1 + i ) 82 i 96 z=\dfrac{(1+i)^{80}-(1+i)^{82}}{i^{96}} z = i 96 ( 1 + i ) 80 − ( 1 + i ) 82
0150 - Soluções
Resolvendo:
a) A fim de provar, devemos apenas efetuar este produto, da mesma forma que resolveríamos um produto notável, assim:
( 1 + i ) 2 = 2 i → 1 2 + 2.1. i + i 2 = 2 i → 1 − 1 + 2 i = 2 i (1+i)^2=2i\to 1^2+2.1.i+i^2=2i\to \cancel{1-1}+2i=2i\,\, ( 1 + i ) 2 = 2 i → 1 2 + 2.1. i + i 2 = 2 i → 1 − 1 + 2 i = 2 i
b) Vamos resolver, efetuando os produtos notáveis que surgirão a partir das expressões (re)escritas, assim:
z = ( 1 + i ) 80 − ( 1 + i ) 82 i 96 → z = [ ( 1 + i ) 2 ] 40 − [ ( 1 + i ) 2 ] 41 ( i 4 1 ) 24 → z=\dfrac{(1+i)^{80}-(1+i)^{82}}{i^{96}}\to z=\dfrac{[(1+i)^{2}]^{40}-[(1+i)^2]^{41}}{(\cancel{i^{4}}^{\,1})^{24}}\to z = i 96 ( 1 + i ) 80 − ( 1 + i ) 82 → z = ( i 4 1 ) 24 [( 1 + i ) 2 ] 40 − [( 1 + i ) 2 ] 41 →
z = ( 2 i ) 40 − ( 2 i ) 41 1 24 → z = 2 40 − i .2 41 1 = 2 40 − i .2 41 z=\dfrac{(2i)^{40}-(2i)^{41}}{1^{24}}\to z=\dfrac{2^{40}-i.2^{41}}{1}=\boxed{2^{40}-i.2^{41}} z = 1 24 ( 2 i ) 40 − ( 2 i ) 41 → z = 1 2 40 − i . 2 41 = 2 40 − i . 2 41
0149
Calcular:
a) ( 3 + 2 i ) 2 (3+2i)^2 ( 3 + 2 i ) 2
b) ( 5 − i ) 2 (5-i)^2 ( 5 − i ) 2
c) ( 1 + i ) 3 (1+i)^3 ( 1 + i ) 3
0149 - Soluções
Calculando:
a) ( 3 + 2 i ) 2 = 3 2 + 2.3.2 i + ( 2 i ) 2 = 9 + 12 i − 4 = 5 + 12 i (3+2i)^2=3^2+2.3.2i+(2i)^2=9+12i-4=\boxed{5+12i} ( 3 + 2 i ) 2 = 3 2 + 2.3.2 i + ( 2 i ) 2 = 9 + 12 i − 4 = 5 + 12 i
b) ( 5 − i ) 2 = 5 2 − 2.5. i − 1 = 24 − 10 i (5-i)^2=5^2-2.5.i-1=\boxed{24-10i} ( 5 − i ) 2 = 5 2 − 2.5. i − 1 = 24 − 10 i
c) ( 1 + i ) 3 = 1 3 + 3.1. i + 3.1. i 2 + i 3 = 1 + 3 i − 3 − i = − 2 + 2 i (1+i)^3=1^3+3.1.i+3.1.i^2+i^3=1+3i-3-i=\boxed{-2+2i} ( 1 + i ) 3 = 1 3 + 3.1. i + 3.1. i 2 + i 3 = 1 + 3 i − 3 − i = − 2 + 2 i
0148
Efetuar:
a) ( 2 − 3 i ) ( 1 + 5 i ) (2-3i)(1+5i) ( 2 − 3 i ) ( 1 + 5 i )
b) ( 1 + 2 i ) ( 2 + i ) (1+2i)(2+i) ( 1 + 2 i ) ( 2 + i )
c) ( 4 − 3 i ) ( 5 − i ) ( 1 + i ) (4-3i)(5-i)(1+i) ( 4 − 3 i ) ( 5 − i ) ( 1 + i )
d) ( 7 + 2 i ) ( 7 − 2 i ) (7+2i)(7-2i) ( 7 + 2 i ) ( 7 − 2 i )
0148 - Soluções
Efetuando as operações indicadas:
a) ( 2 − 3 i ) ( 1 + 5 i ) = 2.1 + 2.5 i − 3 i .1 + 3 i .5 i = 2 + 10 i − 3 i + 15 = 17 + 7 i (2-3i)(1+5i)=2.1+2.5i-3i.1+3i.5i=2+10i-3i+15=\boxed{17+7i} ( 2 − 3 i ) ( 1 + 5 i ) = 2.1 + 2.5 i − 3 i .1 + 3 i .5 i = 2 + 10 i − 3 i + 15 = 17 + 7 i
b) ( 1 + 2 i ) ( 2 + i ) = ( 1.2 + 1. i + 2 i .2 − 1 ) = 1 + 5 i (1+2i)(2+i)=(1.2+1.i+2i.2-1)=\boxed{1+5i} ( 1 + 2 i ) ( 2 + i ) = ( 1.2 + 1. i + 2 i .2 − 1 ) = 1 + 5 i
c) ( 7 + 2 i ) ( 7 − 2 i ) = 7 2 − ( 2 i ) 2 = 49 + 4 = 53 (7+2i)(7-2i)=7^2-(2i)^2=49+4=\boxed{53} ( 7 + 2 i ) ( 7 − 2 i ) = 7 2 − ( 2 i ) 2 = 49 + 4 = 53
d) ( 4 − 3 i ) ( 5 − i ) ( 1 + i ) = [ 4.5 + 4. ( − i ) − 3 i .5 + ( − 3 i ) ( − i ) ] ( 1 + i ) = ( 17 − 19 i ) ( 1 + i ) (4-3i)(5-i)(1+i)=[4.5+4.(-i)-3i.5+(-3i)(-i)](1+i)=(17-19i)(1+i) ( 4 − 3 i ) ( 5 − i ) ( 1 + i ) = [ 4.5 + 4. ( − i ) − 3 i .5 + ( − 3 i ) ( − i )] ( 1 + i ) = ( 17 − 19 i ) ( 1 + i )
0147
Efetuar:
a) ( 3 + 2 i ) + ( 2 − 5 i ) (3+2i)+(2-5i) ( 3 + 2 i ) + ( 2 − 5 i )
b) ( 1 + i ) + ( 1 − i ) − 2 i (1+i)+(1-i)-2i ( 1 + i ) + ( 1 − i ) − 2 i
c) ( 5 − 2 i ) − ( 2 + 8 i ) (5-2i)-(2+8i) ( 5 − 2 i ) − ( 2 + 8 i )
d) ( 6 + 7 i ) − ( 4 + 2 i ) + ( 1 − 10 i ) (6+7i)-(4+2i)+(1-10i) ( 6 + 7 i ) − ( 4 + 2 i ) + ( 1 − 10 i )
0147 - Soluções
Efetuando as operações indicadas:
a) ( 3 + 2 i ) + ( 2 − 5 i ) = ( 3 + 2 ) + ( 2 − 5 ) i = 5 − 3 i (3+2i)+(2-5i)=(3+2)+(2-5)i=\boxed{5-3i} ( 3 + 2 i ) + ( 2 − 5 i ) = ( 3 + 2 ) + ( 2 − 5 ) i = 5 − 3 i
b) ( 1 + i ) + ( 1 − i ) − 2 i = ( 1 + 1 ) + ( 1 − 1 ) i − 2 i = 2 − 2 i (1+i)+(1-i)-2i=(1+1)+\cancel{(1-1)i}-2i=\boxed{2-2i} ( 1 + i ) + ( 1 − i ) − 2 i = ( 1 + 1 ) + ( 1 − 1 ) i − 2 i = 2 − 2 i
c) ( 5 − 2 i ) − ( 2 + 8 i ) = ( 5 − 2 ) + ( − 2 + 8 ) i = 3 + 6 i (5-2i)-(2+8i)=(5-2)+(-2+8)i=\boxed{3+6i} ( 5 − 2 i ) − ( 2 + 8 i ) = ( 5 − 2 ) + ( − 2 + 8 ) i = 3 + 6 i
d) ( 6 + 7 i ) − ( 4 + 2 i ) + ( 1 − 10 i ) = ( 6 − 4 + 1 ) + ( 7 − 2 − 10 ) i = 3 − 5 i (6+7i)-(4+2i)+(1-10i)=(6-4+1)+(7-2-10)i=\boxed{3-5i} ( 6 + 7 i ) − ( 4 + 2 i ) + ( 1 − 10 i ) = ( 6 − 4 + 1 ) + ( 7 − 2 − 10 ) i = 3 − 5 i
0146
Efetuar as operações indicadas:
a) ( 6 + 7 i ) ( 1 + i ) (6+7i)(1+i) ( 6 + 7 i ) ( 1 + i )
b) ( 5 + 4 i ) ( 1 − i ) + ( 2 + i ) i (5+4i)(1-i)+(2+i)i ( 5 + 4 i ) ( 1 − i ) + ( 2 + i ) i
c) ( 1 + 2 i ) 2 − ( 3 + 4 i ) 2 (1+2i)^2-(3+4i)^2 ( 1 + 2 i ) 2 − ( 3 + 4 i ) 2
0146 - Soluções
Efetuando as operações indicadas:
a) ( 6 + 7 i ) ( 1 + i ) = 6 + 7 i + 6 i − 7 = − 1 + 13 i (6+7i)(1+i)=6+7i+6i-7=\boxed{-1+13i} ( 6 + 7 i ) ( 1 + i ) = 6 + 7 i + 6 i − 7 = − 1 + 13 i
b) ( 5 + 4 i ) ( 1 − i ) + ( 2 + i ) i = 5 + 4 i − 5 i − 4 i 2 + 2 i + i 2 = 5 + 4 i − 5 i + 4 + 2 i − 1 = 8 + i (5+4i)(1-i)+(2+i)i=5+4i-5i-4i^2+2i+i^2=5+4i-5i+4+2i-1=\boxed{8+i} ( 5 + 4 i ) ( 1 − i ) + ( 2 + i ) i = 5 + 4 i − 5 i − 4 i 2 + 2 i + i 2 = 5 + 4 i − 5 i + 4 + 2 i − 1 = 8 + i
c) ( 1 + 2 i ) 2 − ( 3 + 4 i ) 2 = 1 + 4 i − 4 − ( 9 + 24 i − 16 ) = − 3 + 4 i + 7 − 24 i = 4 − 20 i (1+2i)^2-(3+4i)^2=1+4i-4-(9+24i-16)=-3+4i+7-24i=\boxed{4-20i} ( 1 + 2 i ) 2 − ( 3 + 4 i ) 2 = 1 + 4 i − 4 − ( 9 + 24 i − 16 ) = − 3 + 4 i + 7 − 24 i = 4 − 20 i
0145
Efetue:
a ) ( 7 + i ) + ( 3 − 4 i ) b ) ( 13 − 2 i ) + ( − 5 + 6 i ) c ) ( 9 − i ) − ( 8 − i ) d ) ( 3 + 2 i ) − ( 6 + 13 i ) e ) ( − 2 + − 8 ) + ( 5 − − 50 ) f ) ( 8 + − 18 ) − ( 4 + 3 i 2 ) g ) 13 i − ( 14 − 7 i ) h ) 25 + ( − 10 + 11 i ) + 15 i i ) − ( 3 2 + 5 2 i ) + ( 5 3 + 11 3 i ) j ) ( 1 , 6 + 3 , 2 i ) + ( − 5 , 8 + 4 , 3 i ) k ) ( 1 + i ) ( 3 − 2 i ) l ) ( 7 − 2 i ) ( 3 − 5 i ) m ) 12 i ( 1 − 9 i ) n ) − 8 i ( 9 + 4 i ) o ) ( 14 + i 10 ) ( 14 − i 10 ) p ) ( 3 + i 15 ) ( 3 − i 15 ) q ) ( 6 + 7 i ) 2 r ) ( 5 − 4 i ) 2 s ) ( 2 + 3 i ) 2 + ( 2 − 3 i ) 2 t ) ( 1 − 2 i ) 2 − ( 1 − 2 i ) 2 u ) [ ( 1 + i ) 2 − ( 1 − i ) 2 ] 5 \begin{array}{ll}
a) (7+i)+(3-4i) & b) (13-2i)+(-5+6i) \\&\\
c) (9-i)-(8-i) & d) (3+2i)-(6+13i) \\&&\\
e) (-2+\sqrt{-8})+(5-\sqrt{-50}) & f) (8+\sqrt{-18})-(4+3i\sqrt{2}) \\&&\\
g) 13i-(14-7i) & h) 25+(-10+11i)+15i\\&&\\
i) -\left(\dfrac{3}{2}+\dfrac{5}{2}i\right)+\left(\dfrac{5}{3}+\dfrac{11}{3}i\right) &
j) (1,6+3,2i)+(-5,8+4,3i)\\&&\\
k) (1+i)(3-2i) & l) (7-2i)(3-5i)\\&&\\
m) 12i(1-9i) & n) -8i(9+4i)\\&&\\
o) (\sqrt{14}+i\sqrt{10})(\sqrt{14}-i\sqrt{10}) &
p) (\sqrt{3}+i\sqrt{15})(\sqrt{3}-i\sqrt{15})\\&&\\
q) (6+7i)^2 & r) (5-4i)^2\\&&\\
s) (2+3i)^2+(2-3i)^2 & t) (1-2i)^2-(1-2i)^2\\&&\\
u) [(1+i)^2-(1-i)^2]^5
\end{array} a ) ( 7 + i ) + ( 3 − 4 i ) c ) ( 9 − i ) − ( 8 − i ) e ) ( − 2 + − 8 ) + ( 5 − − 50 ) g ) 13 i − ( 14 − 7 i ) i ) − ( 2 3 + 2 5 i ) + ( 3 5 + 3 11 i ) k ) ( 1 + i ) ( 3 − 2 i ) m ) 12 i ( 1 − 9 i ) o ) ( 14 + i 10 ) ( 14 − i 10 ) q ) ( 6 + 7 i ) 2 s ) ( 2 + 3 i ) 2 + ( 2 − 3 i ) 2 u ) [( 1 + i ) 2 − ( 1 − i ) 2 ] 5 b ) ( 13 − 2 i ) + ( − 5 + 6 i ) d ) ( 3 + 2 i ) − ( 6 + 13 i ) f ) ( 8 + − 18 ) − ( 4 + 3 i 2 ) h ) 25 + ( − 10 + 11 i ) + 15 i j ) ( 1 , 6 + 3 , 2 i ) + ( − 5 , 8 + 4 , 3 i ) l ) ( 7 − 2 i ) ( 3 − 5 i ) n ) − 8 i ( 9 + 4 i ) p ) ( 3 + i 15 ) ( 3 − i 15 ) r ) ( 5 − 4 i ) 2 t ) ( 1 − 2 i ) 2 − ( 1 − 2 i ) 2
0145 - Soluções
Efetuando, item a item:
a ) ( 7 + i ) + ( 3 − 4 i ) = ( 7 + 3 ) + i ( 1 − 4 ) = 10 − 3 i b ) ( 13 − 2 i ) + ( − 5 + 6 i ) = ( 13 − 5 ) + i ( − 2 + 6 ) = 8 + 4 i c ) ( 9 − i ) − ( 8 − i ) = ( 9 − 8 ) + i ( − 1 + 1 ) = 1 d ) ( 3 + 2 i ) − ( 6 + 13 i ) = ( 3 − 6 ) + i ( 2 + 13 ) = − 3 + 15 i e ) ( − 2 + − 8 ) + ( 5 − − 50 ) = ( − 2 + 5 ) + i ( 2 2 − 5 2 ) = 3 − 3 i 2 f ) ( 8 + − 18 ) − ( 4 + 3 i 2 ) = ( 8 − 4 ) + ( 3 i 2 − 3 i 2 ) = 4 g ) 13 i − ( 14 − 7 i ) = − 14 + ( 13 i + 7 i ) = − 14 + 20 i h ) 25 + ( − 10 + 11 i ) + 15 i = ( 25 − 10 ) + ( 11 i + 15 i ) = 15 + 26 i i ) − ( 3 2 + 5 2 i ) + ( 5 3 + 11 3 i ) = ( − 3 2 + 5 3 ) + ( − 5 2 + 11 3 ) i = 1 6 + 7 6 i j ) ( 1 , 6 + 3 , 2 i ) + ( − 5 , 8 + 4 , 3 i ) = − 4 , 2 + 7 , 5 i k ) ( 1 + i ) ( 3 − 2 i ) = 1.3 − 1.2 i + 3 i − 2 i 2 = 3 − 2 i + 3 i + 2 = 5 + i l ) ( 7 − 2 i ) ( 3 − 5 i ) = 7.3 − 7.5 i − 2 i .3 + 2.5 i 2 = 21 − 35 i − 6 i − 10 = 11 − 41 i m ) 12 i ( 1 − 9 i ) = 12 i .1 − 12 i .9 i = 12 i − 108 i 2 = 108 + 12 i n ) − 8 i ( 9 + 4 i ) = − 72 i − 32 i 2 = 32 − 72 i o ) ( 14 + i 10 ) ( 14 − i 10 ) = ( 14 ) 2 − ( i 15 ) 2 = 14 − 10 i 2 = 14 + 10 = 24 p ) ( 3 + i 15 ) ( 3 − i 15 ) = ( 3 ) 2 − ( i 15 ) 2 = 3 + 15 = 18 q ) ( 6 + 7 i ) 2 = 6 2 + 2.6.7 i + ( 7 i ) 2 = 36 + 84 i − 49 = − 13 + 84 i r ) ( 5 − 4 i ) 2 = 5 2 − 2.5.4 i + ( 4 i ) 2 = 25 − 40 i − 16 = 9 − 40 i s ) ( 2 + 3 i ) 2 + ( 2 − 3 i ) 2 = ( − 5 + 12 i ) + ( − 5 − 12 i ) = − 10 t ) ( 1 − 2 i ) 2 − ( 1 + 2 i ) 2 = ( − 3 − 4 i ) − ( − 3 + 4 i ) = − 8 i u ) [ ( 1 + i ) 2 − ( 1 − i ) 2 ] 5 = [ 2 i − ( − 2 i ) ] 5 = ( 4 i ) 5 = 4 5 × i 5 = 2 10 × i 4 1 . i = 1024 i \begin{array}{ll}
a) & (7+i)+(3-4i)=(7+3)+i(1-4)=\boxed{10-3i}\\
&\\
b) & (13-2i)+(-5+6i)=(13-5)+i(-2+6)=\boxed{8+4i}\\
&\\
c) & (9-i)-(8-i)=(9-8)+i\cancel{(-1+1)}=\boxed{1}\\
&\\
d) & (3+2i)-(6+13i)=(3-6)+i(2+13)=\boxed{-3+15i}\\
&\\
e) & (-2+\sqrt{-8})+(5-\sqrt{-50})=(-2+5)+i(2\sqrt{2}-5\sqrt{2})=\boxed{3-3i\sqrt{2}}\\
&\\
f) & (8+\sqrt{-18})-(4+3i\sqrt{2})=(8-4)+\cancel{(3i\sqrt{2}-3i\sqrt{2})}=\boxed{4}\\
&\\
g) & 13i-(14-7i)=-14+(13i+7i)=\boxed{-14+20i}\\
&\\
h) & 25+(-10+11i)+15i=(25-10)+(11i+15i)=\boxed{15+26i}\\
&\\
i) & -\left(\dfrac{3}{2}+\dfrac{5}{2}i\right)+\left(\dfrac{5}{3}+\dfrac{11}{3}i\right)=\left( -\dfrac{3}{2}+\dfrac{5}{3} \right)+\left( -\dfrac{5}{2}+\dfrac{11}{3} \right)i=\boxed{\dfrac{1}{6}+\dfrac{7}{6}i}\\
&\\
j) & (1,6+3,2i)+(-5,8+4,3i)=\boxed{-4,2+7,5i}\\
&\\
k) & (1+i)(3-2i)=1.3-1.2i+3i-2i^2=3-2i+3i+2=\boxed{5+i}\\
&\\
l) & (7-2i)(3-5i)=7.3-7.5i-2i.3+2.5i^2=21-35i-6i-10=\boxed{11-41i}\\
&\\
m) & 12i(1-9i)=12i.1-12i.9i=12i-108i^2=\boxed{108+12i}\\
&\\
n) & -8i(9+4i)=-72i-32i^2=\boxed{32-72i}\\
&\\
o) & (\sqrt{14}+i\sqrt{10})(\sqrt{14}-i\sqrt{10})=(\sqrt{14})^2-(i\sqrt{15})^2=14-10i^2=14+10=\boxed{24}\\
&\\
p) & (\sqrt{3}+i\sqrt{15})(\sqrt{3}-i\sqrt{15})=(\sqrt{3})^2-(i\sqrt{15})^2=3+15=\boxed{18}\\
&\\
q) & (6+7i)^2=6^2+2.6.7i+(7i)^2=36+84i-49=\boxed{-13+84i}\\
&\\
r) & (5-4i)^2=5^2-2.5.4i+(4i)^2=25-40i-16=\boxed{9-40i}\\
&\\
s) & (2+3i)^2+(2-3i)^2=(-5+\cancel{12i})+(-5-\cancel{12i})=\boxed{-10}\\
&\\
t) & (1-2i)^2-(1+2i)^2=(-3-4i)-(-3+4i)=\boxed{-8i}\\
&\\
u) & [(1+i)^2-(1-i)^2]^5=[2i-(-2i)]^5=(4i)^5=4^5\times i^5=2^{10}\times \cancel{i^4}^{\,1}.i=\boxed{1024i}
\end{array} a ) b ) c ) d ) e ) f ) g ) h ) i ) j ) k ) l ) m ) n ) o ) p ) q ) r ) s ) t ) u ) ( 7 + i ) + ( 3 − 4 i ) = ( 7 + 3 ) + i ( 1 − 4 ) = 10 − 3 i ( 13 − 2 i ) + ( − 5 + 6 i ) = ( 13 − 5 ) + i ( − 2 + 6 ) = 8 + 4 i ( 9 − i ) − ( 8 − i ) = ( 9 − 8 ) + i ( − 1 + 1 ) = 1 ( 3 + 2 i ) − ( 6 + 13 i ) = ( 3 − 6 ) + i ( 2 + 13 ) = − 3 + 15 i ( − 2 + − 8 ) + ( 5 − − 50 ) = ( − 2 + 5 ) + i ( 2 2 − 5 2 ) = 3 − 3 i 2 ( 8 + − 18 ) − ( 4 + 3 i 2 ) = ( 8 − 4 ) + ( 3 i 2 − 3 i 2 ) = 4 13 i − ( 14 − 7 i ) = − 14 + ( 13 i + 7 i ) = − 14 + 20 i 25 + ( − 10 + 11 i ) + 15 i = ( 25 − 10 ) + ( 11 i + 15 i ) = 15 + 26 i − ( 2 3 + 2 5 i ) + ( 3 5 + 3 11 i ) = ( − 2 3 + 3 5 ) + ( − 2 5 + 3 11 ) i = 6 1 + 6 7 i ( 1 , 6 + 3 , 2 i ) + ( − 5 , 8 + 4 , 3 i ) = − 4 , 2 + 7 , 5 i ( 1 + i ) ( 3 − 2 i ) = 1.3 − 1.2 i + 3 i − 2 i 2 = 3 − 2 i + 3 i + 2 = 5 + i ( 7 − 2 i ) ( 3 − 5 i ) = 7.3 − 7.5 i − 2 i .3 + 2.5 i 2 = 21 − 35 i − 6 i − 10 = 11 − 41 i 12 i ( 1 − 9 i ) = 12 i .1 − 12 i .9 i = 12 i − 108 i 2 = 108 + 12 i − 8 i ( 9 + 4 i ) = − 72 i − 32 i 2 = 32 − 72 i ( 14 + i 10 ) ( 14 − i 10 ) = ( 14 ) 2 − ( i 15 ) 2 = 14 − 10 i 2 = 14 + 10 = 24 ( 3 + i 15 ) ( 3 − i 15 ) = ( 3 ) 2 − ( i 15 ) 2 = 3 + 15 = 18 ( 6 + 7 i ) 2 = 6 2 + 2.6.7 i + ( 7 i ) 2 = 36 + 84 i − 49 = − 13 + 84 i ( 5 − 4 i ) 2 = 5 2 − 2.5.4 i + ( 4 i ) 2 = 25 − 40 i − 16 = 9 − 40 i ( 2 + 3 i ) 2 + ( 2 − 3 i ) 2 = ( − 5 + 12 i ) + ( − 5 − 12 i ) = − 10 ( 1 − 2 i ) 2 − ( 1 + 2 i ) 2 = ( − 3 − 4 i ) − ( − 3 + 4 i ) = − 8 i [( 1 + i ) 2 − ( 1 − i ) 2 ] 5 = [ 2 i − ( − 2 i ) ] 5 = ( 4 i ) 5 = 4 5 × i 5 = 2 10 × i 4 1 . i = 1024 i
0144
Escreva os números complexos na sua forma algébrica:
a ) 8 + − 25 b ) 5 + − 36 c ) 2 − − 7 d ) 1 + − 8 e ) − 80 f ) − 4 g ) 14 h ) 75 i ) − 10 i + i 2 j ) − 4 i 2 + 2 i k ) − 0 , 09 l ) − 0 , 0049 \begin{array}{lclclcl}
a) 8+\sqrt{-25} && b) 5+\sqrt{-36} && c) 2-\sqrt{-7} && d) 1+\sqrt{-8}\\\\
e) \sqrt{-80} && f) \sqrt{-4} && g) 14 && h) 75\\\\
i) -10i+i^2 && j) -4i^2+2i && k) \sqrt{-0,09} && l) \sqrt{-0,0049}
\end{array} a ) 8 + − 25 e ) − 80 i ) − 10 i + i 2 b ) 5 + − 36 f ) − 4 j ) − 4 i 2 + 2 i c ) 2 − − 7 g ) 14 k ) − 0 , 09 d ) 1 + − 8 h ) 75 l ) − 0 , 0049
0144 - Solução
Resolvendo, item a item:
a ) 8 + − 25 = 8 + 5 i b ) 5 + − 36 = 5 + 6 i c ) 2 − − 7 = 2 − i 7 d ) 1 + − 8 = 1 + 2 i 2 e ) − 80 = 4 i 5 f ) − 4 = 2 i g ) 14 = 14 h ) 75 = 75 i ) − 10 i + i 2 = − 1 − 10 i j ) − 4 i 2 + 2 i = 4 + 2 i k ) − 0 , 09 = 3 i 10 l ) − 0 , 0049 = 0 , 07 i \begin{array}{llll}
a) & 8+\sqrt{-25}=\boxed{8+5i} & b) & 5+\sqrt{-36}=\boxed{5+6i} \\
& & & \\
c) & 2-\sqrt{-7}=\boxed{2-i\sqrt{7}} & d) & 1+\sqrt{-8}=\boxed{1+2i\sqrt{2}}\\
& & & \\
e) & \sqrt{-80}=\boxed{4i\sqrt{5}} & f) & \sqrt{-4}=\boxed{2i} \\
& & & \\
g) & 14=\boxed{14} & h) & 75=\boxed{75} \\
& & & \\
i) & -10i+i^2=\boxed{-1-10i} & j) & -4i^2+2i=\boxed{4+2i} \\
& & & \\
k) & \sqrt{-0,09}=\boxed{\dfrac{3i}{10}} & l) & \sqrt{-0,0049}=\boxed{0,07i}
\end{array} a ) c ) e ) g ) i ) k ) 8 + − 25 = 8 + 5 i 2 − − 7 = 2 − i 7 − 80 = 4 i 5 14 = 14 − 10 i + i 2 = − 1 − 10 i − 0 , 09 = 10 3 i b ) d ) f ) h ) j ) l ) 5 + − 36 = 5 + 6 i 1 + − 8 = 1 + 2 i 2 − 4 = 2 i 75 = 75 − 4 i 2 + 2 i = 4 + 2 i − 0 , 0049 = 0 , 07 i
0143
Um outdoor medido 1,70m de altura por 4,30m de largura foi pintado de azul com margens brancas. A largura das margens superior e inferior é 40cm e a das margens laterais é 60cm, conforme a imagem ilustrativa abaixo.
Qual a área pintada de branco?
0143 - Solução
A área branca(A b A_{b} A b
A b = ( 4 , 30 m × 1 , 70 m ) − ( 3 , 10 m × 0 , 90 m ) → A_{b}=(4,30\text{m}\times 1,70\text{m})-(3,10\text{m}\times 0,90\text{m})\to A b = ( 4 , 30 m × 1 , 70 m ) − ( 3 , 10 m × 0 , 90 m ) →
A b = 7 , 31 m 2 − 2 , 79 m 2 → A_{b}=7,31\text{m}^2-2,79\text{m}^2\to A b = 7 , 31 m 2 − 2 , 79 m 2 →
A b = 4 , 52 m 2 \boxed{\boldsymbol{A_{b}=4,52\text{m}^2}} A b = 4 , 52 m 2
0142
Encontre os valores "x x x y y y z z z
se { ( x − 4 ) ( y − 4 ) = 16 ( y − 6 ) ( z − 6 ) = 36 ( z − 8 ) ( x − 8 ) = 64 \left\{\begin{array}{lcr}(x-4)(y-4)&=&16\\(y-6)(z-6)&=&36\\(z-8)(x-8)&=&64\end{array}\right. ⎩ ⎨ ⎧ ( x − 4 ) ( y − 4 ) ( y − 6 ) ( z − 6 ) ( z − 8 ) ( x − 8 ) = = = 16 36 64
0142 - Solução
Solução, por etapas:
(I)
[ z − 8 ] = [ ( z − 6 ) − 2 ] → [z-8]=[(z-6)-2]\to [ z − 8 ] = [( z − 6 ) − 2 ] →
[ z − 8 ] = 36 ( y − 6 ) − 2 → [z-8]=\dfrac{36}{(y-6)}-2\to [ z − 8 ] = ( y − 6 ) 36 − 2 →
[ z − 8 ] = 36 − 2 ( y − 6 ) ( y − 6 ) → [z-8]=\dfrac{36-2(y-6)}{(y-6)}\to [ z − 8 ] = ( y − 6 ) 36 − 2 ( y − 6 ) →
[ z − 8 ] = 48 − 2 y y − 6 \boxed{[z-8]=\dfrac{48-2y}{y-6}} [ z − 8 ] = y − 6 48 − 2 y
e
[ x − 8 ] = [ ( x − 4 ) − 4 ] → [x-8]=[(x-4)-4]\to [ x − 8 ] = [( x − 4 ) − 4 ] →
[ x − 8 ] = 16 y − 4 − 4 → [x-8]=\dfrac{16}{y-4}-4\to [ x − 8 ] = y − 4 16 − 4 →
[ x − 8 ] = 16 − 4 ( y − 4 ) y − 4 → [x-8]=\dfrac{16-4(y-4)}{y-4}\to [ x − 8 ] = y − 4 16 − 4 ( y − 4 ) →
[ x − 8 ] = 32 − 4 y y − 4 \boxed{[x-8]=\dfrac{32-4y}{y-4}} [ x − 8 ] = y − 4 32 − 4 y
(II)
48 − 2 y y − 6 ⋅ 32 − 4 y y − 4 = 64 \dfrac{48-2y}{y-6}\cdot\dfrac{32-4y}{y-4}=64 y − 6 48 − 2 y ⋅ y − 4 32 − 4 y = 64
Para y ≠ 6 y\neq 6\,\, y = 6 y ≠ 4 \,\,y\neq 4 y = 4
2 ⋅ ( 24 − y ) ⋅ 4 ⋅ ( 8 − y ) = 64 ⋅ ( y − 6 ) ⋅ ( y − 4 ) 2\cdot(24-y)\cdot 4\cdot(8-y)=64\cdot(y-6)\cdot(y-4) 2 ⋅ ( 24 − y ) ⋅ 4 ⋅ ( 8 − y ) = 64 ⋅ ( y − 6 ) ⋅ ( y − 4 )
8 ⋅ ( y 2 − 32 y + 192 ) = 64 ⋅ ( y 2 − 10 y + 24 ) → \cancel{8}\cdot(y^2-32y+192)=\cancel{64}\cdot(y^2-10y+24)\to 8 ⋅ ( y 2 − 32 y + 192 ) = 64 ⋅ ( y 2 − 10 y + 24 ) →
y 2 − 32 y + 192 = 8 y 2 − 80 y + 192 → y^2-32y+\cancel{192}=8y^2-80y+\cancel{192}\to y 2 − 32 y + 192 = 8 y 2 − 80 y + 192 →
7 y 2 − 48 y = 0 ⇛ y = 0 ✓ \boxed{7y^2-48y=0}\Rrightarrow \boxed{y=0}\checkmark\quad 7 y 2 − 48 y = 0 ⇛ y = 0 ✓ y = 48 7 ✓ \quad\boxed{y=\dfrac{48}{7}}\checkmark y = 7 48 ✓
(III) y y y
vamos obter os respectivos valores de "x x x z z z
⇛ \Rrightarrow ⇛ y = 0 \boxed{y=0} y = 0
Aplicado-o à equação x − 8 = 32 − 4 y y − 4 x-8=\dfrac{32-4y}{y-4} x − 8 = y − 4 32 − 4 y
x − 8 = 32 − 4 ⋅ 0 0 − 4 → x − 8 = − 8 → x = 0 x-8=\dfrac{32-4\cdot 0}{0-4}\to x-8=-8\to\boxed{x=0} x − 8 = 0 − 4 32 − 4 ⋅ 0 → x − 8 = − 8 → x = 0
Aplicando-o à equação z − 8 = 48 − 2 y y − 6 z-8=\dfrac{48-2y}{y-6} z − 8 = y − 6 48 − 2 y
z − 8 = 48 − 2 ⋅ 0 0 − 6 → z − 8 = − 8 → z = 0 z-8=\dfrac{48-2\cdot 0}{0-6}\to z-8=-8\to\boxed{z=0} z − 8 = 0 − 6 48 − 2 ⋅ 0 → z − 8 = − 8 → z = 0
Aqui, a primeira terna final de solução: ( 0 ; 0 ; 0 ) ✓ \boldsymbol{(0;\,0;\,0)}\checkmark ( 0 ; 0 ; 0 ) ✓
⇛ \Rrightarrow ⇛ y = 48 7 \boxed{y=\dfrac{48}{7}} y = 7 48
Aplicado-o à equação x − 8 = 32 − 4 y y − 4 x-8=\dfrac{32-4y}{y-4} x − 8 = y − 4 32 − 4 y
x − 8 = 32 − 4 ⋅ 48 7 48 7 − 4 → x − 8 = 32 7 20 7 → x-8=\dfrac{32-4\cdot\frac{48}{7}}{\frac{48}{7}-4}\to x-8=\dfrac{\frac{32}{\cancel{7}}}{\frac{20}{\cancel{7}}}\to x − 8 = 7 48 − 4 32 − 4 ⋅ 7 48 → x − 8 = 7 20 7 32 →
x − 8 = 8 5 → x = 48 5 x-8=\dfrac{8}{5}\to\boxed{x=\dfrac{48}{5}} x − 8 = 5 8 → x = 5 48
Aplicando-o à equação z − 8 = 48 − 2 y y − 6 z-8=\dfrac{48-2y}{y-6} z − 8 = y − 6 48 − 2 y
z − 8 = 48 − 2 ⋅ 48 7 48 7 − 6 → z − 8 = 240 7 6 7 → z-8=\dfrac{48-2\cdot\frac{48}{7}}{\frac{48}{7}-6}\to z-8=\dfrac{\frac{240}{\cancel{7}}}{\frac{6}{\cancel{7}}}\to z − 8 = 7 48 − 6 48 − 2 ⋅ 7 48 → z − 8 = 7 6 7 240 →
z − 8 = 40 → z = 48 z-8=40\to\boxed{z=48} z − 8 = 40 → z = 48
Aqui, a segunda terna final de solução:( 48 5 ; 48 7 ; 48 ) ✓ \boldsymbol{\left(\dfrac{48}{5};\,\dfrac{48}{7};\,48\right)}\checkmark ( 5 48 ; 7 48 ; 48 ) ✓
0141
Resolva a equação trigonométrica
cos ( 2 x ) = sen ( π 4 − x ) \cos\,(2x) = \text{sen}\,\left(\dfrac{\pi}{4} - x\right)\,\, cos ( 2 x ) = sen ( 4 π − x ) x ∈ [ − π , π ] \,\,x\in[-\pi,\pi] x ∈ [ − π , π ]
0141 - Solução
Por etapas, teremos:
(I)
Lado Esquerdo :
cos ( 2 x ) = cos 2 x − sen 2 x \cos\,(2x)=\cos^2x-\text{sen}^2\,x cos ( 2 x ) = cos 2 x − sen 2 x
cos ( 2 x ) = [ cos ( x ) + sen ( x ) ] × [ cos ( x ) − sen ( x ) ] \cos\,(2x)=[\cos (x)+\text{sen}\,(x)]\times [\cos (x)-\text{sen}\,(x)] cos ( 2 x ) = [ cos ( x ) + sen ( x )] × [ cos ( x ) − sen ( x )]
Lado Direito :
sen ( π 4 − x ) = sen ( π 4 ) × cos ( x ) − sen ( x ) × cos ( π 4 ) → \text{sen}\,\left(\dfrac{\pi}{4} - x\right)=\text{sen}\,\left(\dfrac{\pi}{4}\right)\times\cos(x)-\text{sen}\,(x)\times\cos\left(\dfrac{\pi}{4}\right)\to sen ( 4 π − x ) = sen ( 4 π ) × cos ( x ) − sen ( x ) × cos ( 4 π ) →
sen ( π 4 − x ) = 2 2 × cos ( x ) − sen ( x ) × 2 2 → \text{sen}\,\left(\dfrac{\pi}{4} - x\right)=\dfrac{\sqrt{2}}{2}\times\cos(x)-\text{sen}\,(x)\times\dfrac{\sqrt{2}}{2}\to sen ( 4 π − x ) = 2 2 × cos ( x ) − sen ( x ) × 2 2 →
sen ( π 4 − x ) = 2 2 × [ cos ( x ) − sen ( x ) ] \text{sen}\,\left(\dfrac{\pi}{4} - x\right)=\dfrac{\sqrt{2}}{2}\times[\cos(x)-\text{sen}\,(x)] sen ( 4 π − x ) = 2 2 × [ cos ( x ) − sen ( x )]
(II)
[ cos ( x ) + sen ( x ) ] × [ cos ( x ) − sen ( x ) ] = 2 2 × [ cos ( x ) − sen ( x ) ] [\cos (x)+\text{sen}\,(x)]\times [\cos (x)-\text{sen}\,(x)]=\dfrac{\sqrt{2}}{2}\times[\cos(x)-\text{sen}\,(x)] [ cos ( x ) + sen ( x )] × [ cos ( x ) − sen ( x )] = 2 2 × [ cos ( x ) − sen ( x )]
Para cos ( x ) ≠ sen ( x ) \cos(x)\neq\text{sen}\,(x) cos ( x ) = sen ( x )
[ cos ( x ) + sen ( x ) ] × [ cos ( x ) − sen ( x ) ] = 2 2 × [ cos ( x ) − sen ( x ) ] [\cos (x)+\text{sen}\,(x)]\times\cancel{[\cos (x)-\text{sen}\,(x)]}=\dfrac{\sqrt{2}}{2}\times\cancel{[\cos(x)-\text{sen}\,(x)]} [ cos ( x ) + sen ( x )] × [ cos ( x ) − sen ( x )] = 2 2 × [ cos ( x ) − sen ( x )]
Assim:
cos ( x ) + sen ( x ) = 2 2 \cos (x)+\text{sen}\,(x)=\dfrac{\sqrt{2}}{2} cos ( x ) + sen ( x ) = 2 2
Elevando ambos os lados ao quadrado:
[ cos ( x ) + sen ( x ) ] 2 = ( 2 2 ) 2 → [\cos (x)+\text{sen}\,(x)]^2=\left(\dfrac{\sqrt{2}}{2}\right)^2\to [ cos ( x ) + sen ( x ) ] 2 = ( 2 2 ) 2 →
cos 2 ( x ) + sen 2 ( x ) ⏟ 1 + 2 ⋅ sen ( x ) ⋅ cos ( x ) = 1 2 → \underbrace{\cos^2(x)+\text{sen}^2\,(x)}_{1}+2\cdot\text{sen}\,(x)\cdot\cos(x)=\dfrac{1}{2}\to 1 cos 2 ( x ) + sen 2 ( x ) + 2 ⋅ sen ( x ) ⋅ cos ( x ) = 2 1 →
2 ⋅ sen ( x ) ⋅ cos ( x ) ⏟ = − 1 2 → \underbrace{2\cdot\text{sen}\,(x)\cdot\cos(x)}=-\dfrac{1}{2}\to 2 ⋅ sen ( x ) ⋅ cos ( x ) = − 2 1 →
sen ( 2 x ) = − 1 2 \boxed{\text{sen}\,(2x)=-\dfrac{1}{2}} sen ( 2 x ) = − 2 1
(III) x ∈ [ − π ; π ] \boxed{x\in[-\pi;\,\pi]} x ∈ [ − π ; π ]
2 x = − π 6 → x = − π 12 ✓ 2x=-\dfrac{\pi}{6}\to\boxed{x=-\dfrac{\pi}{12}}\checkmark\quad 2 x = − 6 π → x = − 12 π ✓
2 x = − 5 π 6 → x = − 5 π 12 ✓ 2x=-\dfrac{5\pi}{6}\to\boxed{x=-\dfrac{5\pi}{12}}\checkmark\quad 2 x = − 6 5 π → x = − 12 5 π ✓
0140
Dada a matriz A = [ 1 1 1 1 ] A=\left[\begin{array}{cc}1&1\\1&1\end{array}\right] A = [ 1 1 1 1 ] f f f 2 × 2 2\times 2 2 × 2 f ( X ) = X 2 − 2 X f(X)=X^2-2X f ( X ) = X 2 − 2 X f ( A ) f(A) f ( A )
0140 - Solução
Por etapas, teremos:
1º) Fazendo X 2 = A 2 X^2=A^2 X 2 = A 2 X ⋅ X = A ⋅ A X\cdot X=A\cdot A X ⋅ X = A ⋅ A
A ⋅ A = [ 1 1 1 1 ] × [ 1 1 1 1 ] → A\cdot A=\left[\begin{array}{cc}1&1\\1&1\end{array}\right]\times \left[\begin{array}{cc}1&1\\1&1\end{array}\right]\to A ⋅ A = [ 1 1 1 1 ] × [ 1 1 1 1 ] →
A ⋅ A = [ 1 ⋅ 1 + 1 ⋅ 1 1 ⋅ 1 + 1 ⋅ 1 1 ⋅ 1 + 1 ⋅ 1 1 ⋅ 1 + 1 ⋅ 1 ] → A\cdot A=\left[\begin{array}{ccc}1\cdot 1+1\cdot 1&&1\cdot 1+1\cdot 1\\1\cdot 1+1\cdot 1&&1\cdot 1+1\cdot 1\end{array}\right]\to A ⋅ A = [ 1 ⋅ 1 + 1 ⋅ 1 1 ⋅ 1 + 1 ⋅ 1 1 ⋅ 1 + 1 ⋅ 1 1 ⋅ 1 + 1 ⋅ 1 ] →
A ⋅ A = A 2 = [ 2 2 2 2 ] A\cdot A=A^2=\left[\begin{array}{cc}2&2\\2&2\end{array}\right] A ⋅ A = A 2 = [ 2 2 2 2 ]
2º) Fazendo 2 X = 2 A 2X=2A 2 X = 2 A
2 × [ 1 1 1 1 ] → 2 A = [ 2 2 2 2 ] 2\times\left[\begin{array}{cc}1&1\\1&1\end{array}\right]\to 2A=\left[\begin{array}{cc}2&2\\2&2\end{array}\right] 2 × [ 1 1 1 1 ] → 2 A = [ 2 2 2 2 ]
3º) Obtendo f ( A ) = A 2 − 2 A f(A)=A^2-2A f ( A ) = A 2 − 2 A
f ( A ) = [ 2 2 2 2 ] − [ 2 2 2 2 ] ∴ f ( A ) = 0 ✓ f(A)=\left[\begin{array}{cc}2&2\\2&2\end{array}\right]-\left[\begin{array}{cc}2&2\\2&2\end{array}\right]\therefore \boxed{f(A)=0}\checkmark f ( A ) = [ 2 2 2 2 ] − [ 2 2 2 2 ] ∴ f ( A ) = 0 ✓
0139
Sendo a f ( 4 ) = 2 f(4)=2 f ( 4 ) = 2 f ( 4 ) = 4 a + b f(4)=4a+b f ( 4 ) = 4 a + b 4 a + b = 2 4a+b=2 4 a + b = 2
que f ( 3 ) = 3 a + b f(3)=3a+ b f ( 3 ) = 3 a + b f ( 5 ) = 5 a + b f(5)= 5a+b f ( 5 ) = 5 a + b
0139 - Solução
Reescrevendo f ( 3 ) = − a + 4 a + b ⏟ 2 → f ( 3 ) = − a + 2 f(3)=-a+\underbrace{4a+b}_{2}\to f(3)=-a+2 f ( 3 ) = − a + 2 4 a + b → f ( 3 ) = − a + 2
Reescrevendo f ( 5 ) = a + 4 a + b ⏟ 2 → f ( 5 ) = a + 2 f(5)=a+\underbrace{4a+b}_{2}\to f(5)=a+2 f ( 5 ) = a + 2 4 a + b → f ( 5 ) = a + 2
Escrevendo a função soma das funções, teremos:
f ( 3 ) + f ( 5 ) = − a + 2 + a + 2 → f ( 3 ) + f ( 5 ) = 4 ✓ f(3)+f(5)=-\cancel{a}+2+\cancel{a}+2\to \boxed{f(3)+f(5)=4}\checkmark f ( 3 ) + f ( 5 ) = − a + 2 + a + 2 → f ( 3 ) + f ( 5 ) = 4 ✓
0138
Use a equação racional 4 x + 1 x 2 + 2 x = 3 x + 2 \dfrac{4}{x} + \dfrac{1}{x^{2}+2x} = \dfrac{3}{x+2} x 4 + x 2 + 2 x 1 = x + 2 3
Parte 1: Uma vez que a equação não está mais na forma de uma equação racional, qual método pode ser usado para resolver a equação?
Parte 2: Qual é a solução para a equação racional?
Selecione uma resposta para a Parte 1 e selecione todas as respostas que se aplicam à Parte 2.
PARTE1
A: A equação resultante é uma equação quadrática, onde b = 0; pode ser resolvido isolando a variável e obtendo a raiz quadrada de cada lado.
B: A equação resultante é uma equação quadrática, que pode ser resolvida por fatoração.
C: A equação resultante é uma equação quadrática, que não pode ser resolvida por fatoração; pode ser resolvido pela fórmula quadrática.
D: A equação resultante é uma equação linear, que pode ser resolvida usando as operações inversas e as propriedades da igualdade.
PARTE2
A: 3
B: −3
C: −97
D: −9
E: 9
0138 - Soluções
4 x + 1 x 2 + 2 x = 3 x + 2 → \dfrac{4}{x}+\dfrac{1}{x^{2}+2x}=\dfrac{3}{x+2}\to x 4 + x 2 + 2 x 1 = x + 2 3 →
4 ( x + 2 ) + 1 = 3 x x ( x + 2 ) → \dfrac{4(x+2)+1=3x}{x(x+2)}\to x ( x + 2 ) 4 ( x + 2 ) + 1 = 3 x →
Para x ≠ 0 x\neq 0\,\, x = 0 x ≠ − 2 \,\,x\neq-2 x = − 2 eliminar o denominador; assim, vamos resolver a equação resultante, do tipo linear , apenas do numerador:
4 ( x + 2 ) + 1 = 3 x → 4 x − 3 x = − 1 − 8 → x = − 9 ✓ 4(x+2)+1=3x\to 4x-3x=-1-8\to\boxed{x=-9}\checkmark 4 ( x + 2 ) + 1 = 3 x → 4 x − 3 x = − 1 − 8 → x = − 9 ✓
Respondendo:
Na PARTE 1, a alternativa correta é a "D ".
Na PARTE 2, a alternativa correta é a "D ".
0137
Obtenha o gráfico da função exponencial f ( x ) = 3 x f(x)=3^x f ( x ) = 3 x
0137 - Solução
0136
Qual o valor da ordenada do ponto A ( 3 ; y ) A(3;\,y) A ( 3 ; y ) B B B C ( 3 ; 3 ) C(3;\,3) C ( 3 ; 3 )
0136 - Solução
Questão meramente interpretativa, portanto, desnecessário qualquer cálculo, senão vejamos:
a) O ponto B B B B ( 0 ; 0 ) B(0;\,0) B ( 0 ; 0 )
b) Como os três pontos estão alinhados e C ( 3 ; 3 ) C(3;\,3) C ( 3 ; 3 ) y = x y=x y = x
c) Dessa forma o ponto A A A A ( 3 ; 3 ) A(3;\,3) A ( 3 ; 3 ) y = 3 y=3 y = 3
0135
Em um mapa turístico do Brasil, de escala 1 : 18000000 1:18000000 1 : 18000000 T e a cidade P mede 2cm. Obtenha a distância entre as duas cidades, em quilômetros.
0135 - Solução
De acordo com o texto, cada cm no mapa corresponde a 18.000.000 cm de distância real; logo, 2 c m 2cm 2 c m 36.000.000 c m 36.000.000cm 36.000.000 c m c m cm c m k m km km
36.000.000 c m → 360.000 m → 360 k m ✓ 36.000.000cm\to 360.000m\to\boxed{360km}\checkmark 36.000.000 c m → 360.000 m → 360 km ✓
0134
Resolva, em R \mathbb{R} R
∣ 2 2 3 2 4 x 3 1 2 ∣ = 0 \left|\begin{array}{ccc}2 & 2 & 3\\2 & 4 & x\\3 & 1 & 2\end{array}\right|=0 2 2 3 2 4 1 3 x 2 = 0
0134 - Solução
∣ 2 2 3 2 4 x 3 1 2 ∣ = 0 → \left|\begin{array}{ccc}2 & 2 & 3\\2 & 4 & x\\3 & 1 & 2\end{array}\right|=0\to 2 2 3 2 4 1 3 x 2 = 0 →
( 2 × 4 × 2 ) + ( 2 × 1 × 3 ) + ( 3 × x × 2 ) (2\times 4\times 2)+(2\times 1\times 3)+(3\times x\times 2) ( 2 × 4 × 2 ) + ( 2 × 1 × 3 ) + ( 3 × x × 2 )
− ( 3 × 4 × 3 ) − ( 2 × 2 × 2 ) − ( 2 × 1 × x ) = 0 → -(3\times 4\times 3)-(2\times 2\times 2)-(2\times 1\times x)=0\to − ( 3 × 4 × 3 ) − ( 2 × 2 × 2 ) − ( 2 × 1 × x ) = 0 →
16 + 6 + 6 x − 36 − 8 − 2 x = 0 → 4 x − 22 = 0 → x = 11 2 ✓ 16+6+6x-36-8-2x=0\to 4x-22=0\to\boxed{x=\dfrac{11}{2}}\checkmark 16 + 6 + 6 x − 36 − 8 − 2 x = 0 → 4 x − 22 = 0 → x = 2 11 ✓
0133
Resolva, em R \mathbb{R} R
∣ x 3 5 x + 1 2 1 3 2 4 ∣ = 0 \left|\begin{array}{ccc}x & 3 & 5\\x+1 & 2 & 1\\3 & 2 & 4\end{array}\right|=0 x x + 1 3 3 2 2 5 1 4 = 0
0133 - Solução
∣ x 3 5 x + 1 2 1 3 2 4 ∣ = 0 → \left|\begin{array}{ccc}x & 3 & 5\\x+1 & 2 & 1\\3 & 2 & 4\end{array}\right|=0\to x x + 1 3 3 2 2 5 1 4 = 0 →
[ x × 2 × 4 ] + [ ( x + 1 ) × 2 × 5 ] + [ 3 × 1 × 3 ] [x\times 2\times 4]+[(x+1)\times 2\times 5]+[3\times 1\times 3] [ x × 2 × 4 ] + [( x + 1 ) × 2 × 5 ] + [ 3 × 1 × 3 ]
− [ 5 × 2 × 3 ] − [ 3 × ( x + 1 ) × 4 ] − [ x × 2 × 1 ] = 0 → -[5\times 2\times 3]-[3\times(x+1)\times 4]-[x\times 2\times 1]=0\to − [ 5 × 2 × 3 ] − [ 3 × ( x + 1 ) × 4 ] − [ x × 2 × 1 ] = 0 →
8 x + 10 x + 10 + 9 − 30 − 12 x − 12 − 2 x = 0 → 8x+10x+10+9-30-12x-12-2x=0\to 8 x + 10 x + 10 + 9 − 30 − 12 x − 12 − 2 x = 0 →
4 x − 23 = 0 → x = 23 4 ✓ 4x-23=0\to\boxed{x=\dfrac{23}{4}}\checkmark 4 x − 23 = 0 → x = 4 23 ✓
0132
Encontre os valores reais de a a a b b b
a) a + b i = − 12 + 7 i a+bi=-12+7i a + bi = − 12 + 7 i
b) a + b i = 13 + 4 i a+bi=13+4i a + bi = 13 + 4 i
c) ( a − 1 ) + ( b + 3 ) i = 5 + 8 i (a-1)+(b+3)i=5+8i ( a − 1 ) + ( b + 3 ) i = 5 + 8 i
d) ( a + 6 ) + 2 b i = 6 − 5 i (a+6)+2bi=6-5i ( a + 6 ) + 2 bi = 6 − 5 i
0132 - Soluções
A fim de encontrar os valores reais de a a a b b b
a) a + b i = − 12 + 7 i → a = − 12 a+bi=-12+7i\to\boxed{a=-12} a + bi = − 12 + 7 i → a = − 12 b = 7 \boxed{b=7} b = 7
b) a + b i = 13 + 4 i → a = 13 a+bi=13+4i\to\boxed{a=13} a + bi = 13 + 4 i → a = 13 b = 4 \boxed{b=4} b = 4
c) ( a − 1 ) + ( b + 3 ) i = 5 + 8 i (a-1)+(b+3)i=5+8i ( a − 1 ) + ( b + 3 ) i = 5 + 8 i
{ a − 1 = 5 → a = 6 b + 3 = 8 → b = 5 \left\{\begin{array}{rcrcr}
a & - & 1 & = & 5 \to\boxed{a=6}\\
b & + & 3 & = & 8 \to\boxed{b=5}
\end{array}\right. { a b − + 1 3 = = 5 → a = 6 8 → b = 5
d) ( a + 6 ) + 2 b i = 6 − 5 i (a+6)+2bi=6-5i ( a + 6 ) + 2 bi = 6 − 5 i
{ a + 6 = 6 → a = 0 2 b = − 5 → b = − 5 2 \left\{\begin{array}{rcrcrl}
a & + & 6 & = & 6 &\to\boxed{a=0}\\
2b & & & = & -5 &\to \boxed{b=-\dfrac{5}{2}}
\end{array}\right. ⎩ ⎨ ⎧ a 2 b + 6 = = 6 − 5 → a = 0 → b = − 2 5
0131
Resolva as equações quadráticas:
a) 3 x 2 − 2 x + 5 = 0 3x^2-2x+5=0 3 x 2 − 2 x + 5 = 0
b) 8 x 2 + 14 x + 9 = 0 8x^2+14x+9=0 8 x 2 + 14 x + 9 = 0
0131 - Soluções
Resolveremos cada equação utilizando a fórmula quadrática Bhaskara :
a) 3 x 2 − 2 x + 5 = 0 → B h a s k a r a → x = − ( − 2 ) ± ( − 2 ) 2 − 4 × 3 × 5 2 × 3 3x^2-2x+5=0\to \quad\underrightarrow{Bhaskara}\quad x=\dfrac{-(-2)\pm \sqrt{(-2)^2-4\times 3\times 5}}{2\times 3} 3 x 2 − 2 x + 5 = 0 → B ha s ka r a x = 2 × 3 − ( − 2 ) ± ( − 2 ) 2 − 4 × 3 × 5
x = 2 ± 4 − 60 6 → x = 2 ± − 56 6 → x = 2 ± 2 i 14 6 x = 2 ( 1 ± i 14 ) 6 x=\dfrac{2\pm\sqrt{4-60}}{6}\to x=\dfrac{2\pm\sqrt{-56}}{6}\to x=\dfrac{2\pm2i\sqrt{14}}{6} x=\dfrac{\cancel{2}(1\pm i\sqrt{14})}{\cancel{6}} x = 6 2 ± 4 − 60 → x = 6 2 ± − 56 → x = 6 2 ± 2 i 14 x = 6 2 ( 1 ± i 14 )
x = 1 ± i 14 3 → x 1 = 1 3 − i 14 3 ou x 2 = 1 3 + i 14 3 x=\dfrac{1\pm i\sqrt{14}}{3}\to\boxed{x_{1}=\dfrac{1}{3}-\dfrac{i\sqrt{14}}{3}}\quad\text{ou}\quad\boxed{x_{2}=\dfrac{1}{3}+\dfrac{i\sqrt{14}}{3}} x = 3 1 ± i 14 → x 1 = 3 1 − 3 i 14 ou x 2 = 3 1 + 3 i 14
b) 8 x 2 + 14 x + 9 = 0 → B h a s k a r a → x = − 14 ± 14 2 − 4 × 8 × 9 2 × 8 8x^2+14x+9=0\to \quad\underrightarrow{Bhaskara}\quad x=\dfrac{-14\pm \sqrt{14^2-4\times 8\times 9}}{2\times 8} 8 x 2 + 14 x + 9 = 0 → B ha s ka r a x = 2 × 8 − 14 ± 1 4 2 − 4 × 8 × 9
x = − 14 ± 196 − 288 16 → x = − 14 ± − 92 16 → x = − 14 ± 2 i 23 16 x = 2 ( − 7 ± i 23 ) 16 x=\dfrac{-14\pm\sqrt{196-288}}{16}\to x=\dfrac{-14\pm\sqrt{-92}}{16}\to x=\dfrac{-14\pm2i\sqrt{23}}{16} x=\dfrac{\cancel{2}(-7\pm i\sqrt{23})}{\cancel{16}} x = 16 − 14 ± 196 − 288 → x = 16 − 14 ± − 92 → x = 16 − 14 ± 2 i 23 x = 16 2 ( − 7 ± i 23 )
x = − 7 ± i 23 8 → x 1 = − 7 8 − i 23 8 ou x 2 = − 7 8 + i 23 8 x=\dfrac{-7\pm i\sqrt{23}}{8}\to\boxed{x_{1}=-\dfrac{7}{8}-\dfrac{i\sqrt{23}}{8}}\quad\text{ou}\quad\boxed{x_{2}=-\dfrac{7}{8}+\dfrac{i\sqrt{23}}{8}} x = 8 − 7 ± i 23 → x 1 = − 8 7 − 8 i 23 ou x 2 = − 8 7 + 8 i 23
0130
Efetue as seguintes operações:
a) ( 2 − 4 i ) ( 3 + 7 i ) (2-4i)(3+7i) ( 2 − 4 i ) ( 3 + 7 i )
b) ( 4 − 5 i ) ( − 4 − 5 i ) (4-5i)(-4-5i) ( 4 − 5 i ) ( − 4 − 5 i )
c) ( 3 − 5 i ) 2 (3-5i)^2 ( 3 − 5 i ) 2
d) ( 7 + 2 i ) 2 (7+2i)^2 ( 7 + 2 i ) 2
e) ( 3 − 5 i ) 3 (3-5i)^3 ( 3 − 5 i ) 3
f) ( 7 + 2 i ) 3 (7+2i)^3 ( 7 + 2 i ) 3
g) ( 2 − 11 i ) 4 (2-11i)^4 ( 2 − 11 i ) 4
h) ( 1 + 3 i ) 4 (1+3i)^4 ( 1 + 3 i ) 4
i) ( 1 + i ) 2 (1+i)^2 ( 1 + i ) 2
j) ( 1 − i ) 2 (1-i)^2 ( 1 − i ) 2
0130 - Soluções
Resolvendo cada item:
a) ( 2 − 4 i ) ( 3 + 7 i ) = 2.3 + 2.7 i − 4 i .3 − 4 i .7 i = 6 + 14 i − 12 i + 28 = 34 + 2 i (2-4i)(3+7i)=2.3+2.7i-4i.3-4i.7i=6+14i-12i+28=\boxed{34+2i} ( 2 − 4 i ) ( 3 + 7 i ) = 2.3 + 2.7 i − 4 i .3 − 4 i .7 i = 6 + 14 i − 12 i + 28 = 34 + 2 i
b) ( 4 − 5 i ) ( − 4 − 5 i ) = − ( 4 + 5 i ) ( 4 − 5 i ) = − ( 16 + 25 ) = 41 (4-5i)(-4-5i)=-(4+5i)(4-5i)=-(16+25)=\boxed{41} ( 4 − 5 i ) ( − 4 − 5 i ) = − ( 4 + 5 i ) ( 4 − 5 i ) = − ( 16 + 25 ) = 41
c) ( 3 − 5 i ) 2 = ( 3 − 5 i ) ( 3 − 5 i ) = 9 − 15 i − 15 i − 25 = − 16 − 30 i (3-5i)^2=(3-5i)(3-5i)=9-15i-15i-25=\boxed{-16-30i} ( 3 − 5 i ) 2 = ( 3 − 5 i ) ( 3 − 5 i ) = 9 − 15 i − 15 i − 25 = − 16 − 30 i − 2 ( 8 + 15 i ) \boxed{-2(8+15i)} − 2 ( 8 + 15 i )
d) ( 7 + 2 i ) 2 = 7 2 + 2.7.2 i + ( 2 i ) 2 = 49 + 28 i − 4 = 45 + 28 i (7+2i)^2=7^2+2.7.2i+(2i)^2=49+28i-4=\boxed{45+28i} ( 7 + 2 i ) 2 = 7 2 + 2.7.2 i + ( 2 i ) 2 = 49 + 28 i − 4 = 45 + 28 i
e) ( 3 − 5 i ) 3 = ( 3 − 5 i ) 2 . ( 3 − 5 i ) = ( − 16 − 30 i ) ( 3 − 5 i ) = (3-5i)^3=(3-5i)^2.(3-5i)=(-16-30i)(3-5i)= ( 3 − 5 i ) 3 = ( 3 − 5 i ) 2 . ( 3 − 5 i ) = ( − 16 − 30 i ) ( 3 − 5 i ) =
− 16.3 − 16 ( − 5 i ) − 30 i .3 − 30 i ( − 5 i ) = − 48 + 80 i − 90 i − 150 = − 198 − 10 i -16.3-16(-5i)-30i.3-30i(-5i)=-48+80i-90i-150=\boxed{-198-10i} − 16.3 − 16 ( − 5 i ) − 30 i .3 − 30 i ( − 5 i ) = − 48 + 80 i − 90 i − 150 = − 198 − 10 i − 2 ( 99 + 5 i ) \boxed{-2(99+5i)} − 2 ( 99 + 5 i )
f) ( 7 + 2 i ) 3 = 7 3 + 3.7 2 .2 i + 3.7. ( 2 i ) 2 + ( 2 i ) 3 = 343 + 294 i − 84 − 8 i = 259 + 286 i (7+2i)^3=7^3+3.7^2.2i+3.7.(2i)^2+(2i)^3=343+294i-84-8i=\boxed{259+286i} ( 7 + 2 i ) 3 = 7 3 + 3. 7 2 .2 i + 3.7. ( 2 i ) 2 + ( 2 i ) 3 = 343 + 294 i − 84 − 8 i = 259 + 286 i
g) ( 2 − i ) 4 = [ ( 2 − i ) 2 ] 2 = ( 4 − 4 i − 1 ) 2 = ( 3 − 4 i ) 2 = 9 − 24 i − 16 = − 7 − 24 i (2-i)^4=[(2-i)^2]^2=(4-4i-1)^2=(3-4i)^2=9-24i-16=\boxed{-7-24i} ( 2 − i ) 4 = [( 2 − i ) 2 ] 2 = ( 4 − 4 i − 1 ) 2 = ( 3 − 4 i ) 2 = 9 − 24 i − 16 = − 7 − 24 i
h) ( 1 + 3 i ) 4 = ( 1 + 6 i − 9 ) 2 = ( − 8 + 6 i ) 2 = [ − 2 ( 4 − 3 i ) ] 2 = 4 ( 16 − 24 i − 9 ) = 4 ( 7 − 24 i ) (1+3i)^4=(1+6i-9)^2=(-8+6i)^2=[-2(4-3i)]^2=4(16-24i-9)=\boxed{4(7-24i)} ( 1 + 3 i ) 4 = ( 1 + 6 i − 9 ) 2 = ( − 8 + 6 i ) 2 = [ − 2 ( 4 − 3 i ) ] 2 = 4 ( 16 − 24 i − 9 ) = 4 ( 7 − 24 i )
i) ( 1 + i ) 2 = 1 + 2 i − 1 = 2 i (1+i)^2=1+2i-1=\boxed{2i} ( 1 + i ) 2 = 1 + 2 i − 1 = 2 i
j) ( 1 − i ) 2 = 1 − 2 i − 1 = − 2 i (1-i)^2=1-2i-1=\boxed{-2i} ( 1 − i ) 2 = 1 − 2 i − 1 = − 2 i
0129
Efetue as seguintes operações:
a) ( 8 + 7 i ) + ( 4 − 5 i ) (8+7i)+(4-5i) ( 8 + 7 i ) + ( 4 − 5 i )
b) ( 3 − 4 i ) − ( − 5 + 7 i ) (3-4i)-(-5+7i) ( 3 − 4 i ) − ( − 5 + 7 i )
c) − 3 i + ( 12 i − 7 3 ) − ( − 7 − i 7 3 ) -3i+(12i-\sqrt[3]{7})-(-\sqrt{7}-i\sqrt[3]{7}) − 3 i + ( 12 i − 3 7 ) − ( − 7 − i 3 7 )
d) ( 5 − 3 i ) − ( 3 + 5 i ) + ( 8 + 13 i ) (5-3i)-(3+5i)+(8+13i) ( 5 − 3 i ) − ( 3 + 5 i ) + ( 8 + 13 i )
0129 - Soluções
Resolvendo cada item:
a) ( 8 + 7 i ) + ( 4 − 5 i ) = 8 + 7 i + 4 − 5 i = 12 + 2 i (8+7i)+(4-5i)=8+7i+4-5i=\boxed{12+2i} ( 8 + 7 i ) + ( 4 − 5 i ) = 8 + 7 i + 4 − 5 i = 12 + 2 i
b) ( 3 − 4 i ) − ( − 5 + 7 i ) = 3 − 4 i + 5 − 7 i = 8 − 11 i (3-4i)-(-5+7i)=3-4i+5-7i=\boxed{8-11i} ( 3 − 4 i ) − ( − 5 + 7 i ) = 3 − 4 i + 5 − 7 i = 8 − 11 i
c) − 3 i + ( 12 i − 7 3 ) − ( − 7 − i 7 3 ) = − 3 i + 12 i − 7 3 + 7 + i 7 3 = ( 7 − 7 3 ) + i ( 9 + 7 3 ) -3i+(12i-\sqrt[3]{7})-(-\sqrt{7}-i\sqrt[3]{7})=-3i+12i-\sqrt[3]{7}+\sqrt{7}+i\sqrt[3]{7}=\boxed{(\sqrt{7}-\sqrt[3]{7})+i(9+\sqrt[3]{7})} − 3 i + ( 12 i − 3 7 ) − ( − 7 − i 3 7 ) = − 3 i + 12 i − 3 7 + 7 + i 3 7 = ( 7 − 3 7 ) + i ( 9 + 3 7 )
d) ( 5 − 3 i ) − ( 3 + 5 i ) + ( 8 + 13 i ) = 5 − 3 i − 3 − 5 i + 8 + 13 i = 10 + 5 i ou 5 ( 2 + i ) (5-3i)-(3+5i)+(8+13i)=5-3i-3-5i+8+13i=\boxed{10+5i}\,\,\text{ou}\,\,\boxed{5(2+i)} ( 5 − 3 i ) − ( 3 + 5 i ) + ( 8 + 13 i ) = 5 − 3 i − 3 − 5 i + 8 + 13 i = 10 + 5 i ou 5 ( 2 + i )
0128
Para cada número complexo z z z
Escreva z z z
Determine k ∈ R k\in\mathbb{R} k ∈ R ( R e ) ≥ 0 (Re)\geq 0 ( R e ) ≥ 0
Determine k ∈ R k\in\mathbb{R} k ∈ R ( R e ) < 0 (Re)<0 ( R e ) < 0
Determine k ∈ R k\in\mathbb{R} k ∈ R z z z
Determine k ∈ R k\in\mathbb{R} k ∈ R z z z
a) z = ( k − 3 , 3 − k ) z=(k-3,\,3-k) z = ( k − 3 , 3 − k )
b) z = ( 4 − k , 3 k − 2 ) z=(4-k,\,3k-2) z = ( 4 − k , 3 k − 2 )
c) z = ( k 2 − k + 1 , k − 1 ) z=(k^2-k+1,\,k-1) z = ( k 2 − k + 1 , k − 1 )
d) z = ( k 3 − 729 , 3 k − 81 ) z=(k^{3}-729,\,3k-81) z = ( k 3 − 729 , 3 k − 81 )
e) z = ( k 4 , k 3 ) z=(k^{4},\,k^{3}) z = ( k 4 , k 3 )
0128 - Soluções
Resolvendo cada item:
a) Para z = ( k − 3 , 3 − k ) z=(k-3,\,3-k) z = ( k − 3 , 3 − k )
z z z z = ( k − 3 ) + ( 3 − k ) i \boxed{z=(k-3)+(3-k)i} z = ( k − 3 ) + ( 3 − k ) i
k ∈ R k\in\mathbb{R} k ∈ R ( R e ) ≥ 0 (Re)\geq 0 ( R e ) ≥ 0 k − 3 ≥ 0 → k ≥ 3 k-3\geq 0\to \boxed{k\geq 3} k − 3 ≥ 0 → k ≥ 3
k ∈ R k\in\mathbb{R} k ∈ R ( R e ) < 0 (Re)<0 ( R e ) < 0 k − 3 < 0 → k < 3 k-3<0\to \boxed{k<3} k − 3 < 0 → k < 3
k ∈ R k\in\mathbb{R} k ∈ R z z z R e ( z ) = 0 Re(z)=0 R e ( z ) = 0 k − 3 = 0 → k = 3 k-3=0\to \boxed{k=3} k − 3 = 0 → k = 3
k ∈ R k\in\mathbb{R} k ∈ R z z z I m ( z ) = 0 Im(z)=0 I m ( z ) = 0 3 − k = 0 → k = 3 3-k=0\to\boxed{k=3} 3 − k = 0 → k = 3
Neste caso, quando k = 3 k=3 k = 3 z = 0 z=0 z = 0
b) Para z = ( 4 − k , 3 k − 2 ) z=(4-k,\,3k-2) z = ( 4 − k , 3 k − 2 )
z z z z = ( 4 − k ) + ( 3 k − 2 ) i \boxed{z=(4-k)+(3k-2)i} z = ( 4 − k ) + ( 3 k − 2 ) i
k ∈ R k\in\mathbb{R} k ∈ R ( R e ) ≥ 0 (Re)\geq 0 ( R e ) ≥ 0 4 − k ≥ 0 → k ≤ 4 4-k\geq 0\to \boxed{k\leq4} 4 − k ≥ 0 → k ≤ 4
k ∈ R k\in\mathbb{R} k ∈ R ( R e ) < 0 (Re)<0 ( R e ) < 0 4 − k < 0 → k > 4 4-k<0\to \boxed{k>4} 4 − k < 0 → k > 4
k ∈ R k\in\mathbb{R} k ∈ R z z z R e ( z ) = 0 Re(z)=0 R e ( z ) = 0 4 − k = 0 → k = 4 4-k=0\to \boxed{k=4} 4 − k = 0 → k = 4
k ∈ R k\in\mathbb{R} k ∈ R z z z I m ( z ) = 0 Im(z)=0 I m ( z ) = 0 3 k − 2 = 0 → k = 2 3 3k-2=0\to \boxed{k=\dfrac{2}{3}} 3 k − 2 = 0 → k = 3 2
c) Para z = ( k 2 − k + 1 , k − 1 ) z=(k^2-k+1,\,k-1) z = ( k 2 − k + 1 , k − 1 ) z z z z = ( k 2 − k + 1 ) + ( k − 1 ) i \boxed{z=(k^2-k+1)+(k-1)i} z = ( k 2 − k + 1 ) + ( k − 1 ) i
k ∈ R k\in\mathbb{R} k ∈ R ( R e ) ≥ 0 (Re)\geq 0 ( R e ) ≥ 0 k 2 − k + 1 ≥ 0 k^2-k+1\geq 0 k 2 − k + 1 ≥ 0
Primeiramente, vamos resolver, por Bhaskara a equação:k 2 − k + 1 = 0 → k^2-k+1=0\to k 2 − k + 1 = 0 →
x = − ( − 1 ) ± ( − 1 ) 2 − 4 × 1 × 1 2 × 1 → x = 1 ± − 3 2 , x=\dfrac{-(-1)\pm \sqrt{(-1)^2-4\times 1\times 1}}{2\times 1}\to x=\dfrac{1\pm \sqrt{-3}}{2}, x = 2 × 1 − ( − 1 ) ± ( − 1 ) 2 − 4 × 1 × 1 → x = 2 1 ± − 3 , Δ < 0 \Delta <0 Δ < 0 k k k k k k z z z
k ∈ R k\in\mathbb{R} k ∈ R ( R e ) < 0 (Re)<0 ( R e ) < 0 k 2 − k + 1 < 0 → k^2-k+1<0\to k 2 − k + 1 < 0 → z z z
k ∈ R k\in\mathbb{R} k ∈ R z z z R e ( z ) = 0 Re(z)=0 R e ( z ) = 0 k 2 − k + 1 = 0 k^2-k+1=0 k 2 − k + 1 = 0
k ∈ R k\in\mathbb{R} k ∈ R z z z I m ( z ) = 0 Im(z)=0 I m ( z ) = 0 k − 1 = 0 → k = 1 k-1=0\to \boxed{k=1} k − 1 = 0 → k = 1
d) Para z = ( k 3 − 729 , 3 k − 81 ) z=(k^{3}-729,\,3k-81) z = ( k 3 − 729 , 3 k − 81 )
z z z z = ( k 3 − 729 ) + ( 3 k − 81 ) i \boxed{z=(k^{3}-729)+(3k-81)i} z = ( k 3 − 729 ) + ( 3 k − 81 ) i
k ∈ R k\in\mathbb{R} k ∈ R ( R e ) ≥ 0 (Re)\geq 0 ( R e ) ≥ 0 k 3 − 729 ≥ 0 → k^{3}-729\geq 0\to k 3 − 729 ≥ 0 → f ( k ) = k 3 − 729 f(k)=k^{3}-729 f ( k ) = k 3 − 729 k 3 − 729 = 0 → k = 9 k^{3}-729=0\to \boxed{k=9} k 3 − 729 = 0 → k = 9 k k k z z z
k ∈ R k\in\mathbb{R} k ∈ R ( R e ) < 0 (Re)<0 ( R e ) < 0 z z z k k k
k ∈ R k\in\mathbb{R} k ∈ R z z z k 3 − 729 = 0 → k = 9 k^{3}-729=0\to \boxed{k=9} k 3 − 729 = 0 → k = 9
k ∈ R k\in\mathbb{R} k ∈ R z z z z z z 3 k − 81 = 0 → k = 27 3k-81=0\to \boxed{k=27} 3 k − 81 = 0 → k = 27
e) Para z = ( k 4 , k 3 ) z=(k^{4},\,k^{3}) z = ( k 4 , k 3 )
z z z z = k 4 + i k 3 \boxed{z=k^{4}+ik^{3}} z = k 4 + i k 3
k ∈ R k\in\mathbb{R} k ∈ R ( R e ) ≥ 0 (Re)\geq 0 ( R e ) ≥ 0 k 4 ≥ 0 → ∀ k ∈ R k^{4}\geq 0\to \forall\,k\in\mathbb{R} k 4 ≥ 0 → ∀ k ∈ R
k ∈ R k\in\mathbb{R} k ∈ R ( R e ) < 0 (Re)<0 ( R e ) < 0 k k k
k ∈ R k\in\mathbb{R} k ∈ R z z z k 4 = 0 → k = 0 k^{4}=0\to \boxed{k=0} k 4 = 0 → k = 0 k = 0 k=0 k = 0 k k k
k ∈ R k\in\mathbb{R} k ∈ R z z z z z z k 3 = 0 → k = 0 k^{3}=0\to \boxed{k=0} k 3 = 0 → k = 0
Neste caso, quando k = 0 k=0 k = 0 z = 0 z=0 z = 0
0127
Resolva as seguintes equações:
a) x 2 + 12 = 0 x^{2}+12=0 x 2 + 12 = 0
b) x 2 − 8 x + 41 = 0 x^{2}-8x+41=0 x 2 − 8 x + 41 = 0
c) 2 x ( x − 2 ) = − 3 2x(x-2)=-3 2 x ( x − 2 ) = − 3
d) ( x − 2 ) ( x + 2 ) = − 20 (x-2)(x+2)=-20 ( x − 2 ) ( x + 2 ) = − 20
e) x 4 − 5 x 2 + 6 = 0 x^{4}-5x^{2}+6=0 x 4 − 5 x 2 + 6 = 0
f) 2 x 4 − 3 = x 2 2x^{4}-3=x^{2} 2 x 4 − 3 = x 2
g) x 4 − 2 x 3 + 4 x 2 − 8 x = 0 x^{4}-2x^{3}+4x^{2}-8x=0 x 4 − 2 x 3 + 4 x 2 − 8 x = 0
h) x 2 + 2 x − i = 0 \dfrac{x^{2}+2}{x-i}=0 x − i x 2 + 2 = 0
0127 - Soluções
Importante observar que, quaisquer que sejam as equações a serem resolvidos, os métodos serão exatamente os mesmos até agora utilizados, com o cuidado de calcularmos as soluções imaginárias, quando houver, uma vez que, não sendo citado o conjunto universo, devemos adotar o maior conjunto numérico conhecido, isto é U = C \mathbb{U}=\mathbb{C} U = C
a) x 2 + 12 = 0 → x 2 = − 12 → x 2 = − 1 × 2 2 × 3 → x^{2}+12=0\to x^{2}=-12\to x^{2}=-1\times 2^{2}\times 3\to x 2 + 12 = 0 → x 2 = − 12 → x 2 = − 1 × 2 2 × 3 →
x = ± ( − 1 ⏟ i ⋅ 2 2 ⏟ 2 ⋅ 3 ⏟ 3 ) → x = − 2 i 3 x=\pm (\underbrace{\sqrt{-1}}_{i}\cdot\underbrace{\sqrt{2^{2}}}_{2}\cdot\underbrace{\sqrt3{}}_{\sqrt{3}})\to\boxed{x=-2i\sqrt{3}} x = ± ( i − 1 ⋅ 2 2 2 ⋅ 3 3 ) → x = − 2 i 3 x = 2 i 3 \boxed{x=2i\sqrt{3}} x = 2 i 3
b) x 2 − 8 x + 41 = 0 B h a s k a r a → x = − ( 8 ) ± ( − 8 ) 2 − 4 × 1 × 41 2 × 1 → x^{2}-8x+41=0\quad\underrightarrow{Bhaskara}\quad x=\dfrac{-(8)\pm\sqrt{(-8)^{2}-4\times 1\times 41}}{2\times 1}\to x 2 − 8 x + 41 = 0 B ha s ka r a x = 2 × 1 − ( 8 ) ± ( − 8 ) 2 − 4 × 1 × 41 →
x = 8 ± 64 − 164 2 → x = 8 ± − 100 2 → x = 8 ± 10 i 2 x 1 = 8 − 10 i 2 → x 1 = 2 ( 4 − 5 i ) 2 → x 1 = 4 − 5 i x=\dfrac{8\pm\sqrt{64-164}}{2}\to x=\dfrac{8\pm\sqrt{-100}}{2}\to x=\dfrac{8\pm 10i}{2}\\x_{1}=\dfrac{8-10i}{2}\to x_{1}=\dfrac{\cancel{2}(4-5i)}{\cancel{2}}\to\boxed{x_{1}=4-5i} x = 2 8 ± 64 − 164 → x = 2 8 ± − 100 → x = 2 8 ± 10 i x 1 = 2 8 − 10 i → x 1 = 2 2 ( 4 − 5 i ) → x 1 = 4 − 5 i
x 2 = 8 + 10 i 2 → x 2 = 2 ( 4 + 5 i ) 2 → x 2 = 4 + 5 i x_{2}=\dfrac{8+10i}{2}\to x_{2}=\dfrac{\cancel{2}(4+5i)}{\cancel{2}}\to\boxed{x_{2}=4+5i} x 2 = 2 8 + 10 i → x 2 = 2 2 ( 4 + 5 i ) → x 2 = 4 + 5 i
c) 2 x ( x − 2 ) = − 3 → 2 x 2 − 4 x + 3 = 0 B h a s k a r a → x = − ( − 4 ) ± ( − 4 ) 2 − 4 × 2 × 3 2 × 2 → 2x(x-2)=-3\to 2x^{2}-4x+3=0\quad\underrightarrow{Bhaskara}\quad x=\dfrac{-(-4)\pm \sqrt{(-4)^2-4\times 2\times 3}}{2\times 2}\to 2 x ( x − 2 ) = − 3 → 2 x 2 − 4 x + 3 = 0 B ha s ka r a x = 2 × 2 − ( − 4 ) ± ( − 4 ) 2 − 4 × 2 × 3 →
x = 4 ± 16 − 24 4 → x = 4 ± − 8 4 → x = 4 ± 2 i 2 4 x=\dfrac{4\pm \sqrt{16-24}}{4}\to x=\dfrac{4\pm\sqrt{-8}}{4}\to x=\dfrac{4\pm2i\sqrt{2}}{4} x = 4 4 ± 16 − 24 → x = 4 4 ± − 8 → x = 4 4 ± 2 i 2
x 1 = 4 − 2 i 2 4 → x 1 = 2 ( 2 − i 2 ) 4 2 → x 1 = 2 − i 2 2 ou x 1 = 1 − i 2 2 x_{1}=\dfrac{4-2i\sqrt{2}}{4}\to x_{1}=\dfrac{\cancel{2}(2-i\sqrt{2})}{\cancel{4}^{\,2}}\to\boxed{x_{1}=\dfrac{2-i\sqrt{2}}{2}}\,\,\text{ou}\,\,\boxed{x_{1}=1-i\dfrac{\sqrt{2}}{2}} x 1 = 4 4 − 2 i 2 → x 1 = 4 2 2 ( 2 − i 2 ) → x 1 = 2 2 − i 2 ou x 1 = 1 − i 2 2
x 2 = 4 + 2 i 2 4 → x 2 = 2 ( 2 + i 2 ) 4 2 → x 1 = 2 + i 2 2 ou x 1 = 1 + i 2 2 x_{2}=\dfrac{4+2i\sqrt{2}}{4}\to x_{2}=\dfrac{\cancel{2}(2+i\sqrt{2})}{\cancel{4}^{\,2}}\to\boxed{x_{1}=\dfrac{2+i\sqrt{2}}{2}}\,\,\text{ou}\,\,\boxed{x_{1}=1+i\dfrac{\sqrt{2}}{2}} x 2 = 4 4 + 2 i 2 → x 2 = 4 2 2 ( 2 + i 2 ) → x 1 = 2 2 + i 2 ou x 1 = 1 + i 2 2
d) ( x − 2 ) ( x + 2 ) ⏟ x 2 − 4 = − 20 → x 2 − 4 = − 20 → x 2 = − 16 → x = − 4 i \underbrace{(x-2)(x+2)}_{x^{2}-4}=-20\to x^{2}-4=-20\to x^{2}=-16\to \boxed{x=-4i} x 2 − 4 ( x − 2 ) ( x + 2 ) = − 20 → x 2 − 4 = − 20 → x 2 = − 16 → x = − 4 i x = 4 i \boxed{x=4i} x = 4 i
e) x 4 − 5 x 2 + 6 = 0 x^{4}-5x^{2}+6=0 x 4 − 5 x 2 + 6 = 0 x 4 = k 2 \boxed{x^{4}=k^{2}} x 4 = k 2 x 2 = k \boxed{x^{2}=k} x 2 = k
k 2 − 5 k + 6 = 0 B h a s k a r a → k = − ( − 5 ) ± ( − 5 ) 2 − 4 × 1 × 6 2 × 1 → k^{2}-5k+6=0\quad\underrightarrow{Bhaskara}\quad k=\dfrac{-(-5)\pm\sqrt{(-5)^2-4\times 1\times 6}}{2\times 1}\to k 2 − 5 k + 6 = 0 B ha s ka r a k = 2 × 1 − ( − 5 ) ± ( − 5 ) 2 − 4 × 1 × 6 →
k = 5 ± 25 − 24 2 → k = 5 ± 1 2 → k = 2 k=\dfrac{5\pm\sqrt{25-24}}{2}\to k=\dfrac{5\pm 1}{2}\to\boxed{k=2} k = 2 5 ± 25 − 24 → k = 2 5 ± 1 → k = 2 k = 3 \boxed{k=3} k = 3
Agora, devemos retornar às incógnitas principais:
Para k = 2 \boxed{k=2} k = 2 x 2 = 2 → x = − 2 x^{2}=2\to\boxed{x=-\sqrt{2}} x 2 = 2 → x = − 2 x = 2 \boxed{x=\sqrt{2}} x = 2
Para k = 3 \boxed{k=3} k = 3 x 2 = 3 → x = − 3 x^{2}=3\to \boxed{x=-\sqrt{3}} x 2 = 3 → x = − 3 x = 3 \boxed{x=\sqrt{3}} x = 3
f) 2 x 4 − 3 = x 2 → 2 x 4 − x 2 − 3 = 0 2x^{4}-3=x^{2}\to 2x^{4}-x^{2}-3=0 2 x 4 − 3 = x 2 → 2 x 4 − x 2 − 3 = 0 x 4 = k 2 \boxed{x^{4}=k^{2}} x 4 = k 2 x 2 = k \boxed{x^{2}=k} x 2 = k
2 k 2 − k − 3 = 0 B h a s k a r a → k = − ( − 1 ) ± ( − 1 ) 2 − 4 × 2 × ( − 3 ) 2 × 2 → 2k^{2}-k-3=0\quad\underrightarrow{Bhaskara}\quad k=\dfrac{-(-1)\pm \sqrt{(-1)^2-4\times 2\times (-3)}}{2\times 2}\to 2 k 2 − k − 3 = 0 B ha s ka r a k = 2 × 2 − ( − 1 ) ± ( − 1 ) 2 − 4 × 2 × ( − 3 ) →
k = 1 ± 1 + 24 4 → k = 1 ± 5 4 → k = − 1 k=\dfrac{1\pm\sqrt{1+24}}{4}\to k=\dfrac{1\pm 5}{4}\to\boxed{k=-1} k = 4 1 ± 1 + 24 → k = 4 1 ± 5 → k = − 1 k = 3 2 \boxed{k=\dfrac{3}{2}} k = 2 3
Agora, devemos retornar às incógnitas principais:
Para k = − 1 \boxed{k=-1} k = − 1 x 2 = − 1 → x = − i x^{2}=-1\to\boxed{x=-i} x 2 = − 1 → x = − i x = i \boxed{x=i} x = i
Para k = 3 2 \boxed{k=\dfrac{3}{2}} k = 2 3 x 2 = 3 2 → x = − 6 2 x^{2}=\dfrac{3}{2}\to\boxed{x=-\dfrac{\sqrt{6}}{2}} x 2 = 2 3 → x = − 2 6 x = 6 2 \boxed{x=\dfrac{\sqrt{6}}{2}} x = 2 6
g) x 4 − 2 x 3 + 4 x 2 − 8 x = 0 → ( x 4 + 4 x 2 ) + ( − 2 x 3 − 8 x ) = 0 → x^{4}-2x^{3}+4x^{2}-8x=0\to (x^{4}+4x^{2})+(-2x^{3}-8x)=0\to x 4 − 2 x 3 + 4 x 2 − 8 x = 0 → ( x 4 + 4 x 2 ) + ( − 2 x 3 − 8 x ) = 0 →
x 2 ( x 2 + 4 ) − 2 x ( x 2 + 4 ) = 0 → ( x 2 + 4 ) ⋅ ( x 2 − 2 x ) = 0 → x ⋅ ( x − 2 ) ⋅ ( x 2 + 4 ) = 0 x^{2}(x^{2}+4)-2x(x^{2}+4)=0\to (x^{2}+4)\cdot (x^{2}-2x)=0\to x\cdot(x-2)\cdot(x^{2}+4)=0 x 2 ( x 2 + 4 ) − 2 x ( x 2 + 4 ) = 0 → ( x 2 + 4 ) ⋅ ( x 2 − 2 x ) = 0 → x ⋅ ( x − 2 ) ⋅ ( x 2 + 4 ) = 0
Daí: x = 0 \boxed{x=0} x = 0 x − 2 = 0 → x = 2 x-2=0\to \boxed{x=2} x − 2 = 0 → x = 2 x 2 + 4 = 0 → x = − 2 i x^{2}+4=0\to\boxed{x=-2i} x 2 + 4 = 0 → x = − 2 i x = 2 i \boxed{x=2i} x = 2 i
h) x 2 + 2 x − i = 0 \dfrac{x^{2}+2}{x-i}=0 x − i x 2 + 2 = 0
1)Impondo a condição de existência no denominador: x − i ≠ 0 → x ≠ i x-i\neq 0\to\boxed{x\neq i} x − i = 0 → x = i
2)Para zerar a fração, basta que o numerador seja igual a zero, assim:
x 2 + 2 = 0 → x = − i 2 x^{2}+2=0\to \boxed{x=-i\sqrt{2}} x 2 + 2 = 0 → x = − i 2 x = i 2 \boxed{x=i\sqrt{2}} x = i 2
0126
Calcule o valor de k , k ∈ R k,\,k\in\mathbb{R} k , k ∈ R
a) z = ( 4 − k ) + ( k 2 − 16 ) i z=(4-k)+(k^{2}-16)i z = ( 4 − k ) + ( k 2 − 16 ) i
b) z = ( 2 k 3 − 54 ) + ( k − 3 ) i z=(2k^{3}-54)+(k-3)i z = ( 2 k 3 − 54 ) + ( k − 3 ) i
c) z = ( k 2 − 4 ) + ( 8 − 2 k 2 ) i z=(k^{2}-4)+(8-2k^{2})i z = ( k 2 − 4 ) + ( 8 − 2 k 2 ) i
d) z = ( 3 k − 18 ) − 6 i z=(3k-18)-6i z = ( 3 k − 18 ) − 6 i
e) z = ( 3 k − 18 ) − ( 2 k − 12 ) i z=(3k-18)-(2k-12)i z = ( 3 k − 18 ) − ( 2 k − 12 ) i
0126 - Soluções
Resolvendo, item a item:
a) Para que z = ( 4 − k ) + ( k 2 − 16 ) i z=(4-k)+(k^{2}-16)i z = ( 4 − k ) + ( k 2 − 16 ) i k 2 − 16 = 0 → k 2 = 16 → k = − 4 k^{2}-16=0\to k^{2}=16\to k=-4 k 2 − 16 = 0 → k 2 = 16 → k = − 4 k = 4 k=4 k = 4 k = 4 k=4 k = 4 k = − 4 k=-4 k = − 4
b) Para que z = ( 2 k 3 − 54 ) + ( k − 3 ) i z=(2k^{3}-54)+(k-3)i z = ( 2 k 3 − 54 ) + ( k − 3 ) i 2 k 3 − 54 = 0 → 2 k 3 = 54 → k 3 = 27 → k = 3 2k^{3}-54=0\to 2k^{3}=54\to k^{3}=27\to k=3 2 k 3 − 54 = 0 → 2 k 3 = 54 → k 3 = 27 → k = 3 k = 3 k=3 k = 3 k k k
c) Para que z = ( k 2 − 4 ) + ( 8 − 2 k 2 ) i z=(k^{2}-4)+(8-2k^{2})i z = ( k 2 − 4 ) + ( 8 − 2 k 2 ) i k 2 − 4 = 0 → k 2 = 4 → k = − 2 k^{2}-4=0\to k^{2}=4\to k=-2 k 2 − 4 = 0 → k 2 = 4 → k = − 2 k = 2 k=2 k = 2 k k k k k k
d) Para que z = ( 3 k − 18 ) − 6 i z=(3k-18)-6i z = ( 3 k − 18 ) − 6 i k k k k k k
e) Para que z = ( 3 k − 18 ) − ( 2 k − 12 ) i z=(3k-18)-(2k-12)i z = ( 3 k − 18 ) − ( 2 k − 12 ) i − ( 2 k − 12 ) = 0 → − 2 k + 12 = 0 → − 2 k = − 12 ( − 1 ) → 2 k = 12 → k = 6 -(2k-12)=0\to -2k+12=0\to -2k=-12(-1)\to 2k=12\to k=6 − ( 2 k − 12 ) = 0 → − 2 k + 12 = 0 → − 2 k = − 12 ( − 1 ) → 2 k = 12 → k = 6 k = 6 k=6 k = 6 z = 0 z=0 z = 0 z = ( 0 , 0 ) z=(0,\,0) z = ( 0 , 0 ) seja imaginário puro ", aí sim, não teríamos valor de k k k
