Ensino Fundamental ⇒ Equação Segundo Grau Tópico resolvido

[Angelita]

Se {x1;x2} são raízes da equação do segundo grau x²+bx+c=0; bc [tex3]\neq [/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\neq [/tex3]

0.Calcule [tex3]\frac{(x1+b)^{4}+(x2+b)^{4}}{cx1(x1+b)+b^{4}-4b²c+3c²}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{(x1+b)^{4}+(x2+b)^{4}}{cx1(x1+b)+b^{4}-4b²c+3c²}[/tex3]

a)-1
b)2
c)3
d)4
e)1

Resposta

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e

[mcarvalho]

Por Girard, sabemos: [tex3]x_1+x_2=-b\\x_1x_2=c[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]x_1+x_2=-b\\x_1x_2=c[/tex3]

Substituindo na equação: [tex3]\frac{(x1+b)^{4}+(x2+b)^{4}}{cx1(x1+b)+b^{4}-4b²c+3c²}\\ \frac{(x_1-x_1-x_2)^4+(x_2-x_1-x_2)^4}{x_1^2x_2(-x_2)+(x_1+x_2)^4-4(x_1+x_2)^2(x_1x_2)+3(x_1x_2)^2}\\ \frac{x_1^4+x_2^4}{-(x_1^2x_2^2)+(x_1+x_2)^4-4(x_1+x_2)^2(x_1x_2)+3(x_1x_2)^2}\\ \frac{x_1^4+x_2^4}{2x_1^2x_2^2+(x_1+x_2)^2[(x_1+x_2)^2-4(x_1x_2)]}\\ \frac{x_1^4+x_2^4}{2x_1^2x_2^2+(x_1+x_2)^2(x_1^2+x_2^2-2x_1x_2)}\\ \frac{x_1^4+x_2^4}{2(x_1x_2)^2+(x_1+x_2)^2(x_1-x_2)^2}= \frac{x_1^4+x_2^4}{(x_1^2-x_2^2)^2+2(x_1x_2)^2}\\ \frac{x_1^4+x_2^4}{x_1^4+x^4_2-2x_1^2x_2^2+2x_1^2x_2^2}=1[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{(x1+b)^{4}+(x2+b)^{4}}{cx1(x1+b)+b^{4}-4b²c+3c²}\\ \frac{(x_1-x_1-x_2)^4+(x_2-x_1-x_2)^4}{x_1^2x_2(-x_2)+(x_1+x_2)^4-4(x_1+x_2)^2(x_1x_2)+3(x_1x_2)^2}\\ \frac{x_1^4+x_2^4}{-(x_1^2x_2^2)+(x_1+x_2)^4-4(x_1+x_2)^2(x_1x_2)+3(x_1x_2)^2}\\ \frac{x_1^4+x_2^4}{2x_1^2x_2^2+(x_1+x_2)^2[(x_1+x_2)^2-4(x_1x_2)]}\\ \frac{x_1^4+x_2^4}{2x_1^2x_2^2+(x_1+x_2)^2(x_1^2+x_2^2-2x_1x_2)}\\ \frac{x_1^4+x_2^4}{2(x_1x_2)^2+(x_1+x_2)^2(x_1-x_2)^2}= \frac{x_1^4+x_2^4}{(x_1^2-x_2^2)^2+2(x_1x_2)^2}\\ \frac{x_1^4+x_2^4}{x_1^4+x^4_2-2x_1^2x_2^2+2x_1^2x_2^2}=1[/tex3]

[petras]

Angelita,
[tex3]\mathsf{x_1+x_2=-b\rightarrow x_1+b = -x_2~e~x_2+b=-x_1\\
x_1.x_2=c\rightarrow x_1=\frac{c}{x_2}~e~x_2=\frac{c}{x_1}\\
Substituindo:
\\\frac{(-x_2^4)+(-x_1^4)}{c.\frac{c}{\cancel{x_2}}(\cancel{-x_2})+b^4-4b^2c+3c^2}=\\
\frac{(x_2^4+x_1^4)}{-c^2+b^4-4b^2c+3c^2}=\\
\frac{x_2^4+x_1^4}{b^4-4b^2c+2c^2}=(I)\\
\rightarrow (x_1+x_2)^2=b^2\rightarrow x_1^2+x_2^2+2x_1x_2=b^2\rightarrow x_1^2+x_2^2+2c=b^2\rightarrow x_1^2+x_2^2=b^2-2c\\
(x_1^2+x_2^2)^2 = (b^2-2c)^2\rightarrow x_1^4+x_2^4+2(x_1^2x_2^2)=b^4-4b^2c+4c^2\rightarrow x_1^4+x_2^4+2(x_1x_2)^2=b^4-4b^2c+4c^2\\
x_1^4+x_2^4+2c^2=b^4-4b^2c+4c^2\rightarrow x_1^4+x_2^4=b^4-4b^2c+2c^2\\
Substituindo ~em~ I:
\frac{\cancel {b^4-4b^2c+2c^2}}{\cancel{b^4-4b^2c+2c^2}}=\boxed{\color{red}1}

}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{x_1+x_2=-b\rightarrow x_1+b = -x_2~e~x_2+b=-x_1\\
x_1.x_2=c\rightarrow x_1=\frac{c}{x_2}~e~x_2=\frac{c}{x_1}\\
Substituindo:
\\\frac{(-x_2^4)+(-x_1^4)}{c.\frac{c}{\cancel{x_2}}(\cancel{-x_2})+b^4-4b^2c+3c^2}=\\
\frac{(x_2^4+x_1^4)}{-c^2+b^4-4b^2c+3c^2}=\\
\frac{x_2^4+x_1^4}{b^4-4b^2c+2c^2}=(I)\\
\rightarrow (x_1+x_2)^2=b^2\rightarrow x_1^2+x_2^2+2x_1x_2=b^2\rightarrow x_1^2+x_2^2+2c=b^2\rightarrow x_1^2+x_2^2=b^2-2c\\
(x_1^2+x_2^2)^2 = (b^2-2c)^2\rightarrow x_1^4+x_2^4+2(x_1^2x_2^2)=b^4-4b^2c+4c^2\rightarrow x_1^4+x_2^4+2(x_1x_2)^2=b^4-4b^2c+4c^2\\
x_1^4+x_2^4+2c^2=b^4-4b^2c+4c^2\rightarrow x_1^4+x_2^4=b^4-4b^2c+2c^2\\
Substituindo ~em~ I:
\frac{\cancel {b^4-4b^2c+2c^2}}{\cancel{b^4-4b^2c+2c^2}}=\boxed{\color{red}1}

}[/tex3]