Ensino Fundamental ⇒ Fatoração Tópico resolvido

[botelho]

Sabendo que (a-b)(a-c)+(b-c)(b-a)+(c-a)(c-b)=2(a+b+c)²,onde abc [tex3]\neq [/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\neq [/tex3]

0,calcule o valor de:
[tex3]\frac{a+3b+c}{ac} + \frac{b+3c+a}{ab} + \frac{c+3a+b}{bc}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{a+3b+c}{ac} + \frac{b+3c+a}{ab} + \frac{c+3a+b}{bc}[/tex3]

: [tex3]a^{-1} + b^{-1} + c^{-1}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]a^{-1} + b^{-1} + c^{-1}[/tex3]

a)1
b)-4
c)-6
d)-13
e)-15

Resposta

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d

[deOliveira]

Vamos trabalhar com a condição dada: (eu fiz as contas duma vez pra não ficar grande, mas é só fazer distributiva)
[tex3](a-b)(a-c)+(b-c)(b-a)+(c-a)(c-b)=2(a+b+c)^2\\\implies
a^2+b^2+c^2-ab-ac-bc=2(a^2+b^2+c^2)+4(ab+ac+bc)\\\implies\boxed{-5(ab+ac+bc)=a^2+b^2+c^2}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3](a-b)(a-c)+(b-c)(b-a)+(c-a)(c-b)=2(a+b+c)^2\\\implies
a^2+b^2+c^2-ab-ac-bc=2(a^2+b^2+c^2)+4(ab+ac+bc)\\\implies\boxed{-5(ab+ac+bc)=a^2+b^2+c^2}[/tex3]

Vamos então trabalhar com aquilo que devemos fatorar:
[tex3]\frac{\frac{a+3b+c}{ac}+\frac{b+3c+a}{ab}+\frac{c+3a+b}{bc}}{a^{-1}+b^{-1}+c^{-1}}=\\
\frac{\frac{3b}{ac}+\frac1a+\frac1c+\frac{3c}{ab}+\frac1b+\frac1a+\frac{3a}{bc}+\frac1b+\frac1c}{a^{-1}+b^{-1}+c^{-1}}=\\
\frac{3\left(\frac a{bc}+\frac b{ac}+\frac c{ab}\right)+2\left(\frac1a+\frac1b+\frac1c\right)}{a^{-1}+b^{-1}+c^{-1}}=\\
\frac{3\left(\frac a{bc}+\frac b{ac}+\frac c{ab}\right)}{a^{-1}+b^{-1}+c^{-1}}+\frac{2\left(\frac1a+\frac1b+\frac1c\right)}{a^{-1}+b^{-1}+c^{-1}}=\\
\frac{3\left(\frac a{bc}+\frac b{ac}+\frac c{ab}\right)}{a^{-1}+b^{-1}+c^{-1}}+\frac{2(a^{-1}+b^{-1}+c^{-1})}{a^{-1}+b^{-1}+c^{-1}}=\\
\frac{3\left(\frac a{bc}+\frac b{ac}+\frac c{ab}\right)}{a^{-1}+b^{-1}+c^{-1}}+2=\\
\frac{3\left(\frac{a^2+b^2+c^2}{abc}\right)}{\frac1a+\frac1b+\frac1b}+2=\\
\frac{3\left(\frac{a^2+b^2+c^2}{abc}\right)}{\frac{ab+ac+bc}{abc}}+2=\\
\frac{3(a^2+b^2+c^2)}{ab+ac+bc}+2=[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{\frac{a+3b+c}{ac}+\frac{b+3c+a}{ab}+\frac{c+3a+b}{bc}}{a^{-1}+b^{-1}+c^{-1}}=\\
\frac{\frac{3b}{ac}+\frac1a+\frac1c+\frac{3c}{ab}+\frac1b+\frac1a+\frac{3a}{bc}+\frac1b+\frac1c}{a^{-1}+b^{-1}+c^{-1}}=\\
\frac{3\left(\frac a{bc}+\frac b{ac}+\frac c{ab}\right)+2\left(\frac1a+\frac1b+\frac1c\right)}{a^{-1}+b^{-1}+c^{-1}}=\\
\frac{3\left(\frac a{bc}+\frac b{ac}+\frac c{ab}\right)}{a^{-1}+b^{-1}+c^{-1}}+\frac{2\left(\frac1a+\frac1b+\frac1c\right)}{a^{-1}+b^{-1}+c^{-1}}=\\
\frac{3\left(\frac a{bc}+\frac b{ac}+\frac c{ab}\right)}{a^{-1}+b^{-1}+c^{-1}}+\frac{2(a^{-1}+b^{-1}+c^{-1})}{a^{-1}+b^{-1}+c^{-1}}=\\
\frac{3\left(\frac a{bc}+\frac b{ac}+\frac c{ab}\right)}{a^{-1}+b^{-1}+c^{-1}}+2=\\
\frac{3\left(\frac{a^2+b^2+c^2}{abc}\right)}{\frac1a+\frac1b+\frac1b}+2=\\
\frac{3\left(\frac{a^2+b^2+c^2}{abc}\right)}{\frac{ab+ac+bc}{abc}}+2=\\
\frac{3(a^2+b^2+c^2)}{ab+ac+bc}+2=[/tex3]

Vamos então usar [tex3]\boxed{-5(ab+ac+bc)=a^2+b^2+c^2}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\boxed{-5(ab+ac+bc)=a^2+b^2+c^2}[/tex3]

[tex3]\frac{3(a^2+b^2+c^2)}{ab+ac+bc}+2=\\
\frac{3[-5(ab+ac+bc)]}{ab+ac+bc}+2=\\
-5+2=-13[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{3(a^2+b^2+c^2)}{ab+ac+bc}+2=\\
\frac{3[-5(ab+ac+bc)]}{ab+ac+bc}+2=\\
-5+2=-13[/tex3]

Espero ter ajudado .