Ensino Fundamental

Mensagem não lidapor **botelho** » Ter 22 Out, 2019 08:44

Sejam a,b e c números reais distintos dois a dois e não nulos tais que a+b+c=0.O valor de ([tex3]\frac{b-c}{a} + \frac{c-a}{b} + \frac{a-b}{c}[/tex3] ).([tex3]\frac{a}{b-c} + \frac{b}{c-a} + \frac{c}{a-b}[/tex3] )
a)0
b)1
c)3
d)9
e)27

Mensagem não lidapor **jedi** » Dom 27 Out, 2019 11:59 · Mensagem não lida por **jedi** » Dom 27 Out, 2019 11:59

[tex3]a+c=-b[/tex3]

[tex3]\left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)[/tex3]

[tex3]\left(\frac{-a-c-c}{a}+\frac{c-a}{-a-c}+\frac{a+a+c}{c}\right)\left(\frac{a}{-a-c-c}+\frac{-a-c}{c-a}+\frac{c}{a+a+c}\right)[/tex3]

[tex3]\left(-\frac{a+2c}{a}-\frac{c-a}{a+c}+\frac{2a+c}{c}\right)\left(-\frac{a}{a+2c}-\frac{a+c}{c-a}+\frac{c}{2a+c}\right)[/tex3]

[tex3]\left(\frac{-(a+2c)(a+c)c-ac(c-a)+(a+c)(2a+c)a}{ac(a+c)}\right)\left(\frac{-a(c-a)(2a+c)-(a+c)(2a+c)(2c+a)+c(c-a)(a+2c)}{(2a+c)(a+2c)(c-a)}\right)[/tex3]

[tex3]\left(\frac{(a+c)(-ac-2c^2+2a^2+ac)-ac(c-a)}{ac(a+c)}\right)\left(\frac{(c-a)(-2^2a-ac+ac+2c^2)-(a+c)(2a+c)(2c+a)}{(2a+c)(a+2c)(c-a)}\right)[/tex3]

[tex3]\left(\frac{2(a+c)(a^2-c^2)-ac(c-a)}{ac(a+c)}\right)\left(\frac{2(c-a)(c^2-a^2)-(a+c)(2a+c)(2c+a)}{(2a+c)(a+2c)(c-a)}\right)[/tex3]

[tex3](c-a)\left(\frac{-2(a+c)(a+c)-ac}{ac(a+c)}\right)(a+c)\left(\frac{2(c-a)(c-a)-(2a+c)(2c+a)}{(2a+c)(a+2c)(c-a)}\right)[/tex3]

[tex3]\left(\frac{-2(a+c)(a+c)-ac}{ac}\right)\left(\frac{2(c-a)(c-a)-(2a+c)(2c+a)}{(2a+c)(a+2c)}\right)[/tex3]

[tex3]\left(\frac{-2a^2-4ac-2c^2-ac}{ac}\right)\left(\frac{2c^2-4ac+2a^2-4ac-2a^2-2c^2-ac}{2a^2+4ac+ac+2c^2}\right)[/tex3]

[tex3]\left(-\frac{2a^2+5ac+2c^2}{ac}\right)\left(\frac{-9ac}{2a^2+5ac+2c^2}\right)[/tex3]

[tex3]=9[/tex3]

Ensino Fundamental ⇒ Algebra Tópico resolvido

Algebra

Re: Algebra

Ensino Fundamental ⇒ Algebra Tópico resolvido

Algebra

Re: Algebra

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