Question

Se 2/x +2/y + 2/z +x/yz +y/xz + z/ xy = 8/6
E x+y +z = 16 o produto de 2.x.y.z é:
A) 384
B) 108
C) 48
D)32
E) 10

Answer 1

Olá Mpfg1.


Produto notável usado

$\star~\boxed{\boxed{\mathsf{(a\pm b)^2=a^2\pm 2ab+b^2}}}$

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Organizando as informações e desenvolvendo a expressão

$\mathsf{x+y+z=16}\\\\\\\\\mathsf{\dfrac{2}{x}+\dfrac{2}{y}+\dfrac{2}{z}+\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}=\dfrac{8}{6}}$

Note que os denominadores são x, y e z. Portanto, precisamos deixar todos os denominadores como o produto desses 3 termos, ou seja, xyz. Então por exemplo, onde o denominador for x, basta multiplicarmos e dividirmos por yz, dessa forma o denominador ficará igual a xyz.

$\mathsf{\dfrac{2}{x}+\dfrac{2}{y}+\dfrac{2}{z}+\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}=\dfrac{8}{6}}\\\\\\\\\mathsf{\dfrac{yz}{yz}\cdot\dfrac{2}{x}+\dfrac{xz}{xz}\cdot\dfrac{2}{y}+\dfrac{xy}{xy}\cdot\dfrac{2}{z}+\dfrac{x}{x}\cdot\dfrac{x}{yz}+\dfrac{y}{y}\cdot\dfrac{y}{xz}+\dfrac{z}{z}\cdot\dfrac{z}{xy}=\dfrac{8\div2}{6\div2}}\\\\\\\\\mathsf{\dfrac{2yz+2zx+2xy+x^2+y^2+z^2}{xyz}=\dfrac{4}{3}}\\\\\\\\\mathsf{\dfrac{\boxed{\mathsf{x^2+2xy+z^2}}+2zx+2yz+z^2}{xyz}=\dfrac{4}{3}}$

$\mathsf{\dfrac{(x+y)^2+2zx+2yz+z^2}{xyz}=\dfrac{4}{3}}$

Coloque 2z em evidência

$\mathsf{\dfrac{(x+y)^2+2z\cdot(x+y)+z^2}{xyz}=\dfrac{4}{3}}$


$\mathsf{x+y+z=16~~\Leftrightarrow x+y=16-z}$

Substitua x + y por 16 - z


$\mathsf{\dfrac{(16-z)^2+2z\cdot(16-z)+z^2}{xyz}=\dfrac{4}{3}}\\\\\\\\\mathsf{\dfrac{256-32z+z^2+32z-2z^2+z^2}{xyz}=\dfrac{4}{3}}\\\\\\\\\mathsf{\dfrac{256-\diagup\!\!\!\!\!32z+\diagup\!\!\!\!\!32z+\diagup\!\!\!\!2z^2-\diagup\!\!\!\!2z^2}{xyz}=\dfrac{4}{3}}\\\\\\\\\mathsf{\dfrac{256}{xyz}=\dfrac{4}{3}}$

Multiplique cruzado

$\mathsf{\dfrac{256}{xyz}=\dfrac{4}{3}}\\\\\\\mathsf{256\cdot3=xyz\cdot4}$

Divida ambos os lados por 4

$\mathsf{\dfrac{1}{\diagup\!\!\!\!4}\cdot\diagup\!\!\!\!\!256\cdot3=xyz\cdot\diagup\!\!\!\!4\cdot\dfrac{1}{\diagup\!\!\!\!4}}\\\\\\\mathsf{64\cdot3=xyz}\\\\\\\boxed{\mathsf{384=2xyz}}$


Resposta: alternativa (a)


Dúvidas? comente.