Question

Sabendo que 2x+y=A e x-1/2*y=B, então (4x^2-y^2) é igual a:
a) 2AB
b) AB
c) 2A^2B^2
d) A/2B
e) AB/2

Answer 1

Temos:

$\begin{cases}\textrm{I :}\,\,\,\,\,\,\,2\,x\,+\,y=A\\ \\\textrm{II :}\,\,\,\,\,\,\, x\,-\,\dfrac{y}{2}=B\end{cases}$

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Vamos deixar x e y em função de A e B.

$\bold{II\,:}$

$x=B\,+\,\dfrac{y}{2}\\ \\ \\ x=\dfrac{2\,B\,+\,y}{2}$

Substituindo na $\bold{I}$, temos:

$2\,\bigg(\dfrac{2\,B\,+\,y}{2}\bigg)\,+\,y=A\\ \\ \\ 2\,B\,+\,2y=A\\ \\ \\ \boxed{y=\dfrac{A-2\,B}{2}}$

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$\bold{I\,:}$

$y=A-2x$

Substituindo na $\bold{II\,:}$, temos:

$x-\dfrac{(A-2x)}{2}=B\\ \\ \\ \dfrac{2x-(A-2x)}{2}=B\\ \\ \\ \dfrac{2x-A+2x}{2}=B\\ \\ \\ \dfrac{4x-A}{2}=B\,\,\,\,\,\to\,\,\,\,\,\boxed{x=\dfrac{2B+A}{4}}$

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Fazendo $\bold{4\,x^{2}\,-\,y^{2}}$, teremos:

$4\,.\,\bigg(\dfrac{2B+A}{4}\bigg)^{2}\,-\,\bigg(\dfrac{A-2B}{2}\bigg)^{2}\\ \\ \\ 4\,.\,\dfrac{\big(2B+A\big)^{2}}{4^{2}}\,-\,\dfrac{\big(A-2B\big)^{2}}{2^{2}}\\ \\ \\ \dfrac{\big(4B^{2}+4AB+A^{2}\big)}{4}\,-\,\dfrac{\big(A^{2}-4AB+4B^{2}\big)}{4}\\ \\ \\ \dfrac{4B^{2}+4AB+A^{2}-A^{2}+4AB-4B^{2}}{4}\\ \\ \\ \dfrac{8\,AB}{4}\,\,\,\,\to\,\,\,\,\boxed{\boxed{2\,AB}}$

Desta forma, ficamos com a letra A como resposta.