Do you mean it to be this...

\(\frac{2(x+y)^2}{2(x-y)^2}\)

If so, that would become

\(\frac{2(x^2 + 2xy + y^2)}{2(x^2 - 2xy + y^2)}\)

\(\frac{2x^2 + 4xy + 2y^2}{2x^2 - 4xy + 2y^2}\)

If xy=1, then...

\(\frac{2x^2 + 4 + 2y^2}{2x^2 - 4 + 2y^2}\)

From this, if I did it right, none of the answer choices fit.

Extremepbs wrote:

If xy=1, what is the value of [2(x+y)^2]/[2(x-y)^2]?

A)2

B)4

C)8

D)16

E)32

Is integer x positive ?

1) x > x^3

2) x < x^2

If p,q,r are even numbers such that 2<p<q<r

what is the value of q ?

1) r<10

2)p<6

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J Allen Morris

**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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