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If x -y,

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If x -y, [#permalink] New post 19 May 2018, 15:16
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Question Stats:

96% (00:16) correct 3% (01:06) wrong based on 28 sessions
If \(x \neq -y\), then \(\frac{x^2 + 2xy + y^2}{2(x+y)^2}=\)

(A) \(1\)
(B) \(\frac{1}{2}\)
(C)\(\frac{1}{x+y}\)
(D) \(xy\)
(E) \(2xy\)
[Reveal] Spoiler: OA

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Re: If x -y, [#permalink] New post 21 May 2018, 21:21
The numerator is nothing but x+y whole square. Hence on solving we get 1/2
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Re: If x -y, [#permalink] New post 21 Aug 2020, 07:01
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sandy wrote:
If \(x \neq -y\), then \(\frac{x^2 + 2xy + y^2}{2(x+y)^2}=\)

(A) \(1\)
(B) \(\frac{1}{2}\)
(C)\(\frac{1}{x+y}\)
(D) \(xy\)
(E) \(2xy\)


APPROACH #1: Factoring

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

Factor the numerator to get: \(\frac{(x+y)(x+y)}{2(x+y)^2}\)

In other words: \(\frac{(x+y)^2}{2(x+y)^2}\)

Divide numerator and denominator by \((x+y)^2\) to get: \(\frac{1}{2}\)

Answer: B


APPROACH #2: Plug-in values

If \(x = 0\) and \(y = 1\), we get: \(\frac{x^2 + 2xy + y^2}{2(x+y)^2}=\frac{0^2 + 2(0)(1) + 1^2}{2(0+1)^2}=\frac{1}{2}\)

From here we need to test answer choices C, D and E since there are variables in those answer choices.
So we'll plug in \(x = 0\) and \(y = 1\) to see if any of them also evaluate to be \(\frac{1}{2}\)

(C)\(\frac{1}{x+y}=\frac{1}{0+1}=1\). No good
(D) \(xy=(0)(1)=0\). No good
(E) \(2xy=2(0)(1)=0\). No good

Answer: B
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Re: If x -y,   [#permalink] 21 Aug 2020, 07:01
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