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If x y, what is the value of
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17 Jul 2018, 11:56
17 Jul 2018, 11:56
Expert Reply
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Question Stats:
100% (00:23) correct
0% (00:00) wrong based on 46 sessions
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If x ≠ –y, what is the value of \(\frac{x^2+2xy+y^2}{2(x+y)^2}\)?
(A) \(1\)
(B) \(\frac{1}{2}\)
(C) \(\frac{1}{x+y}\)
(D) \(xy\)
(E) \(2xy\)
(A) \(1\)
(B) \(\frac{1}{2}\)
(C) \(\frac{1}{x+y}\)
(D) \(xy\)
(E) \(2xy\)
Re: If x y, what is the value of
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18 Jul 2018, 07:40
18 Jul 2018, 07:40
1
Expert Reply
Explanation
First, recognize that \(x^2 + 2xy + y^2 = (x + y)^2\). This is one of the “special products” you need to memorize for the exam. Factor the top, then cancel:
\(\frac{x^2 + 2xy + y^2}{2(x+y)^2}=\frac{1}{2}\)
First, recognize that \(x^2 + 2xy + y^2 = (x + y)^2\). This is one of the “special products” you need to memorize for the exam. Factor the top, then cancel:
\(\frac{x^2 + 2xy + y^2}{2(x+y)^2}=\frac{1}{2}\)
Re: If x y, what is the value of
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18 Jul 2018, 12:32
18 Jul 2018, 12:32
1
sandy wrote:
If x ≠ –y, what is the value of \(\frac{x^2+2xy+y^2}{2(x+y)^2}\)?
(A) \(1\)
(B) \(\frac{1}{2}\)
(C) \(\frac{1}{x+y}\)
(D) \(xy\)
(E) \(2xy\)
(A) \(1\)
(B) \(\frac{1}{2}\)
(C) \(\frac{1}{x+y}\)
(D) \(xy\)
(E) \(2xy\)
Given
\(\frac{x^2+2xy+y^2}{2(x+y)^2}\)
= ( x + y)^2 / 2 (x+y)^2
= 1/2
The best answer is B.
