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26 Sep 2010, 02:02
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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
PROPER VERSION OF THIS QUESTION IS HERE: if-xy-1-what-is-the-value-of-2-x-y-2-2-x-y-107238.html
A. 2
B. 4
C. 8
D. 16
E. 32
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1.JPG
PROPER VERSION OF THIS QUESTION IS HERE: if-xy-1-what-is-the-value-of-2-x-y-2-2-x-y-107238.html
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26 Sep 2010, 03:55
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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
\(\frac{2(x+y)^2}{2(x-y)^2}=\frac{x^2+2xy+y^2}{x^2-2xy+y^2}=\frac{x^2+2+y^2}{x^2-2+y^2}\) --> we cannot find the value of this fraction without some additional info.
For example if \(x=2\) and \(y=\frac{1}{2}\) then \(\frac{2(x+y)^2}{2(x-y)^2}=\frac{25}{9}\);
But if if \(x=4\) and \(y=\frac{1}{4}\) then \(\frac{2(x+y)^2}{2(x-y)^2}=\frac{17^2}{15^2}\).
So you are right there is something wrong with this question.
I think the question should be: \(\frac{2^{(x+y)^2}}{2^{(x-y)^2}}=2^{(x+y)^2-(x-y)^2}=2^{(x+y+x-y)(x+y-x+y)}=2^{(2x)(2y)}=2^{4xy}=2^4=16\).
If so, then the answer would be D.
Hope it helps.
A. 2
B. 4
C. 8
D. 16
E. 32
\(\frac{2(x+y)^2}{2(x-y)^2}=\frac{x^2+2xy+y^2}{x^2-2xy+y^2}=\frac{x^2+2+y^2}{x^2-2+y^2}\) --> we cannot find the value of this fraction without some additional info.
For example if \(x=2\) and \(y=\frac{1}{2}\) then \(\frac{2(x+y)^2}{2(x-y)^2}=\frac{25}{9}\);
But if if \(x=4\) and \(y=\frac{1}{4}\) then \(\frac{2(x+y)^2}{2(x-y)^2}=\frac{17^2}{15^2}\).
So you are right there is something wrong with this question.
I think the question should be: \(\frac{2^{(x+y)^2}}{2^{(x-y)^2}}=2^{(x+y)^2-(x-y)^2}=2^{(x+y+x-y)(x+y-x+y)}=2^{(2x)(2y)}=2^{4xy}=2^4=16\).
If so, then the answer would be D.
Hope it helps.
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