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31 Dec 2006, 07:36
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See attached file?
Plugging in, which values for x and y work?
Plugging in, which values for x and y work?
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31 Dec 2006, 09:32
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If you replace y = 1/x, you get:
A = (x + 1/x)^2 / (x - 1/x)^2 = (x^2 + 1)^2 / (x^2 - 1)^2
As x ranges from -infinite to +infinite, A ranges from 0 to +infinite. That is, there´s no single answer.
A = (x + 1/x)^2 / (x - 1/x)^2 = (x^2 + 1)^2 / (x^2 - 1)^2
As x ranges from -infinite to +infinite, A ranges from 0 to +infinite. That is, there´s no single answer.
31 Dec 2006, 10:37
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i am sure there is something wrong in the question as it is put here...
more likely the expression to evaluate is:
2^((x+y)^2)/2^((x-y)^2)
if it is indeed so, the answer would be 16 (2^4)
more likely the expression to evaluate is:
2^((x+y)^2)/2^((x-y)^2)
if it is indeed so, the answer would be 16 (2^4)
