(Algebra) We have \((\frac{x}{2}=\frac{y}{3})\) . What is the value of \(\frac{2x}{x+y}+\frac{3y}{x-y}+\frac{x^2}{x^2-y^2}\) ?
A. -7/2
B. -20/9
C. -9
D. 3/4
E. 7/5
Algebra
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- Max@Math Revolution
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Solution:
When we assume \(\frac{x}{2}=\frac{y}{3}=k\) , we have x = 2k and y = 3k.
\(\frac {2x}{x+y}\) + \(\frac {3y}{x-y}\) +\(\frac{x^2}{x^2-y^2}\)
= \(\frac {2 * 2k}{2k + 3k}\) + \(\frac {3 * 3k}{2k - 3k}\) + \(\frac{2k^2}{(2k)^2 - (3k)^2}\)
= \(\frac {4k}{5k}\) + \(\frac {9k}{ -k}\) + \(\frac{4k^2}{4k^2 - 9k^2}\)
= \(\frac {4}{5}\) - 9 - \(\frac {4}{5}\) = -9
Therefore, C is the correct answer.
Answer: C
When we assume \(\frac{x}{2}=\frac{y}{3}=k\) , we have x = 2k and y = 3k.
\(\frac {2x}{x+y}\) + \(\frac {3y}{x-y}\) +\(\frac{x^2}{x^2-y^2}\)
= \(\frac {2 * 2k}{2k + 3k}\) + \(\frac {3 * 3k}{2k - 3k}\) + \(\frac{2k^2}{(2k)^2 - (3k)^2}\)
= \(\frac {4k}{5k}\) + \(\frac {9k}{ -k}\) + \(\frac{4k^2}{4k^2 - 9k^2}\)
= \(\frac {4}{5}\) - 9 - \(\frac {4}{5}\) = -9
Therefore, C is the correct answer.
Answer: C
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