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I'm trying to simplify: $\frac{x^2}{x^2-4}-\frac{x+1}{x+2}$ but I can't get to the answer: $\frac{1}{x-2}$

How to do it?

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    @Arturo: I agree, which is why I commented on the similarity. I have occasionally found that when a student is struggling with "complicated" algebra with rational functions, pointing out an analogous arithmetic problem clears up what is going on. Incidentally, the example I gave is just a special case of $\frac{x^2-3}{x^2-1}-\frac{x-2}{x-1}=\frac{1}{x+1}$.2011-02-04

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First, do the operation by finding the least common denominator. Since $x^2-4 = (x-2)(x+2)$, the least common denominator is already $x^2-4$. Then do some simple algebra: \begin{align*} \frac{x^2}{x^2-4} - \frac{x+1}{x+2} &= \frac{x^2 - (x+1)(x-2)}{x^2-4} = \frac{x^2-(x^2-x-2)}{x^2-4}\\ &= \frac{x+2}{x^2-4} = \frac{x+2}{(x-2)(x+2)} = \frac{1}{x-2}. \end{align*}

If you didn't realize that $x+2$ already divides $x^2-4$, you probably would get \begin{align*} \frac{x^2}{x^2-4} - \frac{x+1}{x+2} &= \frac{x^2(x+2) - (x+1)(x^2-4)}{(x^2-4)(x+2)}\\ &= \frac{x^3 + 2x^2 - (x^3 +x^2 - 4x - 4)}{(x^2-4)(x+2)}\\ &= \frac{x^2 +4x + 4}{(x^2-4)(x+2)} = \frac{(x+2)^2}{(x^2-4)(x+2)}\\ &= \frac{x+2}{x^2-4} = \frac{1}{x-2}, \end{align*} with the extra work for not noticing.

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    I used the second approach (multiplying the denominators), but I was trying to simplify directly, instead of multiplying all. Actually, I symplified even the denominators before multipling.. A big mental loop..2011-02-04
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HINT $\ \ $ Taking fractional parts (by subtracting $\:1\:$ from both terms) it reduces to the following

$\rm \frac{4}{x^2-4}\ +\ \frac{1}{x+2}\ =\ \frac{1}{x-2}$

which you'll probably find easier to derive.