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?
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?
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.
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.