Seemingly complex

Factor the quadratics on the left hand side:

$\frac{x^2+x-2}{x^2-x-2}=\frac{(x+2)(x-1)}{(x-2)(x+1)}$

We substitute and see

$\frac{(x+2)(x-1)}{(x-2)(x+1)}=\frac{x+2}{x-2}$

This equation takes the form $ab=a$ , where $a=\frac{x+2}{x-2}$ and $\frac{x-1}{x+1}=b$ .

This form of equation has solutions for $b=1$ or $a=0$ . (Verify by subtracting $a$ , factoring and applying the zero product property.)

But if $b=\frac{x-1}{x+1}=1$ , $x-1$ and $x+1$ , which differ by 2, must be equal, $b=1$ does not lead to any solution.

Otherwise we consider $a=\frac{x+2}{x-2}=0$ , which holds if the numerator, $x+2=0$ , giving $\boxed{x=-2}$ as the unique solution.

First, cross multiplying to get $(x-2)(x^2+x-2)=(x+2)(x^2-x-2)$ Simplifying a bit to get $2x^2-8=0$ $2(x-2)(x+2)=0$ Then, we have $x=2$ or $x=-2$ as our solutions. Notice that if we substitute $x=2$ to the equation, the RHS will become $\frac{4}{0}$ which is undefined. Thus, the only solution to this equation is $x=\boxed{-2}$ .