Example 2. Solve the equation
$$
\frac{x-2}{x+2}-\frac{16}{x^{2}-4}=\frac{x+2}{x-2} .
$$
Solution: Multiply both sides of the equation by $(x+2) \quad(x-2)$ to eliminate the denominators, we get
$$
(x-2)^{2}-16=(x+2)^{2} \text {. }
$$

Solving this polynomial equation, we get
$$
x=-2 \text {. }
$$

Verification: When $x=-2$, $(x+2) \cdot(x-2$ , $=0$, so -2 is an extraneous root, hence the original equation has no solution.