B2 Solve the inequality $\frac{1}{x-2}+\frac{1}{x+2} \geq 1-\frac{x}{x^{2}-4}$.
B2. We rearrange the inequality by not equating with 0. We get $\frac{1}{x-2}+\frac{1}{x+2}-1+\frac{x}{x^{2}-4} \geq 0$. We find the common denominator, which is $x^{2}-4$. We expand the numerators and appropriately rearrange the inequality $\frac{x+2+x-2-x^{2}+4+x}{(x-2)(x+2)} \geq 0$. We simplify the numerator $\frac{-x^{2}+3 x+4}{(x-2)(x+2)} \geq 0$. We determine the roots from $-x^{2}+3 x+4=0$, and the poles. The roots are $x_{1}=4$ and $x_{2}=1$, the poles are $x_{1}=2, x_{2}=-2$. We arrange the signs and read off the solution $x \in(-2,1] \cup(2,4]$.

Inequality $\mathrm{z} \frac{1}{x-2}+\frac{1}{x+2}-1+\frac{x}{x^{2}-4} \geq 0$... 1 point
Expansion of numerators $x+2+x-2-x^{2}+4+x$ ..... 1 point
Rearrangement of the inequality $\frac{-\left(x^{2}-3 x-4\right)}{(x-2)(x+2)} \geq 0$ ..... 1 point
Determination of roots $x_{1}=4$ and $x_{2}=1$ ..... 1 point
Determination of poles $x_{1}=2, x_{2}=-2$ ..... 1 point
Arrangement of signs ..... 1 point
Solution written as $x \in(-2,1] \cup(2,4] \ldots$ ..... 1 point