$2.51 \frac{1}{x^{2}}+\frac{1}{(x+2)^{2}}=\frac{10}{9}$.
2.51 Assuming $x+1=z$, we get

$$
\frac{1}{(z-1)^{2}}+\frac{1}{(z+1)^{2}}=\frac{10}{9}, \text { or } \cdot 5 z^{4}-19 z^{2}-4=0
$$

From here, $z_{1}^{2}=4$, i.e., $z= \pm 2$, and $z^{2}=-\frac{1}{5}$ (this equation has no roots).

Answer: $x_{1}=1, x_{2}=-3$.