. Find the real numbers $x>-1, x \neq 0$ such that:

$$
\frac{x^{2}}{(x+1-\sqrt{x+1})^{2}}<\frac{x^{2}+3 x+18}{(x+1)^{2}} .
$$
. Let $y=\sqrt{x+1}$. We have $y \in] 0,1[\cup] 1,+\infty\left[\right.$, and $x=y^{2}-1$. The inequality is equivalent to

$$
\frac{\left(y^{2}-1\right)^{2}}{\left(y^{2}-y\right)^{2}}<\frac{\left(y^{2}-1\right)^{2}+3\left(y^{2}-1\right)+18}{y^{4}}
$$

which is also equivalent to the following inequalities

$$
\frac{(y+1)^{2}}{y^{2}}<\frac{y^{4}+y^{2}+16}{y^{4}} \Leftrightarrow(y+1)^{2} y^{2}<y^{4}+y^{2}+16 \Leftrightarrow 2 y^{3}<16 \Leftrightarrow y<2
$$

The condition is therefore satisfied exactly for $y \in] 0,1[\cup] 1,2[$ and $x \in]-1,0[\cup] 0,3[$.