I1 (2-2, Hungary) For which values of $x$ is the inequality $\frac{4 x^{2}}{(1-\sqrt{1+2 x})^{2}}<2 x+$ 9 satisfied.
Solution 1 For the inequality to hold, it must be true that
$$
\left\{\begin{array}{l}
1+2 x \geqslant 0, \\
1-\sqrt{1+2 x} \neq 0,
\end{array}\right.
$$

So $x \geqslant-\frac{1}{2}$, and $x \neq 0$.
The original inequality is equivalent to
$$
\begin{array}{c}
\frac{4 x^{2}(1+\sqrt{1+2 x})^{2}}{(1-\sqrt{1+2 x})^{2}(1+\sqrt{1+2 x})^{2}}<2 x+9, \\
\frac{4 x^{2}(1+\sqrt{1+2 x})^{2}}{[1-(1+2 x)]^{2}}<2 x+9, \\
(1+\sqrt{1+2 x})^{2}<2 x+9, \\
2 \sqrt{1+2 x}<7, \\
x<\frac{45}{8} .
\end{array}
$$

Therefore, when $-\frac{1}{2} \leqslant x<0$ or $0<x<\frac{45}{8}$, the given inequality holds.
Solution 2 From
$$
\left\{\begin{array}{l}
1+2 x \geqslant 0 \\
1-\sqrt{1+2 x} \neq 0,
\end{array}\right.
$$

So $x \geqslant-\frac{1}{2}$, and $x \neq 0$.
Eliminating the denominator, we get
$$
\begin{array}{c}
4 x^{2}<(2 x+9)(1-\sqrt{1+2 x})^{?} \\
(2 x+9) \sqrt{1+2 x}<11 x+9
\end{array}
$$

Squaring both sides and simplifying, we get
$$
\begin{array}{c}
x^{2}(8 x-45)<0, \\
x<\frac{45}{8} .
\end{array}
$$

Therefore, when $x \in\left[-\frac{1}{2}, 0\right) \cup\left(0, \frac{45}{8}\right)$, the given inequality holds.