4. 1.3 * For which values of $x$ is the inequality $\frac{4 x^{2}}{(1-\sqrt{1+2 x})^{2}}<2 x+9$ satisfied.
To make the inequality hold, it must be 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 $\frac{4 x^{2}(1+\sqrt{1+2 x})^{2}}{(1-\sqrt{1+2 x})^{2}(1+\sqrt{1+2 x})^{2}}<2 x+9$,
$$
(1+\sqrt{1+2 x})^{2}<2 x+9,2 \sqrt{1+2 x}<7, x<\frac{45}{8} \text {. }
$$

Therefore, when $-\frac{1}{2} \leqslant x<0$ or $0<x<\frac{45}{8}$, the original inequality holds.