4. Solve the inequality: $|x+7|-|3 x-4|$ $+\sqrt{3-2 \sqrt{2}}>0$. (78th Shaanxi Province High School Mathematics Competition Question)
$$
\begin{array}{l}
\text { Let }|x+7|-|3 x-4|+\sqrt{2} \\
-1=0 \\
\Rightarrow(x+7+3 x-1+2-1)(x+7 \\
-3 x+4 \div \sqrt{2}-1)(x+7-3 x+4-\sqrt{2} \\
+1)(x+7+3 x-4-\sqrt{2}+1)=0 \\
\Rightarrow(4 x+2+\sqrt{2})(-2 x+10+\sqrt{2}) \\
\text { - }(-2 x+12-\sqrt{2})(4 x+4-\sqrt{2})=0 \\
\Rightarrow x=-\frac{1}{4}(2+\sqrt{2}), \frac{1}{2}(10 \\
+\sqrt{2}), \frac{1}{2}(12-\sqrt{2}),-\frac{1}{4}(4-\sqrt{2}) \text {. } \\
\end{array}
$$

Upon inspection, $\mathrm{x}=-\frac{1}{4}(2+\sqrt{2}), \frac{1}{2}$ (10 $+\sqrt{2}$ ) are solutions to the equation $(\cdot)$.
$$
\begin{aligned}
& \because f(-1)=|6|-|-7|+\sqrt{2}-1 \\
= & -2+\sqrt{2}<0, \\
& f(6)=|13|-|14|+\sqrt{2}-1 \\
= & 2+\sqrt{2}<0,
\end{aligned}
$$
$\therefore$ The solution to the original inequality is $-\frac{1}{4}(2+\sqrt{2})$
$$
<x<\frac{1}{2}(10+\sqrt{2})
$$