7.67 Given $\sqrt{(2-3|x|)^{2}}=2+3 x$, then the range of $x$ is
(A) $-\frac{2}{3} \leqslant x \leqslant \frac{2}{3}$.
(B) $-\frac{2}{3} \leqslant x \leqslant 0$.
(C) $0 \leqslant x \leqslant \frac{2}{3}$.
(D) $x \leqslant-\frac{2}{3}$ or $x \geqslant \frac{2}{3}$.
(Hubei Huanggang Region Junior High School Mathematics Competition, 1991)
[Solution] From the given, we have $|2-3| x||=2+3 x$, hence
$$
2+3 x \geqslant 0 \text{. }
$$

That is, $x \geqslant-\frac{2}{3}$, and when $x \leqslant 0$, we have $2-3|x|=2+3 x$.
When $x>0$, $2-3|x| \neq 2+3 x, \quad 3|x|-2 \neq 2+3 x$.
In summary,
$$
-\frac{2}{3} \leqslant x \leqslant 0 \text{, }
$$

Therefore, the answer is (B).