$1 \cdot 100$ If $\sqrt{2} x>\sqrt{3} x+1$, then
$$
\sqrt[3]{(x+2)^{3}}-\sqrt{(x+3)^{2}}
$$

equals
(A) $2 x+5$.
(B) $-2 x-5$.
(C) 1 .
(D) -1 .
(2nd "Hope Cup" National Mathematics Invitational, 1991)
[Solution] From $\sqrt{2} x>\sqrt{3} x+1$, we have $(\sqrt{3}-\sqrt{2}) x<-1$, which means $x<-(\sqrt{3}+\sqrt{2})$, hence $\quad x<-3$.
$$
\text { Also } \quad \begin{aligned}
& \sqrt[3]{(x+2)^{3}}-\sqrt{(x+3)^{2}} \\
= & x+2-|x+3| \\
= & (x+2)+(x+3) \\
= & 2 x+5 .
\end{aligned}
$$

Therefore, the answer is $(A)$.