$1 \cdot 71 \sqrt{3+2 \sqrt{2}}-\sqrt{3-2 \sqrt{2}}$ equals
(A) 2.
(B) $2 \sqrt{3}$.
(C) $4 \sqrt{2}$.
(D) $\sqrt{6}$.
(E) $2 \sqrt{2}$.
(21st American High School Mathematics Examination, 1970)
$$
\begin{array}{l}
\text { [Solution] Original expression }=\sqrt{2+2 \sqrt{2}+1}-\sqrt{2-2 \sqrt{2}+1} \\
=\sqrt{(1+\sqrt{2})^{2}}-\sqrt{(\sqrt{2}-1)^{2}} \\
=1+\sqrt{2}-\sqrt{2}+1=2 \text {. } \\
\end{array}
$$

Therefore, the answer is $(A)$.