B. If $a=-2+\sqrt{2}$, then the value of $1+\frac{1}{2+\frac{1}{3+a}}$ is ( ).
(A) $-\sqrt{2}$
(B) $\sqrt{2}$
(C) 2
(D) $2 \sqrt{2}$
B. B.

Notice,
$$
\begin{array}{l}
1+\frac{1}{2+\frac{1}{3+a}}=1+\frac{1}{2+\frac{1}{1+\sqrt{2}}} \\
=1+\frac{1}{2+\sqrt{2}-1}=1+\frac{1}{\sqrt{2}+1}=\sqrt{2} .
\end{array}
$$