1. $4 \sqrt{3+2 \sqrt{2}}-\sqrt{41+24 \sqrt{2}}=(\quad)$.
(A) $\sqrt{2}-1$
(B) 1
(C) $\sqrt{2}$
(D) 2
$\begin{array}{l}\text {-1. B. } \\ 4 \sqrt{3+2 \sqrt{2}}-\sqrt{41+24 \sqrt{2}} \\ =4(\sqrt{2}+1)-(3+4 \sqrt{2})=1\end{array}$