2. Let $x>3, M=3 \sqrt{x-1}+\sqrt{x-3}, N=$ $\sqrt{x}+3 \sqrt{x-2}$. Then the size relationship between $M$ and $N$ is ( ).
(A) $M>N$
(B) $M<N$
(C) $M=N$
(D) Any of the three situations is possible
2.B.

Given $x>3$, we have
$$
\begin{array}{l}
\sqrt{x}-\sqrt{x-1}=\frac{1}{\sqrt{x}+\sqrt{x-1}} \\
<\frac{1}{\sqrt{x-2}+\sqrt{x-3}} \\
=\sqrt{x-2}-\sqrt{x-3} .
\end{array}
$$

Thus, $\sqrt{x}+\sqrt{x-3}<\sqrt{x-1}+\sqrt{x-2}$,
which implies $\frac{3}{\sqrt{x}-\sqrt{x-3}}<\frac{1}{\sqrt{x-1}-\sqrt{x-2}}$, or equivalently, $3(\sqrt{x-1}-\sqrt{x-2})<\sqrt{x}-\sqrt{x-3}$.
Rearranging terms, we get $M<N$.