6. For $a>3$, let
$$
\begin{array}{l}
x=3 \sqrt{a-1}+\sqrt{a-3}, \\
y=\sqrt{a}+3 \sqrt{a-2} .
\end{array}
$$

Then the relationship between $x$ and $y$ is ().
(A) $x>y$
(B) $x<y$
(C) $x=y$
(D) cannot be determined
6. B.

From $\sqrt{a}>\sqrt{a-2}, \sqrt{a-1}>\sqrt{a-3}$, adding them we get
$$
\begin{array}{l}
\sqrt{a}+\sqrt{a-1}>\sqrt{a-2}+\sqrt{a-3} \\
\Rightarrow \frac{1}{\sqrt{a}-\sqrt{a-1}}>\frac{1}{\sqrt{a-2}-\sqrt{a-3}}>0 \\
\Rightarrow \sqrt{a}-\sqrt{a-1}<\sqrt{a-2}-\sqrt{a-3} \\
\Rightarrow \sqrt{a}+\sqrt{a-3}<\sqrt{a-1}+\sqrt{a-2} \\
\Rightarrow 0<\frac{3}{\sqrt{a}-\sqrt{a-3}}<\frac{1}{\sqrt{a-1}-\sqrt{a-2}} \\
\Rightarrow 3(\sqrt{a-1}-\sqrt{a-2})<\sqrt{a}-\sqrt{a-3} \\
\Rightarrow 3 \sqrt{a-1}+\sqrt{a-3}<\sqrt{a}+3 \sqrt{a-2} .
\end{array}
$$