2. If $\frac{x}{3 y}=\frac{y}{2 x-5 y}=\frac{6 x-15 y}{x}$, then the value of $\frac{4 x^{2}-5 x y+6 y^{2}}{x^{2}-2 x y+3 y^{2}}$ is ( ).
(A) $\frac{9}{2}$.
(B) $\frac{9}{4}$
(C) 5
(D) 6
2. (A).

From the given conditions, we know that $x \neq 0, y \neq 0$. Transform the given equation and use the ratio to eliminate $y$, we get
$$
\text { - } \begin{aligned}
\frac{25 x}{75 y} & =\frac{15 y}{30 x-75 y}=\frac{6 x-15 y}{x} \\
& =\frac{25 x+15 y+16 x-15}{75}+\frac{1}{x} \\
& =\frac{31 x}{31 x}=1 .
\end{aligned}
$$

Thus, $x=3 y$.
$$
\begin{array}{l}
\text { Therefore, } \frac{4 x^{2}-5 x y+6 y^{2}}{x^{2}-2 x y+3 y^{2}}=\frac{36 y^{2}-15 y^{2}+6 y^{2}}{9 y^{2}-6 y^{2}+3 y^{2}} \\
=\frac{27 y^{2}}{6 y^{2}}=\frac{9}{2} .
\end{array}
$$