7. Given $a>0, b>0$ and
$$
\sqrt{a}(\sqrt{a}+4 \sqrt{b})=3 \sqrt{b}(\sqrt{a}+2 \sqrt{b}) \text {. }
$$

Then the value of $\frac{a+6 \sqrt{a b}-8 b}{2 a-3 \sqrt{a b}+2 b}$ is $(\quad)$.
(A) 1
(B) 2
(C) $\frac{19}{11}$.
(D) $\sqrt{2}$
7. B.

From $\sqrt{a}(\sqrt{a}+4 \sqrt{b})=3 \sqrt{b}(\sqrt{a}+2 \sqrt{b})$, we get $a+\sqrt{a b}-6 b=0$, which is
$$
(\sqrt{a}+3 \sqrt{b})(\sqrt{a}-2 \sqrt{b})=0 \text {. }
$$

Since $\sqrt{a}+3 \sqrt{b} \neq 0$, we have $\sqrt{a}-2 \sqrt{b}=0$, which means $a=4 b$.
Therefore, $\frac{a+6 \sqrt{a b}-8 b}{2 a-3 \sqrt{a b}+2 b}=\frac{4 b+12 b-8 b}{8 b-6 b+2 b}=2$.