Lời giải:
ĐKXĐ: $a\geq 0; a\neq 1; a\neq 9$
\(B=\left[\frac{2\sqrt{a}(\sqrt{a}+3)}{\sqrt{a}+3}+\frac{(\sqrt{a}-3)^2}{\sqrt{a}-3)}\right].\frac{2}{\sqrt{a}-1}\)
\(=(2\sqrt{a}+\sqrt{a}-3).\frac{2}{\sqrt{a}-1}=3(\sqrt{a}-1).\frac{2}{\sqrt{a}-1}=6\)
Điều kiện: \(\left\{{}\begin{matrix}a\ge0\\a\ne1\\a\ne9\end{matrix}\right.\)
\(B=\left[\dfrac{2\sqrt{a}\left(\sqrt{a}+3\right)}{\sqrt{a}+3}+\dfrac{\left(\sqrt{a}-3\right)^2}{\sqrt{a}-3}\right].\dfrac{2}{\sqrt{a}-1}\)
\(=\left[2\sqrt{a}+\left(\sqrt{a}-3\right)\right].\dfrac{2}{\sqrt{a}-1}\)
\(=\left[2\sqrt{a}+\sqrt{a}-3\right].\dfrac{2}{\sqrt{a}-1}\)
\(=\left[3\sqrt{a}-3\right].\dfrac{2}{\sqrt{a}-1}\)
\(=3\left(\sqrt{a}-1\right).\dfrac{2}{\sqrt{a}-1}\)
\(=6\)
