1: Thay \(a=\dfrac{1}{9}\) vào A, ta được:
\(A=\dfrac{6}{\dfrac{1}{9}+2\cdot\dfrac{1}{9}}=18\)
2: Ta có: \(B=\dfrac{\sqrt{a}}{a-4}+\dfrac{2}{2-\sqrt{a}}+\dfrac{1}{\sqrt{a}+2}\)
\(=\dfrac{\sqrt{a}-2\sqrt{a}-4+\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{-6}{a-4}\)




