a: Thay a=16 vào A, ta được:
\(A=\dfrac{4+1}{4-3}=5\)
b: Ta có: \(B=\dfrac{2\sqrt{a}}{\sqrt{a}+3}+\dfrac{\sqrt{a}}{\sqrt{a}-3}+\dfrac{3a+3}{a-9}\)
\(=\dfrac{2a-6\sqrt{a}+a+3\sqrt{a}+3a+3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{6a-3\sqrt{a}+3}{a-9}\)