a, Thay x = 9 ta được \(A=\dfrac{3-2}{3+9}=\dfrac{1}{12}\)
b, Với x > 0 ; x khác 4
\(B=\dfrac{3\sqrt{x}-6+\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3\sqrt{x}-6+x-5\sqrt{x}+6+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)