a) Ta có: \(B=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{2+5\sqrt{x}}{9-x}\)
\(=\dfrac{x+4\sqrt{x}+3+2x-6\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3x-7\sqrt{x}+1}{x-9}\)
b) Để B=3 thì \(3x-7\sqrt{x}+1-3x+27=0\)
\(\Leftrightarrow-7\sqrt{x}=-28\)
\(\Leftrightarrow x=16\)(nhận)