Question

\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x},x\ge0,x\ne9\)

cách làm chi tiết

Answer 1

Với \(x\ge0;x\ne9\)

\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}\)

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{3-11\sqrt{x}}{9-x}\)

\(=\dfrac{2\sqrt{x}+\sqrt{x}+1-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{14\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{14\sqrt{x}-2}{x-9}\)