Question

√x/√x+3 - 3/√x-3 +6√x/x-9 giúp em với

Answer 1

\(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{3}{\sqrt{x}-3}+\dfrac{6\sqrt{x}}{x-9}\left(dkxd:x\ge0,x\ne9\right)\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{3}{\sqrt{x}-3}+\dfrac{6\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-3\left(\sqrt{x}+3\right)+6\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x-3\sqrt{x}-3\sqrt{x}-9+6\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

\(=\dfrac{x-9}{x-9}=1\)