Question

rút gọn biểu thức M=\(\dfrac{x+3}{x-9}\)+\(\dfrac{2}{\sqrt{x}+3}\)-\(\dfrac{1}{\sqrt{x}-3}\) với x≥0,x≠9

Answer 1

Answer 2

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

\(M=\dfrac{x+3+2\sqrt{x}-6-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

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

\(M=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\)

Answer 3

Ta có: \(M=\dfrac{x+3}{x-9}+\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}\)

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

\(=\dfrac{x-\sqrt{x}+2\sqrt{x}-6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

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