Question

cho biểu thức A=\((\dfrac{2}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\)) và B=\((\dfrac{\sqrt{x}+1}{\sqrt{x}-3})\)

1.rút gọn biểu thức M=A:B

2 .so sánh M và M²

Answer 1

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

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

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

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

\(\Rightarrow M=A\div B=\dfrac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\div\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

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

\(=\dfrac{3}{\sqrt{x+3}}\)