Question

cho biểu thức: A=\(\left(\frac{2\sqrt{x}+x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\)

a) rút gọn biểu thức

b) tính giá trị của \(\sqrt{A}\)khi \(x=4+2\sqrt{3}\)

Answer 1

a/ \(A=\left(\frac{2\sqrt{x}+x}{\sqrt{x}^3-1}-\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\)

      \(=\left[\frac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}-1}\right]:\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\)

         \(=\frac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{x+\sqrt{x}+1}{\sqrt{x}+2}\)

          \(=\frac{\sqrt{x}-1}{\sqrt{x}-1}.\frac{1}{\sqrt{x}+2}=\frac{1}{\sqrt{x}+2}\)

b/ Thay \(x=4+2\sqrt{3}\) vào A ta được:

    \(A=\frac{1}{\sqrt{4+2\sqrt{3}}+2}=\frac{1}{\sqrt{\left(\sqrt{3}+1\right)^2}+2}=\frac{1}{\sqrt{3}+3}\)

     \(\Rightarrow\sqrt{A}=\frac{1}{\sqrt{\sqrt{3}+3}}\)