Question

Rút gọn biểu thức: \(B=\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}}\) với  \(x>0,x\ne1.\)                                                                                                
Tính giá trị của B khi \(x=12+8\sqrt{2}\)

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

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

\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt[]{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}\)

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

Thay \(x=12+8\sqrt{2}=\left(2\sqrt{2}+2\right)^2\) vào B, ta được:

\(B=\sqrt{\left(2\sqrt{2}+2\right)^2}-1=2\sqrt{2}+2-1=2\sqrt{2}+1\)