Question

Rút gọn:

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

 

Answer 1

Lời giải:

ĐKXĐ: $x>0; x\neq 1$

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

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

 

Answer 2

ĐKXĐ: \(x>0,x\ne1\)

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

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

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

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

Answer 3

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

Ta có : \(A=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}\right)\)

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

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

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

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