Question

Rút gọn biểu thức:

\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{3\sqrt{x}}{x-\sqrt{x}}\) với x>0;x\(\ne\)1

Answer 1

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

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

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

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

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

Answer 2

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{3}{\sqrt{x}-1}\\ =\dfrac{\sqrt{x}+1-3}{\sqrt{x}-1}=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

Answer 3

\(\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x-1}\right)}-\dfrac{3\sqrt{x}}{\sqrt{x}.\left(\sqrt{x}-1\right)}\) \(=\dfrac{x+\sqrt{x}-3\sqrt{x}}{\sqrt{x}.\left(\sqrt{x-1}\right)}\) \(=\dfrac{x-2\sqrt{x}}{\sqrt{x}.\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}.\left(\sqrt{x}-2\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)