Question

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

Answer 1

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

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

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

Answer 2

Ta có: \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)

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

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

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