Question

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

rút gon

 

Answer 1

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

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

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

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

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