Question

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

rút gọn 

Answer 1

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

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

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