Question

A=(\(\dfrac{1}{1-\sqrt{a}}\)+\(\dfrac{1}{1+\sqrt{a}}\)) . (1- \(\dfrac{1}{\sqrt{a}}\)) 

rút gọn

Answer 1

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

\(=\dfrac{1+\sqrt{a}+1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\cdot\dfrac{\sqrt{a}-1}{\sqrt[]{a}}\)

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