Question

Bài : Rút gọn 
\(\dfrac{\sqrt{a}+\sqrt{ab}}{a-b}\) - \(\dfrac{\sqrt{b}}{\sqrt{a}-\sqrt{b}}\) với a,b ≥ 0 , a≠b

Answer 1

\(\dfrac{\sqrt{a}+\sqrt{ab}}{a-b}-\dfrac{\sqrt{b}}{\sqrt{a}-\sqrt{b}}\left(a,b\ge0;a\ne b\right)\)

\(=\dfrac{\sqrt{a}+\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\dfrac{\sqrt{a}+\sqrt{ab}-\sqrt{ab}-b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\dfrac{\sqrt{a}-b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\dfrac{\sqrt{a}-b}{a-b}\)