Question

cho a, b, a>o, b>o, a\(\ne\)b

Rút gọn biểu thức:

\(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^3-b\sqrt{b}+2a\sqrt{a}}{a\sqrt{a}-b\sqrt{b}}+\dfrac{3a+3\sqrt{ab}}{b-a}\)

Answer 1

\(=\dfrac{a\sqrt{a}+b\sqrt{b}+3\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)-b\sqrt{b}+2a\sqrt{a}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}-\dfrac{3\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\)

\(=\dfrac{3a\sqrt{a}+3\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}-\dfrac{3\sqrt{a}}{\sqrt{a}-\sqrt{b}}\)

\(=\dfrac{3\sqrt{a}\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}-\dfrac{3\sqrt{a}}{\sqrt{a}-\sqrt{b}}\)

=0