Question

Tìm tập xác định và rút gọn \(A=\dfrac{3\left(\sqrt{ab}-b\right)}{a-b}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^3+2a\sqrt{a}+b\sqrt{b}}{a\sqrt{a}+b\sqrt{b}}\)

Answer 1

ĐKXĐ: \(a,b\ge0,a\ne b\)

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

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

= \(\dfrac{3\sqrt{b}}{\sqrt{a}+\sqrt{b}}+\dfrac{3\sqrt{a}}{\sqrt{a}+\sqrt{b}}\)

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