Sửa đề: \(\dfrac{\left(a\sqrt{b}+b\right)\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2\sqrt{ab^3}}{a\left(a+2\sqrt{b}\right)+b}}\)
Đặt \(A=\dfrac{\left(a\sqrt{b}+b\right)\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2\sqrt{ab^3}}{a\left(a+2\sqrt{b}\right)+b}}\)
ĐKXĐ: a>0 và b>0 và a<>b
\(A=\dfrac{\left(a\sqrt{b}+b\right)\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2\sqrt{ab^3}}{a\left(a+2\sqrt{b}\right)+b}}\)
\(=\dfrac{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\cdot\left(\sqrt{a}+\sqrt{b}\right)}\cdot\sqrt{\dfrac{ab+b^2-2\sqrt{ab}\cdot b}{a^2+2a\sqrt{b}+b}}\)
\(=\dfrac{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\sqrt{\dfrac{\left(\sqrt{ab}-b\right)^2}{\left(a+\sqrt{b}\right)^2}}\)
\(=\dfrac{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{ab}-b}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{\sqrt{b}\left(\sqrt{ab}-b\right)}{\sqrt{a}-\sqrt{b}}=\dfrac{b\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\)
=b
