Chứng minh rằng:

- Hướng dẫn giải

Nhận xét rằng:

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

Từ đó, suy ra:

$\begin{array}{l}\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\\ =\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\left(\sqrt{a}+\sqrt{b}\right)}-\sqrt{ab}\\ =a-\sqrt{ab}+b-\sqrt{b}\\ =a-2\sqrt{ab}+b\\ ={\left(\sqrt{a}\right)}^2-2\sqrt{ab}+{\left(\sqrt{b}\right)}^2\\ ={\left(\sqrt{a}-\sqrt{b}\right)}^2\end{array}$