\(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}=a+\sqrt{ab}+b\)
Thiếu đkxđ: \(a,b>0\)
đk: \(a,b\ge0\)
Ta có: \(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}=a-\sqrt{ab}+b\)