\(P=\dfrac{\sqrt{a.b}\left(\sqrt{a}+\sqrt{b}\right)+\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(P=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(P=\dfrac{a+b}{\sqrt{a}-\sqrt{b}}\)
tới đây là tự tìm đc rồi
Lời giải:
Ta có: \(a\sqrt{b}+b\sqrt{a}+a\sqrt{a}+b\sqrt{b}=(a\sqrt{b}+b\sqrt{a})+(a\sqrt{a}+b\sqrt{b})\)
\(=\sqrt{ab}(\sqrt{a}+\sqrt{b})+(\sqrt{a}+\sqrt{b})(a-\sqrt{ab}+b)\)
\(=(\sqrt{a}+\sqrt{b})(\sqrt{ab}+a-\sqrt{ab}+b)=(\sqrt{a}+\sqrt{b})(a+b)\)
Và: \(a-b=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})\)
Do đó: \(P=\frac{a+b}{\sqrt{a}-\sqrt{b}}=\frac{a+b-2\sqrt{ab}+2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)
\(=\frac{(\sqrt{a}-\sqrt{b})^2+2}{\sqrt{a}-\sqrt{b}}=(\sqrt{a}-\sqrt{b})+\frac{2}{\sqrt{a}-\sqrt{b}}\)
\(\geq 2\sqrt{(\sqrt{a}-\sqrt{b}).\frac{2}{\sqrt{a}-\sqrt{b}}}=2\sqrt{2}\) (áp dụng BĐT Cô-si cho 2 số dương)
Vậy \(P_{\min}=2\sqrt{2}\)
Dấu "=" xảy ra khi \((a,b)=(2+\sqrt{3}, 2-\sqrt{3})\)
