Ta có : \(\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}+\sqrt{ab}\right)\)\(\left(\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)=1
⇌ \(\left(\dfrac{\sqrt{a}^3+\sqrt{b}^3}{\sqrt{a}+\sqrt{b}}+\sqrt{ab}\right)\)\(\left(\dfrac{\sqrt{a}+\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right)^2\)=1
⇌ \(\left(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}+\sqrt{ab}\right)\)\(\dfrac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)=1
⇌ \(\left(a+b\right)\)\(\dfrac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)=1
⇌ \(\dfrac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}-1=0\)
⇌ \(\dfrac{a+b-a+\sqrt{ab}-b}{\left(\sqrt{a}-\sqrt{b}\right)^2}=0\)
⇌ \(\sqrt{ab}=0\)
⇌\(\left[{}\begin{matrix}a=0\\b=0\end{matrix}\right.\)(thỏa mãn điều kiện)
Vậy a=0;b=0
\(\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
\(=\dfrac{a\sqrt{a}+b\sqrt{b}-a\sqrt{b}-b\sqrt{a}}{\sqrt{a}+\sqrt{b}}.\left(\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
\(=\dfrac{\sqrt{a}\left(a-b\right)-\sqrt{b}\left(a-b\right)}{\sqrt{a}+\sqrt{b}}.\left[\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}\right]^2\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a-b\right)}{\sqrt{a}+\sqrt{b}}.\left[\dfrac{a-b}{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}\right]^2\)
\(=\left(\sqrt{a}-\sqrt{b}\right)^2.\left(\dfrac{1}{\sqrt{a}-\sqrt{b}}\right)^2\)
\(=\left(\sqrt{a}-\sqrt{b}\right)^2.\dfrac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}=1\)
\(\Rightarrow dpcm\)