Question

[Trang]

\(\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2=1\) (\(a\ge0,b\ge0,a\ne b\))

Answer 1

[Sáng]

Đề bài là rút gọn hả bn?

Answer 2

[cao minh thành]

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