Question

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

Answer 1

Đề bài là rút gọn hả bn?

Answer 2

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

Answer 3

Đề bài là chứng minh rằng nhé. Tớ ghi thiếu.

Answer 4

\(\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)

\(=\dfrac{a\sqrt{a}+b\sqrt{b}-a\sqrt{b}-b\sqrt{a}}{\sqrt{a}+\sqrt{b}}.\left(\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)

\(=\dfrac{\sqrt{a}\left(a-b\right)-\sqrt{b}\left(a-b\right)}{\sqrt{a}+\sqrt{b}}.\left[\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}\right]^2\)

\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a-b\right)}{\sqrt{a}+\sqrt{b}}.\left[\dfrac{a-b}{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}\right]^2\)

\(=\left(\sqrt{a}-\sqrt{b}\right)^2.\left(\dfrac{1}{\sqrt{a}-\sqrt{b}}\right)^2\)

\(=\left(\sqrt{a}-\sqrt{b}\right)^2.\dfrac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}=1\)

\(\Rightarrow dpcm\)