Question

Rút gọn biểu thức

\(\left(a+b-\frac{2a\sqrt{b}-2b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\right):\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

Moinj người giúp em sớm nhất có thể với ạ ? Em cảm ơn :)

 

Answer 1

\(\left(a+b-\frac{2a\sqrt{b}-2b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\right):\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\left(a+b-\frac{2\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\right):\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\left(a+b-2\sqrt{ab}\right):\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

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

\(=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}=1\)