Question

Giúp mình với ạ 

Answer 1

a, \(A=\sqrt{x-\sqrt{x^2-4}}+\sqrt{x+\sqrt{x^2-4}}\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{x-2+x+2-2\sqrt{\left(x-2\right)\left(x+2\right)}}+\sqrt{x-2+x+2+2\sqrt{\left(x-2\right)\left(x+2\right)}}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left[\sqrt{\left(\sqrt{x-2}-\sqrt{x+2}\right)^2}+\sqrt{\left(\sqrt{x-2}+\sqrt{x+2}\right)^2}\right]\)

\(=\dfrac{1}{\sqrt{2}}\left[\sqrt{x+2}-\sqrt{x-2}+\sqrt{x-2}+\sqrt{x+2}\right]\)

\(=\sqrt{2x+4}\)

Answer 2

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

\(=\left[\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a+b-\sqrt{ab}\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right].\dfrac{1}{a-b}+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

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

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

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

Answer 3

b: Ta có: \(\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right):\left(a-b\right)+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

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

\(=1\)