h) \(x-2-\sqrt{4-4x+x^2}\)

\(=x-2-\sqrt{\left(2-x\right)^2}\)

\(=x-2-\left|2-x\right|\)

\(=x-2-2+x\)

\(=2x-4\)

g) \(x-2-\sqrt{4-4x+x^2}\)

\(=x-2-\sqrt{\left(2-x\right)^2}\)

\(=x-2-\left|2-x\right|\)

\(=x-2-\left[-\left(2-x\right)\right]\)

\(=x-2+2-x\)

\(=0\)

i) \(3-x+\sqrt{9+6x+x^2}\)

\(=3-x+\sqrt{\left(3+x\right)^2}\)

\(=3-x+\left|3+x\right|\)

\(=3-x-3-x\)

\(=-2x\)

\(=\sqrt{\left(x-2\right)^2}=\left|x-2\right|=x-2\)

\(a)C=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\dfrac{x-4}{\sqrt{4x}}\\ =\left(\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}+\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-4}\right)\dfrac{x-4}{2\sqrt{x}}\\ =\left(\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}\right)\dfrac{x-4}{2\sqrt{x}}\\ =\dfrac{2x}{x-4}\cdot\dfrac{x-4}{2\sqrt{x}}\\ =\dfrac{2x\left(x-4\right)}{2\sqrt{x}\left(x-4\right)}\\ =\sqrt{x}\)

b) C>3

\(\Rightarrow\sqrt{x}>3\\ \Leftrightarrow x>9\)

a) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right).\dfrac{x-4}{\sqrt{4x}}=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}.\dfrac{x-4}{2\sqrt{x}}=\dfrac{2x}{2\sqrt{x}}=\sqrt{x}\)

b) \(P=\sqrt{x}>3\Leftrightarrow x>9\)

a: Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\cdot\dfrac{x-4}{2\sqrt{x}}\)

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

\(=\sqrt{x}\)

b: Để P>3 thì x>9

\(\dfrac{\sqrt{x^2-4x+4}}{x-2}=\dfrac{\left|x-2\right|}{x-2}=\left[{}\begin{matrix}1\left(x\ge2\right)\\-1\left(x< 2\right)\end{matrix}\right.\)

\(\dfrac{\sqrt{x^2-4x+4}}{x-2}=\dfrac{\left|x-2\right|}{x-2}=\pm1\)

\(A=\dfrac{2}{x-1}\sqrt{\dfrac{\left(x-1\right)^2}{4x^2}}=\dfrac{2}{x-1}\left|\dfrac{x-1}{2x}\right|=\dfrac{\left|x-1\right|}{\left(x-1\right)\left|x\right|}\)

\(B=\left(x^2-4\right)\sqrt{\dfrac{9}{x^2-4x+4}}=\dfrac{3\left(x^2-4\right)}{\left|x-2\right|}\)

a) Ta có: \(A=\dfrac{2}{x-1}\cdot\sqrt{\dfrac{x^2-2x+1}{4x^2}}\)

\(=\dfrac{2}{x-1}\cdot\dfrac{x-1}{2x}\)

\(=\dfrac{1}{x}\)

b) Ta có: \(\left(x^2-4\right)\cdot\sqrt{\dfrac{9}{x^2-4x+4}}\)

\(=\dfrac{\left(x-2\right)\left(x+2\right)\cdot3}{\left(x-2\right)^2}\)

\(=\dfrac{3x+6}{x-2}\)

`đk:x-\sqrt{x^2-4x+4}>=0`

`<=>x>=\sqrt{x^2-4x+4}`

`<=>x^2>=x^2-4x+4(x>=0)`

`<=>4x-4>=0`

`<=>4x>=4<=>x>=1`

`b)A=sqrt{x-sqrt{(x-2)^2}}`

`=sqrt{x-|x-2|}`

`x>=2=>|x-2|=x-2`

`=>A=sqrt{x-x+2}=sqrt2`

`1<=x<=2=>|x-2|=2x-`

`=>A=\sqrt{x+x-2}=sqrt{2x-2}`

Bài 1: 

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

b) \(\dfrac{\sqrt{\left(x-3\right)^2}}{3-x}=\dfrac{\left|x-3\right|}{3-x}=\pm1\)

Bài 2: 

a) \(\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}=\dfrac{\left|3x-1\right|}{\left(3x-1\right)\left(3x+1\right)}=\pm\dfrac{1}{3x+1}\)

b) \(4-x-\sqrt{x^2-4x+4}=4-x-\left|x-2\right|=\left[{}\begin{matrix}6-2x\left(x\ge2\right)\\2\left(x< 2\right)\end{matrix}\right.\)