\(A=1-\left(\dfrac{2}{1+2\sqrt{x}}-\dfrac{5\sqrt{x}}{4x-1}-\dfrac{1}{1-2\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{4x+4\sqrt{x}+1}\)

\(A=1-\dfrac{2\left(2\sqrt{x}-1\right)-5\sqrt{x}+\left(2\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-1}{\left(2\sqrt{x}+1\right)^2}\)

\(A=1-\dfrac{4\sqrt{x}-2-5\sqrt{x}+2\sqrt{x}+1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}\cdot\dfrac{\left(2\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)

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

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

Tick hộ nha

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

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

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

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

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

\(=\sqrt{4x-1-2\sqrt{4x-1}+1}+\sqrt{4x-1+2\sqrt{4x-1}+1}\)

\(=\sqrt{\left(\sqrt{4x-1}-1\right)^2}+\sqrt{\left(\sqrt{4x-1}+1\right)^2}\)

\(=\left|\sqrt{4x-1}-1\right|+\sqrt{4x-1}+1\)

\(=\left[{}\begin{matrix}2\sqrt{4x-1}\text{ nếu }x\ge\dfrac{1}{2}\\2\text{ nếu }\dfrac{1}{4}\le x< \dfrac{1}{2}\end{matrix}\right.\)

a: \(=\dfrac{\left|x+2\right|}{x-1}\)

b: \(=x-2y-\left|x-2y\right|\)\(=\left[{}\begin{matrix}x-2y-x+2y=0\\x-2y+x-2y=2x-4y\end{matrix}\right.\)

c: \(=\dfrac{\left|x+2\right|}{\left(x+2\right)\left(x-2\right)}=\pm\dfrac{1}{x-2}\)

`[2x+\sqrt{2}]/[4x^2+4\sqrt{2}x+\sqrt{2}]`

`=[\sqrt{2}(\sqrt{2}x+1)]/[\sqrt{2}(2\sqrt{2}x^2+4x+1)]`

`=[\sqrt{2}x+1]/[2\sqrt{2}x^2+4x+1]`

\(\dfrac{2x+\sqrt{2}}{4x^{2^{ }}4\sqrt{2}x^{2^{ }}+\sqrt{2}}\)

= \(\dfrac{\sqrt{2}\left(\sqrt{2}x+1\right)}{\sqrt{2}\left(2\sqrt{2}x^2+4x+1\right)}\)

= \(\dfrac{\sqrt{2}x+1}{2\sqrt{2}x^24x+1}\)

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

`a)ĐKXĐ:{(x > 0),(x \ne 4):}`

`b)` Với `x > 0,x \ne 4` có:

`A=[\sqrt{x}(\sqrt{x}+2)+\sqrt{x}(\sqrt{x}-2)]/[x-4].[x-4]/[\sqrt{4x}]`

`A=[x-2\sqrt{x}+x-2\sqrt{x}]/[2\sqrt{x}]`

`A=[2\sqrt{x}(\sqrt{x}-2)]/[2\sqrt{x}]=\sqrt{x}-2`

`c)` Với `x > 0,x \ne 4` có:

`A < 3 <=>\sqrt{x}-2 < 3<=>\sqrt{x} < 5<=>x < 25`

           Kết hợp đk

 `=>0 < x < 25 ,x \ne 4`

\(=\dfrac{\left(\sqrt{x}-\sqrt{3}\right)^2}{\left(\sqrt{x}-\sqrt{3}\right).\left(\sqrt{x}+\sqrt{3}\right)}.\left(2\sqrt{x}+\sqrt{12}\right)\)

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

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

Với `x ne 3;x >= 0` có:

`[(\sqrt{x}-3)^2]/[(\sqrt{x}-3)(\sqrt{x}+3)].(2\sqrt{x}+2\sqrt{3})`

`=[\sqrt{x}-3]/[\sqrt{x}+3].2(\sqrt{x}+3)`

`=2(\sqrt{x}-3)`

`=2\sqrt{3}-6`