ĐK: \(x\ge\frac{1}{4}\)

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

Với \(x\ge\frac{1}{2},A=4x+2\left(2x-1\right)=8x-2\)

Do \(A\ge0\) nên \(A=\sqrt{8x-2}\)

Với \(\frac{1}{4}\le x< \frac{1}{2},A^2=4x+2\left(1-2x\right)=2\)

Do \(A\ge0\) nên \(A=\sqrt{2}\)

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

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

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

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

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

\(\frac{2x-x-1}{3x^2-3x-x+1}\)

\(=\frac{x-1}{\left(x-1\right)\left(3x-1\right)}\)

\(=\frac{1}{3x-1}\)

`[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]`

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

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

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