a) ĐKXĐ: \(x\ge0\); \(1-4x\ne\)0; \(2\sqrt{x}-1\ne0\); \(\frac{1+2x}{1-4x}-\frac{2\sqrt{x}}{2\sqrt{x}-1}-1\ne\)0

<=> \(x\ge0\); x \(\ne\)1/4

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

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

\(A=\frac{\sqrt{x}-1}{1-4x}\cdot\frac{1-4x}{6x+4x+2\sqrt{x}}\)

\(A=\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\)

b)Với x \(\ge\)0 và x \(\ne\)1/4

Ta có: A > A² <=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}>\left(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\right)^2\)

<=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\cdot\left(1-\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\right)>0\)

<=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\cdot\frac{10x+2\sqrt{x}-\sqrt{x}+1}{10x+2\sqrt{x}}>0\)

<=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\cdot\frac{10+\sqrt{x}+1}{10x+2\sqrt{x}}>0\)

<=> \(\sqrt{x}-1>0\) <=> \(x>1\)

c) Với x\(\ge\)0 và x \(\ne\)1/4 (1)

Ta có: \(\left|A\right|>\frac{1}{4}\) <=> \(\orbr{\begin{cases}A>\frac{1}{4}\\A< -\frac{1}{4}\end{cases}}\)

TH1: \(A>\frac{1}{4}\) <=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}>\frac{1}{4}\)

<=> \(4\left(\sqrt{x}-1\right)>10x+2\sqrt{x}\)

<=> \(4\sqrt{x}-4>10x+2\sqrt{x}\)

<=> \(10x-2\sqrt{x}+4< 0\)(vô liia  vì \(10x-2\sqrt{x}+4>0\))

TH2: \(A< -\frac{1}{4}\) <=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}< -\frac{1}{4}\)

<=> \(4\left(\sqrt{x}-1\right)< -10x-2\sqrt{x}\)

<=> \(4\sqrt{x}-4+10x+2\sqrt{x}< 0\)

<=> \(10x+6\sqrt{x}-4< 0\)

<=> \(5x+3\sqrt{x}-2< 0\)

<=> \(\left(5\sqrt{x}-2\right)\left(\sqrt{x}+1\right)< 0\)

<=> \(x< \frac{4}{25}\) (2)

Từ (1) và (2) => \(0\le x< \frac{4}{25}\)

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

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

\(=\frac{\sqrt{x}-1}{1-4x}:\frac{2x-4\sqrt{x}}{1-4x}=\frac{\sqrt{x}-1}{1-4x}.\frac{1-4x}{2\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{1}{2\sqrt{x}}\)

b, \(A>A^2\Rightarrow\frac{1}{2\sqrt{x}}>\left(\frac{1}{2\sqrt{x}}\right)^2\Rightarrow\frac{1}{2\sqrt{x}}>\frac{1}{4x}\Rightarrow\frac{1}{2\sqrt{x}}-\frac{1}{4x}>0\Rightarrow\frac{2\sqrt{x}-1}{4x}>0\)

\(2\sqrt{x}-1>0\);\(4x>0\)

\(\Rightarrow x>0\)thì \(A>A^2\)

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

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

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

để A là số nguyên thì \(1-2\sqrt{x}\) là ước của 2 khi đó ta tìm được \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

nhanh hộ mình