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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. ĐKXĐ : \(x\ne\frac{1}{2};\frac{5}{2};4;-\frac{3}{2};\frac{1\pm\sqrt{43}}{2}\)
\(A=\left(\frac{2x-3}{4x^2-12x+5}+\frac{3x-8}{13x-2x^2-20}-\frac{3}{2x-1}\right):\frac{21+2x-2x^2}{4x^2+4x-3}+\)
\(=\left(\frac{2x-3}{\left(2x-1\right)\left(2x-5\right)}-\frac{3x-8}{\left(2x-5\right)\left(x-4\right)}-\frac{3}{2x-1}\right).\frac{\left(2x-1\right)\left(2x+3\right)}{21+2x-2x^2}+1\)
\(=\frac{\left(2x-3\right)\left(x-4\right)-\left(3x-8\right)\left(2x-1\right)-3\left(2x-5\right)\left(x-4\right)}{\left(2x-1\right)\left(2x-5\right)\left(x-4\right)}.\frac{\left(2x-1\right)\left(2x+3\right)}{21+2x-2x^2}+1\)
\(=\frac{-10x^2+47x-56}{\left(2x-5\right)\left(x-4\right)}.\frac{2x+3}{-2x^2+2x+21}+1\) số to wa