a: ĐKXĐ: x>=0; x∉{1/4}
Ta có: \(\frac{\sqrt{x}-4x}{1-4x}-1\)
\(=\frac{-4x+\sqrt{x}}{1-4x}-1=\frac{-4x+\sqrt{x}-1+4x}{1-4x}\)
\(=\frac{\sqrt{x}-1}{1-4x}=\frac{-\left(\sqrt{x}-1\right)}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}\)
Ta có: \(\frac{1+2x}{1-4x}-\frac{2\sqrt{x}}{2\sqrt{x}-1}-1\)
\(=\frac{-2x-1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}-\frac{2\sqrt{x}}{2\sqrt{x}-1}-1\)
\(=\frac{-2x-1-2\sqrt{x}\left(2\sqrt{x}+1\right)-\left(4x-1\right)}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}\)
\(=\frac{-2x-1-4x-2\sqrt{x}-4x+1}{\left(2\sqrt{x}-1\right)\cdot\left(2\sqrt{x}+1\right)}=\frac{-10x-2\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}\)
Ta có: \(A=\left(\frac{\sqrt{x}-4x}{1-4x}-1\right):\left(\frac{1+2x}{1-4x}-\frac{2\sqrt{x}}{2\sqrt{x}-1}-1\right)\)
\(=\frac{-\left(\sqrt{x}-1\right)}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}:\frac{-10x-2\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}\)
\(=\frac{-\left(\sqrt{x}-1\right)}{-2\sqrt{x}\left(5\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{2\sqrt{x}\left(5\sqrt{x}+1\right)}\)
b: \(A>A^2\)
=>\(A^2
=>\(A^2-A<0\)
=>A(A-1)<0
=>\(\frac{\sqrt{x}-1}{2\sqrt{x}\left(5\sqrt{x}+1\right)}\left(\frac{\sqrt{x}-1}{2\sqrt{x}\left(5\sqrt{x}+1\right)}-1\right)<0\)
=>\(\left(\sqrt{x}-1\right)\left(\frac{\sqrt{x}-1-2\sqrt{x}\left(5\sqrt{x}+1\right)}{2\sqrt{x}\left(5\sqrt{x}+1\right)}\right)<0\)
=>\(\left(\sqrt{x}-1\right)\left(\sqrt{x}-1-10x-2\sqrt{x}\right)<0\)
=>\(\left(\sqrt{x}-1\right)\left(-10x-\sqrt{x}-1\right)<0\)
=>\(\left(\sqrt{x}-1\right)\left(10x+\sqrt{x}+1\right)>0\)
=>\(\left(\sqrt{x}-1\right)\cdot10\cdot\left(x+\frac{1}{10}\sqrt{x}+\frac{1}{10}\right)>0\)
=>\(\left(\sqrt{x}-1\right)\left(x+\frac{1}{10}\sqrt{x}+\frac{1}{10}\right)>0\)
=>\(\sqrt{x}-1>0\)
=>x>1
c: \(\left|A\right|>\frac14\)
=>\(\left[\begin{array}{l}A>\frac14\\ A<-\frac14\end{array}\right.\)
TH1: \(A>\frac14\)
=>\(\frac{\sqrt{x}-1}{2\sqrt{x}\left(5\sqrt{x}+1\right)}>\frac14\)
=>\(\frac{2\sqrt{x}-2}{4\sqrt{x}\left(5\sqrt{x}+1\right)}>\frac{\sqrt{x}\left(5\sqrt{x}+1\right)}{4\sqrt{x}\left(5\sqrt{x}+1\right)}\)
=>\(2\sqrt{x}-2>\sqrt{x}\left(5\sqrt{x}+1\right)\)
=>\(5x+\sqrt{x}-2\sqrt{x}+2<0\)
=>\(5x-\sqrt{x}+2<0\)
=>\(x-\frac15\sqrt{x}+\frac25<0\)
=>\(x-2\cdot\sqrt{x}\cdot\frac{1}{10}+\frac{1}{100}+\frac{39}{100}<0\)
=>\(\left(\sqrt{x}-\frac{1}{10}\right)^2+\frac{39}{100}<0\) (vô lý)
TH2: \(A<-\frac14\)
=>\(\frac{\sqrt{x}-1}{2\sqrt{x}\left(5\sqrt{x}+1\right)}<\frac{-1}{4}\)
=>\(\frac{2\sqrt{x}-2}{4\sqrt{x}\left(5\sqrt{x}+1\right)}<\frac{-\sqrt{x}\left(5\sqrt{x}+1\right)}{4\sqrt{x}\left(5\sqrt{x}+1\right)}\)
=>\(2\sqrt{x}-2<-\sqrt{x}\left(5\sqrt{x}+1\right)\)
=>\(2\sqrt{x}-2+5x+5\sqrt{x}<0\)
=>\(5x+7\sqrt{x}-2<0\)
=>\(x+\frac75\sqrt{x}-\frac25<0\)
=>\(x+2\cdot\sqrt{x}\cdot\frac{7}{10}+\frac{49}{100}<\frac{89}{100}\)
=>\(\left(\sqrt{x}+\frac{7}{10}\right)^2<\frac{89}{100}\)
=>\(\sqrt{x}+\frac{7}{10}<\frac{\sqrt{89}}{10}\)
=>\(\sqrt{x}<\frac{\sqrt{89}-7}{10}\)
=>\(x<\frac{\left(\sqrt{89}-7\right)^2}{100}=\frac{138-14\sqrt{89}}{100}=\frac{69-7\sqrt{89}}{50}\)
