a: \(\sqrt{4x^2-4x+1}=2x\)
=>\(\sqrt{\left(2x-1\right)^2}=2x\)
=>|2x-1|=2x
=>\(\begin{cases}2x\ge0\\ \left(2x-1\right)^2=\left(2x\right)^2\end{cases}\Rightarrow\begin{cases}x\ge0\\ \left(2x-1-2x\right)\left(2x-1+2x\right)=0\end{cases}\)
=>\(\begin{cases}x\ge0\\ -1\left(4x-1\right)=0\end{cases}\Rightarrow4x-1=0\)
=>x=1/4
b: ĐKXĐ: x∉{0;1}
\(\frac{x-1}{x}-\frac{x+1}{x-1}=2\)
=>\(\frac{\left(x-1\right)^2-x\left(x+1\right)}{x\left(x-1\right)}=2\)
=>\(2x\left(x-1\right)=x^2-2x+1-x^2-x=-3x+1\)
=>\(2x^2-2x+3x-1=0\)
=>\(2x^2+x-1=0\)
=>\(2x^2+2x-x-1=0\)
=>(x+1)(2x-1)=0
=>\(\left[\begin{array}{l}x+1=0\\ 2x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-1\left(nhận\right)\\ x=\frac12\left(nhận\right)\end{array}\right.\)
