\(a,ĐK:\dfrac{x-1}{x+1}\ge0\Leftrightarrow x\ge1;-1>x\\ \Leftrightarrow\dfrac{x-1}{x+1}=4\\ 4\left(x+1\right)=x-1\\ \Leftrightarrow3x=-5\Leftrightarrow x=-\dfrac{5}{3}\left(tm\right)\)
\(b,ĐK:x\ge5\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-3\sqrt{x-5}=4\\ \Leftrightarrow0\sqrt{x-5}=4\Leftrightarrow x\in\varnothing\)
\(c,ĐK:x\in R\\ \Leftrightarrow\sqrt{\left(5x-1\right)^2}+2x=1\\ \Leftrightarrow\left|5x-1\right|=1-2x\\ \Leftrightarrow\left[{}\begin{matrix}5x-1=1-2x\left(x\ge\dfrac{1}{5}\right)\\5x-1=2x-1\left(x< \dfrac{1}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{7}\left(x\ge\dfrac{1}{5}\right)\left(tm\right)\\x=0\left(x< \dfrac{1}{5}\right)\left(tm\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{7}\\x=0\end{matrix}\right.\)
