Sửa lại nha :
\(\left|x-1\right|+3x=1\)
\(\Leftrightarrow\left|x-1\right|=1-3x\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=1-3x\\1-x=1-3x\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+3x=1+1\\-x+3x=1-1\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}4x=2\\-2x=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{2}\left(KTMĐK\right)\\x=0\end{array}\right.\)
Vậy \(x=0\)
\(\left|x-1\right|+3x=1\)
\(\left|x-1\right|=1-3x\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=1-3x\\1-x=1-3x\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+3x=1+1\\-x+3x=1-1\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}4x=2\\-2x=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{2}\\x=0\end{array}\right.\)
Vậy \(x\in\left\{\frac{1}{2};0\right\}\)
