\(\left|3x-2\right|=x+1\left(1\right)\)
\(ĐK:x+1\ge0\Rightarrow x\ge-1\)
Với \(x\ge-1\) thì \(\left(1\right)\Rightarrow\left[{}\begin{matrix}3x-2=x+1\\3x-2=-\left(x+1\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-x=1+2\\3x-2=-x-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=3\\3x+x=-1+2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1,5\\4x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1,5\\x=0,25\end{matrix}\right.\)
Vậy \(x\in\left\{1,5;0,25\right\}\)
\(\left|2x-1\right|-3+5x=7x-2\)
\(\Rightarrow\left|2x-1\right|=7x-2+3-5x\)
\(\Rightarrow\left|2x-1\right|=\left(7x-5x\right)-2+3\)
\(\Rightarrow\left|2x-1\right|=2x+1\left(2\right)\)
\(ĐK:2x+1\ge0\Rightarrow2x\ge-1\Rightarrow x\ge\dfrac{-1}{2}\)
Với \(x\ge\dfrac{-1}{2}\) thì
\(\left(2\right)\Rightarrow\left[{}\begin{matrix}2x-1=2x+1\\2x-1=-\left(2x+1\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-2x=1+1\\2x-1=-2x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}0=2(loại)\\2x+2x=-1+1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}0=2\left(loại\right)\\4x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}0=2\left(loại\right)\\x=0\left(tm\right)\end{matrix}\right.\)
Vậy \(x=0\)
