\(\left|2x+1\right|+\left|x+2\right|=x\)
\(\Rightarrow\left[{}\begin{matrix}2x+1+x+2=x\\-2x-1-x-2=x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x+3=x\\-3x-3=x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-2x=3\Rightarrow x=\dfrac{-2}{3}\\-4x=3\Rightarrow x=\dfrac{-4}{3}\end{matrix}\right.\)
Ta có :
\(\left|2x+1\right|+\left|x+2\right|=x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1+x+2=x\\-2x-1-x-2=x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+3=x\\-3x-3=x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=3\\-4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{-3}{4}\end{matrix}\right.\)
Vậy ....
Ta có VT là tổng 2 giá trị tuyệt đối => \(VT\ge0\)
\(\Rightarrow x\ge0\)
Với \(x\ge0\) thì \(\left\{{}\begin{matrix}\left|2x+1\right|=2x+1\\\left|x+2\right|=x+2\end{matrix}\right.\)
Do đó
\(\left|2x+1\right|+\left|x+2\right|=x\)
\(\Leftrightarrow\left(2x+1\right)+\left(x+2\right)=x\)
\(\Leftrightarrow x=\dfrac{-3}{2}\) (vô lý vì \(x\ge0\))
Vậy pt vô nghiệm.