\(\Leftrightarrow\left[{}\begin{matrix}5x-4=x+2\\4-5x=x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=6\\6x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
|5x-4|=|x+2|
TH1: 5x-4=x+2⇒4x=6⇒x=3/2
TH2: 5x-4=-x-2⇒ 6x=2⇒x=1/3
Vậy ....
\(\left|5x-4\right|=\left|x+2\right|\)
\(\Rightarrow\left[{}\begin{matrix}5x-4=x+2\\5x-4=-x-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}5x-x=4+2\\5x+x=4-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x=6\\6x=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x=\dfrac{3}{2}\) hoặc \(x=\dfrac{1}{3}\)
