\(\left|2x-5\right|=x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=x+1\\2x-5=-\left(x+1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=x+1\\2x-5=-x+\left(-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-x=1+5\\2x+x=5-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=6\\3x=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=\dfrac{4}{3}\end{matrix}\right.\)\(\left(TM\right)\)
Vậy \(x\in\left\{6;\dfrac{3}{4}\right\}\) là giá trị cần tìm
\(\left|2x-5\right|=x+1\)
\(\left|2x-5\right|\ge0\Rightarrow x+1\ge0\)
\(\Rightarrow2x-5=x+1\)
\(2x=x+6\)
\(x=6\)
Tập hợp 1 pt
