\(\text{|x−1|=2x+3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2x+3\\x-1=-2x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-\dfrac{2}{3}\end{matrix}\right.\)
\(|x-1|=2x+3\Rightarrow|x-1|=5x\Rightarrow x-1=\pm5x\)
\(x-1=5x\Rightarrow x=5x+1\Rightarrow x=6x\Rightarrow x^2=6\Rightarrow x=3\)
\(x-1=-5x\Rightarrow x=-5x+1\Rightarrow x=-4x\Rightarrow x^2=-4\Rightarrow x=...\)
\(\left|x-1\right|=2x+3\)
Với \(x-1\ge0\Leftrightarrow x\ge1\)
\(\Leftrightarrow x-1=2x+3\)
\(\Leftrightarrow-x=4\)
\(\Leftrightarrow x=-4\) ( loại )
Với \(x-1< 0\Leftrightarrow x< 1\)
\(\Leftrightarrow-x+1=2x+3\)
\(\Leftrightarrow-3x=2\)
\(\Leftrightarrow x=\dfrac{-2}{3}\) ( nhận )
Vậy \(S=\left\{\dfrac{-2}{3}\right\}\)
Ta có: \(\left|x-1\right|=2x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2x+3\left(x\ge1\right)\\1-x=2x+3\left(x< 1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=4\\-3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\left(loại\right)\\x=-\dfrac{2}{3}\left(nhận\right)\end{matrix}\right.\)
