\(TH1\\ \left|x-5\right|=x-5\\ x-5=1-2x\\ x+2x=1+5\\ 3x=6\\ x=6:3\\ x=2\\ TH2\left|x-5\right|=5-x\\ 5-x=1-2x\\ -x+2x=1-5\\ x=-4\)
\(\left|x-5\right|=1-2x\)
ĐKXĐ: \(1-2x\ge0\Leftrightarrow x\le\dfrac{1}{2}\)
\(x-5=1-2x\); \(x-5=2x-1\)
*) \(x-5=1-2x\)
\(x+2x=1+5\)
\(3x=6\)
\(x=2\) (loại)
*) \(x-5=2x-1\)
\(x-2x=-1+5\)
\(-x=4\)
\(x=-4\) (nhận)
Vậy \(x=-4\)
\(\left|x-5\right|=1-2x\) \(ĐK:1-2x\ge0\Rightarrow x\le\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x-5=1-2x\\x-5=2x-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x+x=1+5\\x-2x=5-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
\(\Rightarrow x=-4\)
x-5=1-2x
x-5+5=-2x+1+5
x=-2x+1+5
x=-2x+6
x+2x=-2x+6+2x
x+2x=6
3x=6
x=6:3
x=2
