a) \(\left|2x-1\right|=\left|x+1\right|\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=x+1\\2x-1=-x-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{0;2\right\}\)
b) \(\left|2x+7\right|=\left|-x+6\right|\)
\(\Rightarrow\left[{}\begin{matrix}2x+7=-x+6\\2x+7=x-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=-1\\x=-13\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-13\end{matrix}\right.\)
Vậy \(x\in\left\{-\dfrac{1}{3};-13\right\}\)
a) | 2x - 1 | = | x + 1 |
2x-1=x+1
2x-1 -x-1=0
x -2 =0
x= 2
vậy x=2
b) | 2x + 7 | = | -x + 6 |
2x+7 =-x+6
2x+7+x-6 = 0
3x+1=0
3x=-1
x=-1/3
