|2x-1|=5
=> 2x-1=5 hoặc 2x-1=-5
+, 2x-1=5
2x=5+1
2x=6
x=6:2
x=3
+, 2x-1=-5
2x=-5+1
2x=-4
x=(-4):2
x=-2
Vậy x ∈ {3;-2}
|2x-1|=5
\(\Rightarrow\)\(\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\left(5+1\right):2=3\\x=\left(-5+1\right):2=-2\end{matrix}\right.\)
Vậy: \(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
|2x−1|=5
=> 2x - 1 = 5 hay 2x - 1 = -5
2x = 5+1 hay 2x = -5 + 1
2x = 6 hay 2x = -4
x = 3 hay x = -2
=> x \(\in\) { 3 , -2 }