`#3107.101107`
`|3x - 1| = x + 2`
`\Rightarrow` TH1: `3x - 1 = x + 2`
`\Rightarrow 3x - x = 2 + 1`
`\Rightarrow 2x = 3`
`\Rightarrow x =` $\dfrac{3}2$
TH2: `3x - 1 = -(x + 2)`
`\Rightarrow 3x - 1 = -x - 2`
`\Rightarrow 3x + x = -2 + 1`
`\Rightarrow 4x = -1`
`\Rightarrow x =` $\dfrac{-1}4$
Vậy, \(x\in\left\{-\dfrac{1}{4};\dfrac{3}{2}\right\}.\)
|3x - 1| = x + 2
*) TH1: x ≥ 1/3, ta có:
|3x - 1| = x + 2
3x + 1 = x + 2
3x - x = 2 - 1
2x = 1
x = 1/2 (nhận)
*) TH2: x < 1/3, ta có:
|3x - 1| = x + 1
1 - 3x = x + 1
-3x - x = 1 - 1
-4x = 0
x = 0 (nhận)
Vậy x = 0; x = 1/2