2 câu trả lời
`3x - | 2x + 1 | = 3`
`⇒ | 2x + 1 | = 3x - 3`
Do `| 2x + 1 | ≥ 0` nên `3x - 3 ≥ 0 ⇒ x ≥ 1`
Khi đó `: | 2x + 1 | = 3x - 3`
`⇒` \(\left[ \begin{array}{l}2x + 1 = 3x - 3\\2x + 1 = - ( 3x - 3 )\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}3x - 2x = 3 + 1\\2x + 1 = - 3x + 3\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x = 4\\- 3x - 2x = 1 - 3\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x = 4\\- 5x = - 2\end{array} \right.\)
`⇒` \(\left[ \begin{array}{l}x = 4\\x = \frac{2}{5} ( Loại )\end{array} \right.\)
Vậy `, x = 4 .`
Answer
`3x - |2x + 1| = 3`
`=> |2x + 1| = 3x - 3`
Ta có:
`|2x + 1| >= 0 AA x`
`=> 3x - 3 >= 0 AA x`
`=> 3x >= 3`
`=> x >= 1`
`=>` `[(2x + 1 = 3x - 3),(2x + 1 = 3 - 3x):}`
`=>` `[(2x - 3x = -3 - 1),(2x + 3x = 3 - 1):}`
`=>` `[(-x = -4),(5x = 2):}`
`=>` `[(x = 4),(x = 2 : 5):}`
`=>` $\left[\begin{matrix} x = 4 \ \text{(TM)}\\ x = \dfrac{2}{5} \ \text{(KTM)}\end{matrix}\right.$
Vậy `x = 4`
