| x + 2 | + | x + 4 | + | x + 6 | = 4x – 1 12/07/2021 Bởi Charlie | x + 2 | + | x + 4 | + | x + 6 | = 4x – 1
$| x + 2 | + | x + 4 | + | x + 6 | = 4x – 1$ Vì : $|x+2|;|x+4|;|x+6| ≥ 0 ∀ x$ $⇔ |x+2| + |x+4| + |x+6| ≥ 0 ∀ x$ $⇒ 4x – 1 ≥ 0$ $⇔ x ≥ \dfrac{1}{4}$ $⇒ x+2+x+4+x+6=4x-1$ $⇔ 3x + 12 = 4x – 1$ $⇔ 12 + 1 = 4x – 3x$ $⇔ x = 13$ Vậy $x=13$. Bình luận
`| x + 2 | + | x + 4 | + | x + 6 | = 4x – 1` Vì : `|x+2|, |x+4|, |x+6| ≥ 0 ∀x` `⇔ |x+2| + |x+4| + |x+6| ≥ 0 ∀x` `⇒ 4x – 1 ≥ 0` `⇔ x ≥ 1/4` `⇒ x+2+x+4+x+6=4x-1` `⇔ 3x + 12 = 4x – 1` `⇔ 12 + 1 = 4x – 3x` `⇔ x = 13` Vậy `x=13` Bình luận
$| x + 2 | + | x + 4 | + | x + 6 | = 4x – 1$
Vì : $|x+2|;|x+4|;|x+6| ≥ 0 ∀ x$
$⇔ |x+2| + |x+4| + |x+6| ≥ 0 ∀ x$
$⇒ 4x – 1 ≥ 0$
$⇔ x ≥ \dfrac{1}{4}$
$⇒ x+2+x+4+x+6=4x-1$
$⇔ 3x + 12 = 4x – 1$
$⇔ 12 + 1 = 4x – 3x$
$⇔ x = 13$
Vậy $x=13$.
`| x + 2 | + | x + 4 | + | x + 6 | = 4x – 1`
Vì : `|x+2|, |x+4|, |x+6| ≥ 0 ∀x`
`⇔ |x+2| + |x+4| + |x+6| ≥ 0 ∀x`
`⇒ 4x – 1 ≥ 0`
`⇔ x ≥ 1/4`
`⇒ x+2+x+4+x+6=4x-1`
`⇔ 3x + 12 = 4x – 1`
`⇔ 12 + 1 = 4x – 3x`
`⇔ x = 13`
Vậy `x=13`