$\huge\text{|}$`|x-1|+|x-3|`$\huge\text{|}$`=4`
`⇔ |x-1|+|x-3| = 4`
TH1 : Nếu `x < 1`, ta có :
`-(x-1) - (x-3) = 4`
`⇔ -x + 1 - x + 3 = 4`
`⇔ -x - x = 4 - 1 - 3`
`⇔ -2x = 0`
`⇔ x = 0` $(T/m)$
TH2 : Nếu `1 ≤ x < 3`, ta có :
`(x-1) - (x-3) = 4`
`⇔ x - 1 - x + 3 = 4`
`⇔ x - x = 4 + 1 - 3`
`⇔ 0x = 2` `(KTM)`
TH3 : Nếu `x ≥ 3`, ta có :
`(x-1) + (x-3) = 4`
`⇔ x - 1 + x - 3 = 4`
`⇔ x + x = 4 + 1 + 3`
`⇔ 2x = 8`
`⇔ x = 4` $(T/m)$
Vậy `x = 0` hoặc `x=4`
Xin hay nhất !
