Tìm `x` biết : $\huge\text{|}$`|x-1|+|x-3|`$\huge\text{|}$

$\huge\text{|}$`|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 !