Xét 2 trường hợp :
TH1 : x ≥ 0 => |x - 3| = x - 3
=> 3(x - 1) - 2|x - 3| = 3(x - 1) - 2(x - 3)
= 3x - 3 - 2x + 6
= x + 3
TH2 : x < 0 => |x - 3| = 3 - x
=> 3(x - 1) - 2|x - 3| = 3(x - 1) - 2(3 - x)
= 3x - 3 - 6 + 2x
= 5x - 9
Vậy 3(x - 1) - 2|x - 3| = x + 3 hoặc 3(x - 1) - 2|x - 3| = 5x - 9
3(x-1) - 2|x-3| (1)
* Nếu x>3 => x-3 >0
(1) = 3x-3-2x+6 = x+3
*Nếu x=3 => |x-3| =0
(1) = 3x-3
*Nếu x<3 => x-3<0 => |x-3| = 3-x
(1) =3x-3 - 6+ 2x = 5x -9