|x+3|=1-4x-2|x-1|
TH1: x+3 = 1 -4x-2|x-1|
<=>|x-1|=$\frac{-5x-2}{2}$
TH a: x-1=$\frac{-5x-2}{2}$
<=>2x-2=-5x-2
<=>7x=0
=> x=0
TH b: x-1=- $\frac{-5x-2}{2}$
<=> 2x-2=5x+2
<=> -3x=4
<=> x= -4/3
TH 2: x+3=-1+4x+2|x-1|
<=> |x-1|=$\frac{-3x+4}{2}$
TH a: x-1=$\frac{-3x+4}{2}$
,<=>2x-2=-3x+4
<=>x=1,2
TH b: x-1 = - $\frac{-3x+4}{2}$
<=> 2x-2= 3x-4
<=>x=2
Câu thứ 2 làm tương tự
