|x+3| = 1 – 4x – 2|x-1| |x-2015|+|x-2017|=2

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1. |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

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