*)Nếu x\(\ge\)1 ta có:
3(x-1)-6x=0
<=>3x-3-6x=0
<=>-3x=3
<=>x=-1(L)
*)Nếu 1>x\(\ge\)0 ta có:
3(1-x)-6x=0
<=>3-3x-6x=0
<=>3-9x=0
<=>9x=3
<=>x=\(\dfrac{1}{3}\)(TM)
*)Nếu x<0 ta có:
3(1-x)+6x=0
<=>3-3x+6x=0
<=>3+3x=0
<=>3x=-3
<=>x=-1(TM)
Vậy tập nghiệm S={\(\dfrac{1}{3};-1\)}
