| 3-x | -2=x+2
=) | 3-x | =x+2+2=x+4
=) 3-x = x+4 hoặc = -(x+4)=-x-4
TH1 : 3-x=x+4
=) 3-4=x+x
=) -1 = 2x
=) \(x=\frac{-1}{2}\)
TH2 : 3-x=-x-4
=) 3+4=-x+x
=) 7=0 (Loại vì \(7\ne0\))
=) \(x\in\)rỗng
Vậy \(x=\frac{-1}{2}\)
b) | 2x+1 | + x = 2x-5
=) | 2x+1 | = 2x-5-x
=) 2x+1 = 2x-5-x hoặc = -(2x-5-x) = -2x+5+x
TH1 : 2x+1 = 2x-5-x
=) 2x-2x=-5-x-1
=) 0 = -5-1-x = -6-x
=) x = (-6)-0=-6
TH2 : 2x+1 = -2x+5+x
=) 2x+2x = 5+x-1
=) 4x = 4+x
=) 4x-x = 4
=) 3x = 4
=) \(x=\frac{4}{3}\)
Vậy \(x=\left\{-6,\frac{4}{3}\right\}\)