II3x-3I+2x+1I=3x+2021^0
II3x-3I+2x+1I=3x+1
\(\)ĐK:3x+1\(\ge\)0
3x\(\ge\)-1
x\(\ge\frac{-1}{3}\)
\(\Rightarrow\)I3x-3I+2x+1=3x+1
I3x-3I=x
\(\Rightarrow\)3x-3=\(\pm\)x
TH1:3x-3=x TH2:3x-3=-x
2x=3 4x=3
x=\(\frac{3}{2}\) x=\(\frac{3}{4}\)
Vậy x=\(\frac{3}{2}\); x=\(\frac{3}{4}\)