TH1: x<-3/2
=>BPT vô nghiệm
TH2: x>=-3/2
\(\Leftrightarrow\left(x-1\right)^2< \left(2x+3\right)^2\)
\(\Leftrightarrow\left(2x+3\right)^2-\left(x-1\right)^2>0\)
\(\Leftrightarrow\left(2x+3-x+1\right)\left(2x+3+x-1\right)>0\)
=>(x+4)(3x+2)>0
=>x>-2/3 hoặc x<-4