Ta có: \(\left(x+3\right)\left(x-1\right)-\left(4x+3\right)^2=9\)
=>\(x^2-x+3x-3-\left(16x^2+24x+9\right)=9\)
=>\(x^2+2x-3-16x^2-24x-9-9=0\)
=>\(-15x^2-22x-21=0\)
=>\(15x^2+22x+21=0\)
\(\Delta=22^2-4\cdot15\cdot21=-776<0\)
=>Phương trình vô nghiệm
=>x∈∅


