Sửa đề: \(\left(x-1\right)^2-\left(3x+2\right)\left(x-12\right)=\left(x^2+1\right)\left(x-2\right)-x^2\)
\(\Leftrightarrow x^3-3x^2+3x-1-\left(3x^2-36x+2x-24\right)=x^3-2x^2+x-2-x^2\)
=>\(x^3-3x^2+3x-1-3x^2+34x+24=x^3-3x^2+x-2\)
=>\(x^3-6x^2+37x+23-x^3+3x^2-x+2=0\)
=>\(-3x^2+36x+25=0\)
=>\(x=\dfrac{18\pm\sqrt{399}}{3}\)