Ta có: \(4\left(x+3\right)\left(3x-2\right)-3\left(x-1\right)\left(4x-1\right)=-27\)
\(\Leftrightarrow4\left(3x^2+7x-6\right)-3\left(4x^2-5x+1\right)=-27\)
\(\Leftrightarrow12x^2+28x-24-12x^2+15x-3=-27\)
\(\Leftrightarrow43x=-27+24+3=0\)
hay x=0
a)\(4\left(x+3\right)\left(3x-2\right)-3\left(x-1\right)\left(4x-1\right)=-27\)
\(\Leftrightarrow12x^2+28x-24-12x^2+15x-3=-27\)
\(\Leftrightarrow43x=0\\ \Leftrightarrow x=0\)