\(A=3\left(2x-3\right)\left(3x+2\right)-\left(2x+4\right)\left(4x-3\right)+9x\left(4-x\right)\)
\(=\left(6x-9\right)\left(3x+2\right)-8x^2+6x-16x+12+36x-9x^2\)
\(=18x^2+12x-27x-18-17x^2+26x+12\)
\(=x^2+11x-6\)
Để A = 0
\(\Leftrightarrow x^2+11x-6=0\)
\(\Leftrightarrow\left(x^2+11x+\frac{121}{4}\right)-\frac{145}{4}=0\)
\(\Leftrightarrow\left(x+\frac{11}{2}\right)^2-\left(\frac{\sqrt{145}}{2}\right)^2=0\)
\(\Leftrightarrow\left(x+\frac{11}{2}-\frac{\sqrt{145}}{2}\right)\left(x+\frac{11}{2}+\frac{\sqrt{145}}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{145}-11}{2}\\x=\frac{-\sqrt{145}-11}{2}\end{matrix}\right.\)
Vậy..................
