\(6\left(3x^2-2x\right)-6\left(3x^2-3\right)=18\\ 6\left(3x^2-2x-3x^2+3\right)=18\\ -2x+3=3\\ -2x=0\\ x=0\)
Vậy x = 0
\(6.\left(3x^2-2x\right)-6.\left(3x^2-3\right)=18\)
\(\Rightarrow6.\left[3x^2-2x-\left(3x^2-3\right)\right]=18\)
\(\Rightarrow6.\left(3x^2-2x-3x^2+3\right)=18\)
\(\Rightarrow6.\left(-2x+3\right)=18\)
\(\Rightarrow-12x+18=18\)
\(\Rightarrow-12x=18-18\)
\(\Rightarrow-12x=0\)
\(\Rightarrow x=0:\left(-12\right)\)
\(\Rightarrow x=0\)
Vậy \(x=0.\)
Chúc bạn học tốt!