câu 2 đã qus dễ mk giải câu 1 thôi
\(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(27x^3+18x^2+12x-18x^2-12x-8-3x\left(9x^2-3x+1\right)+\left(9x^2-3x+1\right)=x-4\)
\(27x^3-8-3x\left(9x^2-3x+1\right)+9x^2-3x+1=x-4\)
\(-7+18x^2-6x=x-4\)
\(7+18x^2+6x+x-4=0\)
\(3-18x^2+7x=0\)
\(x=\frac{-7+\sqrt{265}}{36};-\frac{-7-\sqrt{265}}{36}\)