\(x^2\left(3x-2\right)+2x\left(3x-2\right)+3\left(3x-2\right)=0\)
\(\left(3x-2\right)\left(x^2+2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\\left(x+1\right)^2+2=0\left(VL\right)\end{matrix}\right.\)
Vậy ..................
\(3x^3+4x^2+5x-6=0\\ \Leftrightarrow\left(3x^3-2x^2\right)+\left(6x^2-4x\right)+\left(9x-6\right)\\ \Leftrightarrow x^2\left(3x-2\right)+2x\left(3x-2\right)+3\left(3x-2\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x^2+2x+3\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x^2+2x+1+2\right)=0\\ \Leftrightarrow\left(3x-2\right)\left[\left(x+1\right)^2+2\right]=0\)
Vì \(\left(x+1\right)^2+2>0\forall x\\ \Rightarrow3x-2=0\)
\(\Rightarrow x=\frac{2}{3}\)
Vậy tập nghiệm của phương trình là S=\(\left\{\frac{2}{3}\right\}\)
\(3x^3+4x^2+5x-6=0\\ \Leftrightarrow3x^3-2x^2+6x^2-4x+x-6=0\\ \Leftrightarrow x^2\left(3x-2\right)+2x\left(3x-2\right)+3\left(3x-2\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x^2+2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-2=0\\\left(x^2+2x+3\right)=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\\left(x+1\right)^2+2=0\left(voly\right)\end{matrix}\right.\)
Vậy \(S=\left\{\frac{2}{3}\right\}\)
