\(1,\\ a,=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\\ b,=3\left(x^2+3x-10\right)=3\left(x^2-2x+5x-10\right)=3\left(x-2\right)\left(x+5\right)\\ c,=x^2-x-2x+2=\left(x-2\right)\left(x-1\right)\\ d,=x^2-3x-6x+18=\left(x-3\right)\left(x-6\right)\\ e,=x^2-2x-4x+8=\left(x-2\right)\left(x-4\right)\\ f,=x^2+2x-7x-14=\left(x+2\right)\left(x-7\right)\\ 2,\\ a,\Rightarrow3x^2+2x=0\Rightarrow x\left(3x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{3}\end{matrix}\right.\\ b,\Rightarrow\left(5x-0,8\right)\left(5x+0,8\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{0,8}{5}=\dfrac{4}{25}\\x=-\dfrac{0,8}{5}=-\dfrac{4}{25}\end{matrix}\right.\)
\(c,\Rightarrow x^2\left(x-4\right)\left(x+4\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\\ d,\Rightarrow x^2+x-6=0\Rightarrow x^2+2x-3x-6=0\\ \Rightarrow\left(x+2\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ e,\Rightarrow x^2-7x+12=0\\ \Rightarrow x^2-3x-4x+12=0\\ \Rightarrow\left(x-3\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\\ f,\Rightarrow x^3-x^2+x=0\\ \Rightarrow x\left(x^2-x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\end{matrix}\right.\\ \Rightarrow x=0\)
