\(=x^4-x^3+3x^3-3x^2+3x^2-3x+9x-9\\ =\left(x-1\right)\left(x^3+3x^2+3x+9\right)\\ =\left(x-1\right)\left(x+3\right)\left(x^2+3\right)\)
\(x^4+2x^3+6x-9=x^3\left(x-1\right)+3x^2\left(x-1\right)+3x\left(x-1\right)+9\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3+3x^2+3x+9\right)\)
\(=\left(x-1\right)\left[x^2\left(x+3\right)+3\left(x+3\right)\right]\)
\(=\left(x-1\right)\left(x+3\right)\left(x^2+3\right)\)
\(x^4+2x^3+6x-9=0\)\(\)
\(\Rightarrow x^4+3x^3-x^3+9x-3x-9+3x^2-3x^2=0\)
\(\Rightarrow\left(x^4+3x^3+3x^2+9x\right)-\left(x^3+3x^2+3x+9\right)=0\)
\(\Rightarrow x\left(x^3+3x^2+3x+9\right)-\left(x^3+3x^2+3x+9\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x^3+3x^2+3x+9\right)=0\)
\(\Rightarrow\left(x-1\right)\left[x^2\left(x+3\right)+3\left(x+3\right)\right]=0\)
\(\Rightarrow\left(x-1\right)\left(x^2+3\right)\left(x+3\right)=0\)
Mà \(x^2+3>0\) với mọi x\(\in\)R.
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
