\(x^3-6x^2-x+30=0\)
\(\Leftrightarrow x^3-5x^2-x^2+5x-6x+30=0\)
\(\Leftrightarrow x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)=0\)
\(\Leftrightarrow\left(x^2-x-6\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left(x^2-3x+2x-6\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[x\left(x-3\right)+2\left(x-3\right)\right]\left(x-5\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\x=5\end{matrix}\right.\)
Vậy...
\(x^3-6x^2-x+30=0\)
\(x^3+2x^2-8x^2-16x+15x+30=0\)
\(\left(x^2+2x^2\right)-\left(8x^2+16x\right)+\left(15x+30\right)=0\)
\(x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)=0\)
\(\left(x+2\right)\left(x^2-8x+15\right)=0\)
TH1: \(x+2=0\Leftrightarrow x=-2\) (1)
TH2: \(x^2-8x+15=0\)
\(x^2-8x=-15\)
\(x^2-2x.4+16=-15+16\)
\(\left(x-4\right)^2=1\)
\(\Rightarrow\left[{}\begin{matrix}x-4=1\\x-4=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\left(2\right)\)
Từ (1) và (2) \(\Rightarrow x\in\left\{-2;5;3\right\}\)
