\(x^3-6x^2-19x+84=0\)
\(\Leftrightarrow\left(x^3-3x^2\right)-\left(3x^2-9x\right)-\left(28x-84\right)=0\)
\(\Leftrightarrow x^2\left(x-3\right)-3x\left(x-3\right)-28\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-3x-28\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x^2-3x-28=0\end{cases}}\)
Ta có : \(x^2-3x-28=0\)
\(\Leftrightarrow\left(x^2-7x\right)+\left(4x-28\right)=0\)
\(\Leftrightarrow x\left(x-7\right)+4\left(x-7\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=7\end{cases}}\)
Vậy phương trình có tập nghiệm \(S=\left\{3;-4;7\right\}\)
