\(x^3+4x^2+x+6=0\)
\(\Leftrightarrow\text{ (x + 3).(x + 2).(x - 1) = 0 }\)
<=>
Tự làm nhé mk nhẩm ko nhầm là dậy :D
\(x^3+4x^2+x-6=x^3-x^2+5x^2-5x+6x-6=x^2\left(x-1\right)+5x\left(x-1\right)+6\left(x-1\right)=\left(x-1\right)\left(x^2+5x+6\right)=\left(x-1\right)\left(x^2+2x+3x+6\right)=\left(x-1\right)\left(x+2\right)\left(x+3\right)\)
chúc bạn học tốt
\(x^3+4x^2+x-6=0\)
\(\Leftrightarrow x^3-x^2+5x^2-5x+6x-6=0\)
\(\Leftrightarrow x^2\left(x-1\right)+5x\left(x-1\right)+6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2+5x+6\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2+2x+3x+6\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[x\left(x+2\right)+3\left(x+2\right)\right]\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\\x=1\end{matrix}\right.\)
Vậy \(S=\left\{1;-2;-3\right\}\)
