\(x^3-5x^2+6x\)
\(=x\left(x^2-5x+6\right)\)
\(=x\left(x^2+x-6x+6\right)\)
\(=x\left[x\left(x+1\right)-6\left(x+1\right)\right]\)
\(=x\left(x+1\right)\left(x-6\right)\)
\(x^3-5x^2+6x=0\)
\(\Rightarrow x^3-2x^2-3x^3+6x\) = 0
\(\Rightarrow x^2\left(x-2\right)-3x\left(x-2\right)=0\Rightarrow x\left(x-3\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)
x3 - 5x2 + 6x = 0
⇔ x (x2 - 5x + 6) = 0
⇔ x(x - 3)(x - 2)
⇔ x ϵ {0; 2; 3}
Vậy tập nghiệm của pt là : S = {0; 2; 3}
Ta có: \(x^3-5x^2+6x=0\)
\(\Leftrightarrow\left(x^3-3x^2\right)-\left(2x^2-6x\right)=0\)
\(\Leftrightarrow x^2\left(x-3\right)-2x\left(x-3\right)=0\)
\(\Leftrightarrow\left(x^2-2x\right)\left(x-3\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x-2=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{0;2;3\right\}\)
