Ta có: \(x^3-5x^2+6x=0\)
\(\Leftrightarrow x\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)
Vậy: S={0;2;3}
\(x^3-5x^2+6x=0\)
\(\Leftrightarrow x^3-2x^2-3x^2+6x=0\)
\(\Leftrightarrow x^2\left(x-2\right)-3x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^2-3x\right)\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)
\(S=\left\{0,2,3\right\}\)
x^3 -5x^2 + 6x=0
\(< =>x\left(x^2-5x+6\right)=0\)
\(< =>x=0;x^2-5x+6=0\)
+/ \(x^2-5x+6=0\)
\(< =>x^2-2x-3x+6=0\)
\(< =>\left(x^2-2x\right)-\left(3x-6\right)=0\)
\(< =>x\left(x-2\right)-3\left(x-2\right)=0\)
\(< =>\left(x-2\right)\left(x-3\right)=0\)
=> \(x-2=0;x-3=0\)
=> \(x=2;3\)
Vayayj tập nghiệm của......
