Ta có: \(2x^3-5x^2+3x=0\)
\(\Leftrightarrow x\left(2x^2-2x-3x+3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x-1\right)-3\left(x-1\right)\right]=0\)
\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;1;\frac{3}{2}\right\}\)
Cách khác:
\(2x^3-5x^2+3x=0\)
\(\Leftrightarrow2x^3-2x^2-3x^2+3x=0\)
\(\Leftrightarrow\left(2x^3-2x^2\right)-\left(3x^2-3x\right)=0\)
\(\Leftrightarrow2x^2.\left(x-1\right)-3x.\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right).\left(2x^2-3x\right)=0\)
\(\Leftrightarrow\left(x-1\right).x.\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{1;0;\frac{3}{2}\right\}.\)
Chúc bạn học tốt!
