\(2x^3+5x^2-3x=0\)
\(\Leftrightarrow x.\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x.\left(2x^2+6x-x-3\right)=0\)
\(\Leftrightarrow x.\left[2x.\left(x+3\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow x.\left(x+3\right).\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy : \(S=\left\{-3;0;\frac{1}{2}\right\}\)
2x3 + 5x2 - 3x = 0
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x^2+5x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{2}\\x=-3\end{matrix}\right.\)
\(2x^{3}+5x^{2}-3x=0\)
⇔ \(2x^{3}-x^{2}+6x^{2}-3x=0\)
⇔ \(x^{2}.(2x-1)+3x.(2x-1)=0\)
⇔ \((2x-1).(x^2+3x)=0\)
⇔ \((2x-1).x.(x+3)=0\)
\(TH1: 2x-1=0\iff 2x=1\iff x=\dfrac{1}{2} \)
\(TH2:x=0 \)
\(TH3:x+3=0\iff x=-3\)
