a, \(3x^2-3x+2x^3-2x^2=0\)
\(\Rightarrow3x.\left(x-1\right)+2x^2.\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right).\left(3x+2x^2\right)=0\)
\(\Rightarrow\left(x-1\right).x.\left(3+2x\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\x=0\\3+2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=0\\x=\dfrac{-3}{2}\end{matrix}\right.\)
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Câu b tương tự!!
a, \(3x^2-3x+2x^3-2x^2=0\)
\(\Leftrightarrow3x\left(x-1\right)+2x^2\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x+2x^2\right)\left(x-1\right)=0\)
\(\Leftrightarrow x\left(3+2x\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3+2x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-2}{3}\\x=1\end{matrix}\right.\)
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