(4x+2)($x^{2}$ +1)=0
⇔\(\left[ \begin{array}{l}4x+2=0\\x^2+1=0\end{array} \right.\) \(\left[ \begin{array}{l}x=-1/2\\x^2=-1(loại)\end{array} \right.\)
⇒x=-1/2
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$x^{3}$ -3$x^{2}$ +3x -1 =0
(x-1)($x^{2}$ +x+1)-3x(x-1) =0
(x-1)($x^{2}$ +x+1-3x)=0
\(\left[ \begin{array}{l}x-1=0\\x^2-2x+1=0\end{array} \right.\) \(\left[ \begin{array}{l}x=1\\(x-1)^2=0\end{array} \right.\) \(\left[ \begin{array}{l}x=1\\x=1\end{array} \right.\)
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