\(x^3+2x^2-3=0\)
\(\Leftrightarrow x^3-x^2+3x^2-3=0\)
\(\Leftrightarrow x^2\left(x-1\right)+3\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2+3\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+3x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2+3x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2+3x+\dfrac{9}{4}+\dfrac{3}{4}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+\dfrac{3}{2}\right)^2=-\dfrac{3}{4}\left(VL\right)\end{matrix}\right.\)
Vậy \(x=1\)
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