\(a^3-2a+4=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+2\right)\)(chia hoocner)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x^2-2x+2=0\left(VN\right)\end{matrix}\right.\)
\(\Leftrightarrow x=-2\)
Cách khác: Bằng cách thêm bớt hạng tử.
(Truong Viet Truong) cho a mà tìm x
\(a^3-2a+4=0\\ \Leftrightarrow a^2+2a^2-2a^2-4a+2a+4=0\\ \Leftrightarrow a^2\left(a+2\right)-2a\left(a+2\right)+2\left(a+2\right)=0\\ \Leftrightarrow\left(a+2\right)\left(a^2-2a+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a+2=0\\a^2-2a+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=-2\\a^2-2a+2=0\left(VN\right)\end{matrix}\right.\)
Vậy $a=-2$
