\(A=x^2-x+\frac{12}{x}+2016\)
\(=\left(x^2-x+\frac{12}{x}-8\right)+2024\)
\(=\left(\frac{x^3}{x}-\frac{x^2}{x}+\frac{12}{x}-\frac{8x}{x}\right)+2024\)
\(=\left(\frac{x^3-x^2+12-8x}{x}\right)+2024\)
\(=\frac{\left(x-2\right)^2\left(x+3\right)}{x}+2024\ge2024\)
Dấu "=" xảy ra khi \(\left[\begin{matrix}\left(x-2\right)^2=0\\x+3=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy \(Min_A=2024\) khi \(\left[\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
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