\(A=\left(x-\frac{2}{5}\right)^2+\left(y+20\right)^{10}+2010\)
+ Ta có:
\(\left\{{}\begin{matrix}\left(x-\frac{2}{5}\right)^2\ge0\\\left(y+20\right)^{10}\ge0\end{matrix}\right.\forall x,y.\)
\(\Rightarrow\left(x-\frac{2}{5}\right)^2+\left(y+20\right)^{10}\ge0\) \(\forall x,y.\)
\(\Rightarrow\left(x-\frac{2}{5}\right)^2+\left(y+20\right)^{10}+2010\ge2010\) \(\forall x,y.\)
\(\Rightarrow A\ge2010.\)
Dấu '' = '' xảy ra khi:
\(\left\{{}\begin{matrix}\left(x-\frac{2}{5}\right)^2=0\\\left(y+20\right)^{10}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-\frac{2}{5}=0\\y+20=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0+\frac{2}{5}\\y=0-20\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{2}{5}\\y=-20\end{matrix}\right.\)
Vậy \(MIN_A=2010\) khi \(x=\frac{2}{5}\) và \(y=-20.\)
Chúc bạn học tốt!
