Question

Tìm MIN

P=(x-2y)^2+(y-2015)^2016

Answer 1

Answer 2

Với \(\forall x;y\) ta có :

\(\left\{{}\begin{matrix}\left(x-2y\right)^2\ge0\\\left(y-2015\right)^{2016}\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left(x-2y\right)^2+\left(y-2015\right)^{2016}\ge0\)

Dấu "=" xảy ra khi vào chỉ khi :

\(\left\{{}\begin{matrix}\left(x-2y\right)^2=0\\\left(y-2015\right)^{2016}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-2y=0\\y-2015=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2015}{2}\\y=2015\end{matrix}\right.\)

Vậy \(P_{Min}=0\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2015}{2}\\y=2015\end{matrix}\right.\)