$x^2+5y^2-2xy+2x+2y=-2$
$⇔x^2-2xy+y^2+2x-2y+4y^2+4y+2=0$
$⇔(x-y)^2+2(x-2)+1+(4y^2+4y+1)=0$
$⇔(x-y+1)^2+(2y+1)^2=0$
Vì $(x-y+1)^2 \geq 0 ∀ x ; y$
$(2y+1)^2 \geq 0 ∀ y$
$⇒(x-y+1)^2+(2y+1)^2 \geq 0 ∀ x ; y$
Dấu $"="$ xảy ra khi và chỉ khi :
$\left\{ \begin{matrix}x-y+1=0\\2y+1=0\end{matrix} \right. ⇔ \left\{ \begin{matrix}x+1=y\\y=\dfrac{-1}{2}\end{matrix} \right. ⇔ \left\{ \begin{matrix}x=\dfrac{-3}{2}\\y=\dfrac{-1}{2}\end{matrix} \right.$
Ta có :
$M=\left [ 2.\left (\dfrac{-3}{2} \right )+2.\left (\dfrac{-1}{2} \right )+5 \right ]^{2020}+99$
$M=[(-3)+(-1)+5]^{2020}+99$
$M=1^{2020}+99$
$M=100$
Vậy $M=100$ với $x ; y$ thảo mãn $x^2+5y^2-2xy+2x+2y=-2$
