Ta có
5x^2+5y^2+8xy−2x+2y+2=0⇔4x^2+8xy+4y^2+x^2−2x+1+y^2+2y+1=0

<=>4(x+y)^2+(x−1)^2+(y+1)^2=0

$$\begin{gathered} \mathrm{mà} \left\{\begin{array}{l}4{(\mathrm{x}+\mathrm{y})}^2\ge 0\\ {\left(\mathrm{y}+1\right)}^2\ge 0\Rightarrow 4{(\mathrm{x}+\mathrm{y})}^2+{\left(\mathrm{y}+1\right)}^2+{(\mathrm{x}-1)}^2\ge 0\\ {(\mathrm{x}-1)}^2\ge 0\end{array}\right. \\ \mathrm{Dấu} "=" \mathrm{xảy} \mathrm{ra} \mathrm{khi} \left\{\begin{array}{l}\mathrm{x}=1\\ \mathrm{y}=-1\end{array}\right. \\ \mathrm{Thay} \mathrm{vào} \mathrm{M} \mathrm{ta} \mathrm{được} \\ \mathrm{M}={(\mathrm{x}+\mathrm{y})}^{2015}+{(\mathrm{x}-2)}^{2016}+{(\mathrm{y}+1)}^{2017}={(1-1)}^{2015}+{(1-2)}^{2016}+{(-1+1)}^{2017}=1 \end{gathered}$$