toan

I

iceghost

$5x^2+5y^2+8xy-2x+2y+2=0 \\
\iff (4x^2+8xy+4y^2) + (x^2-2x+1) + (y^2+2y+1) = 0 \\
\iff 4(x+y)^2 + (x-1)^2 + (y+1)^2 = 0$
Mà $4(x+y)^2 \ge 0 \\
(x-1)^2 \ge 0 \\
(y+1)^2 \ge 0$
$\implies 4(x+y)^2 + (x-1)^2 + (y+1)^2 \ge 0$
Dấu '=' xảy ra khi $\left\{ \begin{array}{l} {} x+y=0 \\ x-1=0 \\ y+1=0 \end{array} \right.$
Lại có :$M = (x+y)^{2015}+(x-2)^{2016}+(y-1)^{2017}$
$= 0^{2015} + (-1)^{2016} + 0^{2017} \\
= 1$
 
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