Bài toán GTNN , GTLN

E

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Đặt $A=x+y$

GT \Leftrightarrow $(x+y)^2+6(x+y)+8=-y^2 \le 0$

\Leftrightarrow $A^2+6A+8 \le 0$

\Leftrightarrow $(A+3)^2 \le 1$

\Leftrightarrow $-1 \le A+3 \le 1$

\Leftrightarrow $-4 \le A \le -2$

\Rightarrow $2005 \le A+2009=x+y+2009 \le 2007$
 
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