timf maxx -X^2 – 2y^2 + 2xy – 2x + 6y + 5

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1. $-x^2-2y^2+2xy-2x+6y+5$

   =$-(x^2+2y^2-2xy+2x-6y-5)$

   =$-[x^2-2x(y-1)+y^2-2y+1+y^2-4x+4-10]$

   =$-[(x-y+1)^2+(y-2)^2-10]$

   Ta có:

   $(x-y+1)^2≥0$

   $(y-2)^2≥0$

   ⇒$(x-y+1)^2+(y-2)^2≥0$

   ⇒$(x-y+1)^2+(y-2)^2-10≥-10$

   ⇒$-[(x-y+1)^2+(y-2)^2-10]≤10$

   Dấu = xảy ra⇔$x-y+1=0;y-2=0$⇔$x=1;y=2$

   Vậy $Max=10$ tại $x=1;y=2$

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2. Ta có

   $-x^2+2xy-2y^2-2x+6y+5\\
   =-x^2+2x\left(y-1\right)-2y^2+6y+5$

   $=-x^2+2x\left(y-1\right)-y^2+2y-1-y^2+4y-4+10$

   $=-\left[x^2-2x\left(y-1\right)+\left(y^2-2y+1\right)\right]-\left(y^2-4y+4\right)+10$

   $=-\left[x^2-2x\left(y-1\right)+\left(y-1\right)^2\right]-\left(y-2\right)^2+10\\
   =-\left(x-y+1\right)^2-\left(y-2\right)^2+10$

   Ta thấy 

   $-\left(x-y+1\right)^2\le 0\\
   -\left(y-2\right)^2\le 0\\
   10\ge 10>0$

   $\Rightarrow -\left(x-y+1\right)^2-\left(y-2\right)^2+10\le 10$

   Dấu bằng xảy ra $\Leftrightarrow x=1;y=2$

   Vậy GTLN=10 tại x=1;y=2

    

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