Tìm GTLN K=4xy-10x+22y-x^2-5y^2-3

1 bình luận về “Tìm GTLN K=4xy-10x+22y-x^2-5y^2-3”

1. $Khang$ 
   $K=4xy-10x+22y-x^2-5y^2-3$
   $K=-x^2+4xy-4y^2-10x+20y-25-y^2+2y-1+23$
   $K=-(x-2y{)}^2-10(x-2y)-25-(y-1{)}^2+23$
   $K=-[(x-2y{)}^2+10(x-2y)+25]-(y-1{)}^2+23$
   $K=-(x-2y+5{)}^2-(y-1{)}^2+23$
   $Vì$ $-(x-2y+5{)}^2$ $\le$ $0$ $\mathrm{\forall }$ $x,y$
   $-(y-1{)}^2$ $\le$ $0$ $\mathrm{\forall }$ $y$
   $Nên$ $-(x-2y+5{)}^2-(y-1{)}^2$ $\le$ $0$ $\mathrm{\forall }$ $x,y$
   $Do$ $đó$ $-(x-2y+5{)}^2-(y-1{)}^2+23$ $\le$ $0+23$
   $Hay$ $K$ $\le$ $23$
   $\text{Dấu “=” xảy ra }\iff x-2y+5=0vày-1=0$
   $x=-3$ $và$ $y=1$
   $\text{Vậy }GTLN\text{ của }K\text{ là }23\text{ khi }x=-3vày=1$

   Trả lời