tìm GTNN của bt...khó khó

N

nguyenbahiep1

[laTEX]A =x^2 + x(y -3) + y^2- 3y + 2006 \\ \\ A= (x +\frac{y-3}{2})^2+y^2- 3y + 2006 - \frac{y^2-6y+9}{4} \\ \\ A= (x +\frac{y-3}{2})^2 +\frac{3y^2-6y+8015}{4} \\ \\ A = (x +\frac{y-3}{2})^2 +\frac{3(y-1)^2}{4}+ 2003 \geq 2003 \\ \\ \Rightarrow GTNN_ A = 2003 \\ \\ x = y = 1 [/laTEX]
 
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P

popstar1102


$x^2+y^2+xy-3x-3y+2006$
\Leftrightarrow$x^2-2x+1+y^2-2y+1+xy-x-y+2004$
\Leftrightarrow$(x-1)^2+(y-1)^2-(y-1)(x-1)+2003$
\Leftrightarrow$((x-1)+\frac{y-1}{2})^2$+$\frac{3}{4}(y-1)^2$+2003\geq 2003

vậy GTNN là 2003\Leftrightarrowx=y=1
 
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