[TEX]M=x^2+2y^2+2xy-2x-6y+2015[/TEX]
[TEX]= x^2 + y^2 +1 +2xy-2x-2y +y^2-3y+2014[/TEX][TEX]=(x+y-1)^2+(y^2-2.y.\frac{3}{2}+\frac{9}{4})+\frac{8047}{4}[/TEX]
[TEX]=(x+y-1)^2+(y-\frac{3}{2})^2+\frac{8047}{4}[/TEX]
Ta có :
[TEX](x+y-1)^2 \ \geq \ 0 \ \forall \ x;y[/TEX]
[TEX](y-\frac{3}{2})^2 \ \geq \ 0 \ x;y[/TEX]
[TEX]\Rightarrow (x+y-1)^2+(y-\frac{3}{2})^2 \ \geq \ 0 \ \forall \ x;y [/TEX]
[TEX]\Rightarrow (x+y-1)^2+(y-\frac{3}{2})^2+\frac{8047}{4} \ \geq \ \frac{8047}{4} \ \forall \ x;y[/TEX]
Dấu " = " xảy ra khi và chỉ khi :
[TEX]\left{\begin{x+y-1=0}\\{y-\frac{3}{2}=0} [/TEX]
[TEX]\Rightarrow \left{\begin{x= \ - \ \frac{1}{2}}\\{y=\frac{3}{2}}[/TEX]
Vậy : [TEX]Min \ M = \frac{8047}{4} \ \Leftrightarrow \ x = - \ \frac{1}{2} \ ; \ y = \frac{3}{2}[/TEX]
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