[toan 8] tim min ne ae...

P

pandahieu

Ta có $4A=4x^2+4xy+4y^2-12x-12y+8024=(2x)^2+4x(y-3)+(y-3)^2+3(y-1)^2+8012$

\Rightarrow $4A=(2x+y-3)^2+3(y-1)^2+8012$

\Rightarrow $4A \ge 8012$ $A=2003$ \Leftrightarrow $x=1;y=1$

Vậy $A min=2003$ \Leftrightarrow $x=y=1$
 
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V

vipboycodon

ban hinh nhu sai roi... nhung van cam on ban

[TEX]C=x^2-2x+1+y^2-2y+1+xy-x-y+2003[/TEX]

[TEX]=(x-1)^2+(y-1)^2+x(y-1)-(y-1)+2003[/TEX]

[TEX]=(x-1)^2+(y-1)^2+(x-1)(y-1)+2003[/TEX]

[TEX]=(x-1)^2+2(x-1)\frac{(y-1)}{2}+(\frac{y-1}{2})^2+\frac{3(y-1)^2}{4}+2003[/TEX]

[TEX]=[(x-1)+\frac{y-1}{2}]^2+\frac{3}{4}(y-1)^2+2003\geq2003[/TEX]
\Rightarrow C\geq2003
Vay min C=2003 khi (+)[TEX]x-1+\frac{y-1}{2}=0[/TEX]
(+)y-1=0
(+)x=1
(+)y=1
\Leftrightarrow x=y=1
 
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