A= 2x2 + 2xy + y2 - 2x + 2y + 2
C= x2 + xy + y2 - 3x - 3y
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#2
Posted 03-02-2016 - 13:03
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$A=(x^2+y^2+1+2xy+2x+2y)+(x^2-4x+4)-3=(x+y+1)^2+(x-2)^2-3\geq -3$
Min A=-3 khi x=-y-1=2
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#3
Posted 03-02-2016 - 13:11
tpdtthltvp
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Trung úy
C= x2 + xy + y2 - 3x - 3y
$2C=2x^2+2xy+2y^2-6x-6y=(x^2+y^2+4-4x-4y+2xy)+(x^2-2x+1)+(y^2-2y+1)-2=(x+y-2)^2+(x-1)^2+(y-1)^2-6\geq -6\Rightarrow C\geq -3$
Dấu "=" xảy ra khi $x=y=1$
Edited by tpdtthltvp, 03-02-2016 - 16:58.
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#4
Posted 03-02-2016 - 13:13
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$C=\left(x^2+\frac{y^2}{4}+\frac{9}{4}+xy-3x-\frac{3}{2}y\right)+\frac{3}{4}(y^2-2y+1)-3=\left(x+\frac{y}{2}-\frac{3}{2}\right)^2+\frac{3}{4}(y-1)^2-3\geq -3$
Min C=-3 khi $x=y=1$
Even when you had two eyes, you'd see only half the picture.
#5
Posted 03-02-2016 - 13:13
khanhto012
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$2C=2x^2+2xy+2y^2-6x-6y=(x^2+y^2+4-4x-4y+2xy)+(x^2-2x+1)+(y^2-2y+1)-2=(x+y-2)^2+(x-1)^2+(y-1)^2-2\geq -2\Rightarrow C\geq -1$
Dấu "=" xảy ra khi $x=y=1$
bạn ơibài nè sai rùi
#6
Posted 03-02-2016 - 13:15
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Trung sĩ
$2C=2x^2+2xy+2y^2-6x-6y=(x^2+y^2+4-4x-4y+2xy)+(x^2-2x+1)+(y^2-2y+1)-2=(x+y-2)^2+(x-1)^2+(y-1)^2-2\geq -2\Rightarrow C\geq -1$
Dấu "=" xảy ra khi $x=y=1$
là $-6$
Min C=-3
Even when you had two eyes, you'd see only half the picture.
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