[toán 8] cực trị (khẩn cấp)

H

hellangel98

A=(x^2+y^2+xy-3x-3y+3)-3

=[(2x+y-3)^2 + 3(y-1)^2]/4 -3\geq-3
minA=-3\Leftrightarrowx=y=1
 
H

hellangel98

x^2+5y^2+2x-4xy-10y+14
=[x^2+2x(1-2y)+(1-2y)^2]+y^2-6y+13
=(x+1-2y)^2+(y^2-2y.3+9)+4
=(x+1-2y)^2+(y-3)^2+4.
đến đây tự tìm x,y(ngại vít:D)
 
H

hellangel98

x^2-4xy+5y^2+10x-22y+28
=x^2-4xy+4y^2+25+10x-20y+y^2-2y+1+2
=(x-2y+5)^2+(y-1)^2+2>=2
=>min =2<=>........................................
 
H

hellangel98

phần D tớ đoán minD=2008 nhưng dễ sai lắm:D
................................................................................
 
L

luffy_1998

[TEX]D = (x-5)(x-2) . (x-1)(x-6) + 2012[/TEX]
[TEX] = (x^2 - 7x + 10)(x^2 - 7x + 6) + 2012[/TEX]
[TEX]y = x^2 - 7x + 8 \Rightarrow D = (y - 2)(y+2) + 2012 = y^2 - 4 + 2012 = y^2 + 2008 \geq 2008. [/TEX]
[TEX]D_{min} = 2008 \Leftrightarrow y = 0 \Leftrightarrow x^2 - 7x + 8 = 0 \Leftrightarrow (x - 3.5)^2 - \frac{17}{4} = 0 \Leftrightarrow (x - 3.5 - \frac{sqrt{17}}{2})(x - 3.5 + \frac{sqrt{17}}{2}) \Leftrightarrow \left[\begin{x = 3.5 + \frac{sqrt{17}}{2}}\\{x = 3.5 - {\frac{sqrt{17}}{2}} [/TEX]
 
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