giải bất phương trình

H

hoangtubongdem5

[TEX]y = min{x,y,z} y \leq 1[/TEX]

[TEX]0 \leq z = 3 - x - y \leq 2 \Rightarrow 1 \leq x + y \leq 3 \Rightarrow 1 - y \leq x \leq 3 - y[/TEX]

[TEX]P = (x+y+z)^2 - 3(xy+yz+zx) = 9 - 3(xy+yz+ zx) - (xy+yz+zx) = -x(y+z) -y(3-x-y) = (x^2 + y^2 ) - 3(x+y) + xy = x^2 +(y-3)x + y^2 - 3y = f(x)[/TEX]

[TEX]1 - y \leq x \leq 2 \Rightarrow f(x) \leq max{f(1-y), f(2)} = -y^2 +y + 2 = y(1-y) + 2\geq 2[/TEX]

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

[TEX]0 \leq P \leq 3; P(1,1,1) = 0, P(2,1,0) = 3[/TEX]

Min P = 0; Max P = 3

 
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T

tahoangthaovy

Đặt y=min{x,y,z}, y \leq1

0 \leq z=3−x−y \leq 2\Rightarrow1\leq x+y \leq 3\Rightarrow 1−y \leq x \leq 3−y

P=$(x+y+z)^2-3(xy+yz+zx)$
=$9-3(xy+yz+zx) -(xy+yz+zx)$
=$-x(y+z)-y(3-x-y)$
=$(x^2+y^2)-3(x+y)+xy$
=$x^2+(y-3)x+y^2-3y$
=$f(x)$

1−y \leq x \leq 2
\Rightarrow f(x)\leq max{f(1−y),f(2)}=$-y^2+y+2=y(1−y)+2$ \geq 2
f(x)=$-(x+\frac{y-3}{2})^2 - \frac{3(y-1)^2}{4}+3$ \leq 3

0\leqP\leq3; P(1,1,1)=0, P(2,1,0)=3

minP=0;maxP=3
 
H

huynhbachkhoa23

Làm theo cách của cấp 3:

Thay $z=3-x-y$

$\dfrac{1}{3}P=x^2+(y-3)x+y^2-3y+3=f$

$\dfrac{d f}{d x}=2x+(y-3)=0\leftrightarrow x=\dfrac{3-y}{2}$

Lại có $\dfrac{d^2 f}{dx^2}>0$

Thay $x=\dfrac{3-y}{2}$, $P \ge 0$

Thay $x=0 \rightarrow y\in [1;2]: P=3(y^2-3y+3) \le 3$

Thay $x=2 \rightarrow y\in [0;1]: P=3(y^2-y+1) \le 3$

Kết luận:

$\text{minP}=0 \leftrightarrow x=y=z=1$

$\text{maxP}=3 \leftrightarrow (x;y;z)=(0;1;2), (1;2;0), (2;0;1)$
 
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