Toan 8 Bai toan nang cao

P

phamhuy20011801

$1, 2A=x^2+2xy+y^2+y^2+2yz+z^2+z^2-2zx+2x^2-14 \ge (x+y)^2+(y+z)^2+(z-x)^2-14 \ge -14 \leftrightarrow A \ge -7 \leftrightarrow x=y=z=0$
$3, $ PT $\leftrightarrow (x-1)^2+4(y+1)^2+(z-3)^2+1=0$, điều này vô lí.
 
Last edited by a moderator:
V

vanmanh2001

Bài 2
$A = x-y+z-x^2-y^2-z^2-7$
$-A = x^2 + y^2 + z^2 - 2x + 2y - 2z + 7$
$- A = (x-1)^2 + (y+1)^2 + (z-1)^2 + 4 \geq 4 $
$\Rightarrow A \leq 4 $
$Max A = 4 \Leftrightarrow x = z = 1 , y=-1$
 
T

transformers123

vanmanh2001 said:
Bài 2
$A = x-y+z-x^2-y^2-z^2-7$
$-A = x^2 + y^2 + z^2 - 2x + 2y - 2z + 7$
$- A = (x-1)^2 + (y+1)^2 + (z-1)^2 + 4 \ge 4 $
\Rightarrow $A \le 4 $
$Max A = 4$ \Leftrightarrow $x = z = 1 , y=-1$
Bấm để xem đầy đủ nội dung ...

Sai rồi nhé :D

$A = x-y+z-x^2-y^2-z^2-7$

$\iff A=-(x^2-x+\dfrac{1}{4}-(y^2+y+\dfrac{1}{4})-(z^2-z+\dfrac{1}{4})-\dfrac{25}{4}$

$\iff A=-(x-\dfrac{1}{2})^2-(y+\dfrac{1}{2})^2-(z-\dfrac{1}{2})^2-\dfrac{25}{4}$

$\iff A \le \dfrac{-25}{4}$

Dấu "=" xảy ra khi $(x;y;z)=(\dfrac{1}{2};\dfrac{-1}{2};\dfrac{1}{2})$
 
You must log in or register to reply here.
Top Bottom