Toán 8

H

huuthuyenrop2

3) tìm max
$B = -4-x^2+6x = -(x^2-6x+4)= -(x^2-6x+9-5)= -(x-3)^2+5 \leq 5$
VẬy MAX B= 5 khi x=3
 
Last edited by a moderator:
V

vipboycodon

bài 3:
B=[TEX] -4-x^2+6x[/TEX]
=[TEX] -(x^2-6x+4)[/TEX]
=[TEX] -(x^2-6x+9-5)[/TEX]
=[TEX] -[(x-3)^2-5][/TEX]
=[TEX] -(x-3)^2+5 \leq 5[/TEX]
\Rightarrow B \leq 5
vậy max = 5 khi x=3

C=|x-3|(2-|x-3|)
=(x-3)(2-x+3)
=(x-3)(5-x)
=[TEX]5x-x^2-15+3x[/TEX]
=[TEX] -(x^2-8x+15)[/TEX]
=[TEX] -(x^2-8x+16-1)[/TEX]
=[TEX] -[(x-4)^2-1][/TEX]
=[TEX] -(x-4)^2+1 \leq 1[/TEX]
\Rightarrow C \leq 1
Vậy max C=1 khi x=4
 
Last edited by a moderator:
C

congchuaanhsang

1, Ta có: $10x^2-10y^2-z^2$=0\Leftrightarrow$4z^2=40x^2-40y^2$
VT=$(7x-3y)^2-4z^2$=$(7x-3y)^2-40x^2+40y^2$
Khai triển rồi thu gọn ta được
VT=$9x^2-2.3.7xy+49y^2$=$(3x-7y)^2$
 
C

congchuaanhsang

2, A=$(x^2+4x-5)(x^2+4x+5)$=$(x^2+4x)^2-25$\geq-25
Vậy $A_{min}$=-25\Leftrightarrowx=0 hoặc x=-4
 
You must log in or register to reply here.
Top Bottom