|
|
发表于 11-10-2011 08:39 AM
|
显示全部楼层
|
|
|
|
|
|
|
|
|
|
|
发表于 13-10-2011 04:16 PM
|
显示全部楼层
我想知道如果题目是 by using the laws of set algebra, prove (A∩B)-(A∩C)=A∩(B-C）
我直接写(A∩B)-(A∩C)=A∩(B-C）（distributive law)可以的吗？ |
|
|
|
|
|
|
|
|
|
|
|
发表于 13-10-2011 06:40 PM
|
显示全部楼层
你如何distribute法？ |
|
|
|
|
|
|
|
|
|
|
|
发表于 13-10-2011 07:34 PM
|
显示全部楼层
回复 2803# cheesit92
我是想说如果题目有 （-） 的话可以直接用别的law直接做吗？ |
|
|
|
|
|
|
|
|
|
|
|
发表于 13-10-2011 08:59 PM
|
显示全部楼层
|
不能，因为law只能apply去∩跟U, 有（-）的话必须换去∩跟U才能apply law~因为law就只写到∩跟U |
|
|
|
|
|
|
|
|
|
|
|
发表于 13-10-2011 09:56 PM
|
显示全部楼层
回复 2805# cheesit92
好，谢谢。我得五分就酱飞走了
就当作一个教训好了.. |
|
|
|
|
|
|
|
|
|
|
|
发表于 24-10-2011 04:53 PM
|
显示全部楼层
A chord of the parabola y^2=8(x-1), with gradient 2 passes through point A(3,0) and meets the parabola at P(2p+1 , 4p) and Q(2q+1 , 4q).Find the coordinates of the midpoint of PQ.
谁会的帮个忙，谢谢 |
|
|
|
|
|
|
|
|
|
|
|
发表于 24-10-2011 09:09 PM
|
显示全部楼层
A chord of the parabola y^2=8(x-1), with gradient 2 passes through point A(3,0) and meets the parabo ...
red_snail 发表于 24-10-2011 04:53 PM 
题目有误， 除非你的P 和 Q是 (2p^2+1 , 4p) 和 (2q^2+1 , 4q)。 |
|
|
|
|
|
|
|
|
|
|
|
发表于 24-10-2011 09:59 PM
|
显示全部楼层
|
题目没错。。。就用你所说的吧。。。我看看你的答案。。。参考参考。。。 |
|
|
|
|
|
|
|
|
|
|
|
发表于 24-10-2011 10:00 PM
|
显示全部楼层
回复 2808# Allmaths
题目没错。。。就用你所说的吧。。。看看你的答案。。。参考参考。。。 |
|
|
|
|
|
|
|
|
|
|
|
发表于 28-10-2011 08:13 AM
|
显示全部楼层
|
|
|
|
|
|
|
|
|
|
|
发表于 4-11-2011 02:08 PM
|
显示全部楼层
请问
3(x-1)^2(3-x)-(x-1)^3
怎样变去
2(x-1)^2(5-2x)
感激 |
|
|
|
|
|
|
|
|
|
|
|
发表于 4-11-2011 06:37 PM
|
显示全部楼层
|
3(x-1)^2(3-x)-(x-1)^3 = (x-1)^2(3(3-x)-(x-1)) <<----把(x-1)^2 拉出来 |
|
|
|
|
|
|
|
|
|
|
|
发表于 5-11-2011 12:07 PM
|
显示全部楼层
就是把它expand了,再用long division ?
谢谢。
做到了。
感激不尽=) |
|
|
|
|
|
|
|
|
|
|
|
发表于 10-11-2011 11:51 AM
|
显示全部楼层
set of values of b given tat the given equation hav 3 real and distinct roots. one of the root is 2.
2x^3 -10x+4=b(x-2)
答案是b>-4, not equal to 14
我用factor theorem, 可是一直form不到（x-2）后面的equation…… |
|
|
|
|
|
|
|
|
|
|
|
发表于 10-11-2011 09:25 PM
|
显示全部楼层
set of values of b given tat the given equation hav 3 real and distinct roots. one of the root is 2. ...
红姐 发表于 10-11-2011 11:51 AM
不是很明白你form 不到（x-2）后面的equation的意思。。。
2x^3-10x+4=b(x-2)
2x^2+4x-2=b
2x^2+4x-2-b=0
b^2-4ac>0
4^2-4(2)(-2-b)>0
b>-4
but equation 2x^3-10x+4=b(x-2) has 3 distinct roots, so x≠2
sub x=2 into 2x^2+4x-2=b,
2(2)^2+4(2)-2=b
b=14
∴{b : b>-4, b≠14} |
|
|
|
|
|
|
|
|
|
|
|
发表于 13-11-2011 11:44 PM
|
显示全部楼层
C5 coordinate geometry
P and Q are two variable points lying on the hyperbola X=3t Y=3/t. The tangents at P and Q me at T. If PQ passes through the point (6,2), find the equation of the locus of T as PQ varies.
Please (show the graph) |
|
|
|
|
|
|
|
|
|
|
|
发表于 28-11-2011 12:36 PM
|
显示全部楼层
HI , 初次报到...
请问有谁知道这题怎样做？
希望有人会知道。
MATH T
Given that p^1/2 --p^-1/2 = 2,show that p+p^-1=6 |
|
|
|
|
|
|
|
|
|
|
|
发表于 28-11-2011 06:59 PM
|
显示全部楼层
HI , 初次报到...
请问有谁知道这题怎样做？
希望有人会知道。
MATH T
Given that p^1/2 --p^-1/2 = 2, ...
恩媛 发表于 28-11-2011 12:36 PM 
p^0.5 - p^-0.5 = 2
[p^0.5 - p^-0.5]^2 = 2^2
p - 2(p^0.5)(p^-0.5) + p^-1 = 4
p - 2(p^0) + p^-1 = 4
p + p^-1 = 6 |
|
|
|
|
|
|
|
|
|
|
|
发表于 28-11-2011 10:21 PM
|
显示全部楼层
回复 2819# siewlam
thanks... |
|
|
|
|
|
|
|
|
|
| |
 本周最热论坛帖子
|
3489 
112
