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【费费数学】第六部分(1-5)

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51#
发表于 2004-9-19 09:37:00 | 只看该作者
well, the problem is that you don't care whether xyz > 0; the problem asks for x(y + z) > 0, since we know that x, y, and z all have the same sign, that means y + z have the same sign as x thus x * (y + z) > 0
[此贴子已经被作者于2004-9-19 9:38:01编辑过]
52#
发表于 2004-9-19 11:06:00 | 只看该作者
以下是引用tiger1234在2004-9-19 9:37:00的发言:
well, the problem is that you don't care whether xyz > 0; the problem asks for x(y + z) > 0, since we know that x, y, and z all have the same sign, that means y + z have the same sign as x thus x * (y + z) > 0

sorry, 刚刚我把问题写错了.我的问题:由条件(1)可知XY 是同号,由条件(2)可知YZ 也是同号,XYZ 都是同号

我不能肯定是否得到这个结论:XYZ 都是同号


[此贴子已经被作者于2004-9-19 11:07:12编辑过]
53#
发表于 2004-9-19 11:09:00 | 只看该作者
well if xy > 0 and yz > 0 => xy * yz > 0 = x * y^2 * z > 0 since y^2 > 0 => x*z > 0 thus x and z have the same sign as well, since x and y have the same sign, y and z have the same sign, and x and z have the same sign, thus x, y, z all have the same sign
54#
发表于 2004-9-19 11:52:00 | 只看该作者
以下是引用tiger1234在2004-9-19 11:09:00的发言:
well if xy > 0 and yz > 0 => xy * yz > 0 = x * y^2 * z > 0 since y^2 > 0 => x*z > 0 thus x and z have the same sign as well, since x and y have the same sign, y and z have the same sign, and x and z have the same sign, thus x, y, z all have the same sign

说服我了. 非常谢谢! 将来有机会姐姐请你吃饭表示感谢!

55#
发表于 2004-9-19 12:06:00 | 只看该作者

YAY!!!

56#
发表于 2004-9-21 22:35:00 | 只看该作者
以下是引用耳朵在2004-2-10 4:07:00的发言:
第二题,费费详解里面的答案是B,应该是错误的。正确答案应该是D。
因为可以从展开后的等式得出:A=2C并且B=C^2,那么第一条B=0,则C=0,因此A=0是可以推出的。因此两个statement都独立有效,选D。


贊成
57#
发表于 2004-11-9 01:22:00 | 只看该作者

3. A

(1) N =3

C(2, N) / C(2,10) = 1/15

(2) N has so solution.

C(1,2)P(1-P) = 7/15

or C(1,2)C(1, N)C(1, 1-N) / C(2,10) = 7/15


[此贴子已经被作者于2004-11-9 1:23:01编辑过]
58#
发表于 2004-11-10 01:54:00 | 只看该作者
以下是引用xyl927在2004-11-9 1:22:00的发言:

3. A


(1) N =3


C(2, N) / C(2,10) = 1/15


(2) N has so solution.


C(1,2)P(1-P) = 7/15


or C(1,2)C(1, N)C(1, 1-N) / C(2,10) = 7/15




(2)用不着排列,只要C(1, N)C(1, 1-N) / C(2,10) = 7/15
59#
发表于 2005-1-5 14:05:00 | 只看该作者

第一题:我认为答案是91%。60% of incorrect paper record(P')=75% of incorrect electric record(E')=3% of all, 即0.6P'=0.75E'=0.03=P'E',求出P'=0.05,E'=0.04,所以

所以P(correct paper )=1-P'=1-0.05=0.95

E( correct electric record)=1-E'=1-0.04=0.96

所以PE(both correct paper and correct electric record)=P+E-1=0.91

0.94的答案是指correct paper or correct electric paper,而不是PE(both correct paper and correct electric record)=1-P'-E'+P'E'=0.94

60#
发表于 2005-1-28 10:41:00 | 只看该作者

5、1/a+1/b+1/c+1/d+1/e=1,a…e全是不同的正整数,问:a+b+c+d+e的least possible value?=3+4+5+6+20

好像还是没有人说怎么解出这个答案的????

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