帮忙做这几题数学

prove
1)
   (AUB)"n[(AuC)-(Buc)]=empty set
2)    (AnB)-(AnC)=An(B-C)
3)(B-C)u(C-B)=(BuC)n(BnC)"
4)determine the value of a if [(square of 3)-ai]/[1-(square of 3)i]
请会做的帮帮忙

walterchoo

1a)
      LHS
    =(AUB)''n{(AnC)-(BUC)}
    =(AUB)n{(AnC)n(BUC)'
    =(AUB)n{(AnC)n(B'nC')
     ={ (AnAnAnB'nCnC') U (AnBnBnB'nCnC')}
     =(AnBn empty set) U (A n empty set)
     = A n empty set
     = empty set (proved)

2 LHS
= (AnB) n (AnC)'
= (AnB) n (A'UC')
=(AnBnA') U (AnBnC')
= (EMPTY SET n B) U (AnBnC')
= AnBnC'
= An(BnC')
= An (B-C)

3
LHS
= (BnC') U (CnB')
= (BUC) n (C'nB')
= (BUC) n (BnC)''

cgcy

1......
=(AUB)n{(AnC)n(B'nC')
={ (AnAnAnB'nCnC') U (AnBnBnB'nCnC')}....

i don't understand for this step.. Can you show a clearer step? Thank you

junchung2003

(AUB)'n[(AUC)-(BUC)]=empty set



(A' n B') n [ (A U C) n (B U C)']
(A' n B') n [ (A U C) n (B' n C')]

你rearrange的话，就会变成

B' n (A' n C') n (A U C)
B' n (A U C)' n (A U C)
B' n empty set

因为(A U C)' n (A U C) = empty set

B' n empty set = empty set (proved)

----------------------------------------------------------------------------------------------------------------------------------------------------------------------------
 (AnB)-(AnC)=An(B-C)

(A n B) n (A n C)'
(A n B) n (A' U C')
[A' n (A n B)] U [C' n (A n B)]
empty set U [A n (B n C')]
A n (B - C) (proved)

----------------------------------------------------------------------------------------------------------------------------------------------------------------------------
(B-C)u(C-B)=(BuC)n(BnC)"

(B n C') U (C n B')
[B U (C n B')] n [C' U (C n B')]
[(B U C) n (B U B')] n [(C' U C) n (C' U B')]

As (B U B') and (C' U C) = universal set
Then,

(B U C) n (C' U B')
(B U C) n (B n C)' (Proved)

----------------------------------------------------------------------------------------------------------------------------------------------------------------------------
determine the value of a if [(square of 3)-ai]/[1-(square of 3)i]

什么是square of 3？

夜神锋

(√3)  -ai
1-(√3)i
determine the value of a if it is a real number and find this real number

楼主是这样才对吧
ace ahead，exam practice1,no.17
solution:

[(√3)  -ai][1+(√3)i]
[1-(√3)i][1+(√3)i]

√3+3i-ai-√3ai2
12+(√3)2

√3+√3a+3i-ai
         4

√3+√3a+(3-a)i
         4
√3+√3a      +(3-a)i
         4             4
by comparing the real part and imaginary part

(3-a)     =   0
   4

(3-a)  =  0

a=3

let the real number be  x

x   =     √3+√3a
                 4
     =      √3+√3(3)
                  4
      =      √3+(3√3)
                   4
      =       4√3
                   4
       =       √3


so the real number is √3   , and  a=3




[ 本帖最后由 夜神锋 于 2-7-2009 04:52 PM 编辑 ]

harry_lim

这个有详细的答案哦!
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