maths F4

1.
{(4^n)[3^(2n-1)]} / [36^(n+1)]
= {(4^n)[3^(-1)][3^(2n)]} / [(36)(36^n)]
= [(4^n)[3^(-1)](9^n)] / [(36)(36^n)]
= {[3^(-1)](36^n)} / [(36)(36^n)]
= [3^(-1)] / (36)
= 1 / 108

2.
(m x m x m x ....... x m) / (m + m + m + ...... + m)
= (m^m) / (m^2)
= m^(m – 2)

3.
2 [y^(a-b)] x 3 [y^(b-c)] x 4 [y^(c-a)]
= 24 [y^(a-b+b-c+c-a)]
= 24 y^0
= 24

4.
2[9^(m/2)] = a^3
2(3^m) = a^3
3^m = (a^3)/2
(3^m)^3 = [ (a^3)/2 ]^3
3^(3m) = (a^9)/8
27^m = (a^9)/8

5.
4^(a+2) = 144
(4^a)(4^2) = 144
4^a = 144/(4^2)
4^a = 9
4^(-a/2) = 9^(-1/2)
4^(-a/2) = 1/3

有錯請指正! 有問題歡迎提出!

2007-12-23 15:23:50 補充：
你第2題的問題是甚麼?請問是不是:[m x m x m x ....... x m / m + m + m + ...... + m] 的 (2m) 次方?假設是, 答案是:[m x m x m x ....... x m / m + m + m + ...... + m] ^(2m)= [(m^m)/(m^2)]^(2m)= [m^(m-2)]^(2m)= m^(2m^2 - 4m)請檢查清楚問題, 謝謝!

2007-12-25 21:40:15 補充：
我似乎明你的問題是甚麼了.[(m x m x m x ....... x m) / (m + m + m + ...... + m)] 上下分別有 2m 次.(m x m x m x ....... x m) <-- 有 2m 次= m^(2m)(m + m + m + ...... + m) <-- 有 2m 次= 2m (m)= 2m^2所以, 問題變成= m^(2m) / 2m^2= (1/2) x [m^(2m-2)]= (1/2) x {m^[2(m-1)]} 即是你的答案中的C.希望幫到. :)

1)
4^n x 3^2n-1/36^n+1
=2^2n*3^2n-(1/6)^2n+1^2n
=6^2n-6^(-1)(2n)+1^2n
=6^2n-6^-2n+1^2n (不是*不能約)

2)
(XmXmXmXm....Xm/m+m+m.....+m)2M times
=(X^2M) (m^2M)/ (m*2M) (都係不能約= =)

3)ｙ不等如０　２ｙ^a-b x 3y^b-c x 4y^c-a

你想問２ｙ^a-b x 3y^b-c x 4y^c-a 定係２ｙ^(a-b) x 3y^(b-c) x 4y^(c-a)=.=?
前者係24y^(a+b+c)-8cy^(a+c)-12by^(a+c)+4bcy^c-6ay^(a+b)+2acy^a+3aby^b-bc
設２ｙ^(a-b) x 3y^(b-c) x 4y^(c-a)=O
12y^(a-b) x 12y^(b-c) x 12y^(c-a)=6*4*3 O
12y^0=72 O
12=72 O
O=1/6

4)2(9^m/2) =a^3
2(3^m)=a^3
3^m=(a^3)/2
(3^m)^3=[ (a^3)/2 ]^3
3^3m=(a^9)/8

27^m
=(3^3)^m
=3^3m
=(a^9)/8

5)4^a+2 =144
4^a=142
4^-a=1/142
4^-a/2=<1/142> < >為開方根