F.3數學問題~~(急)

1.
[ (4)3^(n+2)-(5)3^(n+1) ] / [ (10)3^n-(8) 3^(n+1) ]
= [ (36)3^n- (15)3^n ] / [ (10)3^n-(24) 3^n ]
= [ (21)3^n ] / [ (-14)3^n ]
= -21/14
= -3/2

2.
睇唔明打乜

Given that n is a positive integer when n>2, simplify the following expressions
3.
[a^n b^(n+1)]^2/[ a^(n-1)b^(n+2) ]
= [a^(2n) b^(2(n+1)) ] / [ a^(n-1)b^(n+2) ]
= [a^(2n) b^(2n+2) ] / [ a^(n-1)b^(n+2) ]
= a^(2n-(n-1) ) b^(2n+2- (n+2))
= a^(n+1) b^n

4.
(a^2 b^3)^n (ab)^(n+1)
= a^(2n) b^(3n) a^(n+1) b^(n+1)
= a^(2n+n+1) b^(3n+n+1)
= a^(3n+1) b^(4n+1)

5.
[x^(2n+1)+x^(2n+2) ]/ [x^(2n+3)+x^(2n+1) ]
= [x x^(2n)+ x^2 x^(2n) ]/ [x^3 x^(2n)+ x x^(2n) ]
= (x + x^2) x^(2n) / [(x^3 + x) x^(2n) ]
= (x + x^2) / (x^3 + x)
= x (1 + x) / [ x (x^2 +1) ]
= (1 + x) / (x^2 +1)

6.
[ 2x^(n+3)-x^(2n) ] / [ 2^(n+3)-2x^(n+6) ]
有無打錯野? 又唔用括號, 邊個分子分母都睇唔清,點做?

Arrange the following numbers in descending order
7.
4^100 = (2^2)^100 = 2^200
8^68 = (2^3)^68 = 2^204
16^49 = (2^4)^49 = 2^196
32^41 = (2^5)^41 = 2^205

32^41 > 8^68 > 4^100 > 16^49

8.
(-1/8)^40 = 1/8^40 = 1/2^120
(1/4)^61 = 1/4^61 = 1/ 2^122
(1/16)^32 = 1/16^32 = 1/ 2^128
1/2^123
(-1/8)^40 > (1/4)^61 > 1/2^123 > (1/16)^32