我有DF.1Maths唔識做,可唔可以教下我..

1a)

The next 3 terms: 3/4, -3/8, 3/16

1b)

1st term: 12 = -24*(-1/2)^1

2nd term: -6 = -24*(-1/2)^2

3rd term: 3 = -24*(-1/2)^4

4th term: -3/2 = -24*(-1/2)^4

Therfore, the nth term = -24*(-1/2)^n

1c)

10th term = -24*(-1/2)^10 = -24/1024 = -3/128



2a)

(k+2x)(5x+x^2-3)

=k(5x+x^2-3) + 2x(5x+x^2-3)

=5kx + kx^2 - 3k + 10x^2 + 2x^3 - 6x

=2x^3 + (k+10)x^2 + (5k-6)x - 3k

2b)

coefficient of x = 5k-6

5k-6 = -11

5k = -11 + 6

5k = -5

k = -1

2c)

constant term = -3k

= -3(-1)

=3

2d)

i) y=(k+2x)(5x+x^2-3)

=[-1+2(-2)] [5(-2)+(-2)^2-3]

=(-1-4)(-10+4-3)

=(-5)*(-9)

=45

ii) y=(k+2x)(5x+x^2-3)

=[-1+2(4)] [5(4)+(4)^2-3]

=(-1+8)(20+16-3)

=7(33)

=231

1)
(a)-3/4,3/8,3/16
b)12x-1/2^(n-1)
Reason:This sequence is a geometric sequence.So,the formula is:
First term x common ratio ^(n-1)
c)Arccording to the formula,so the answer is:
12 x -1/2^(10-1)=12 x -1/512=- 3/128
2)
(a)k(5x+x^2-3)+2x(5x+x^2-3)
=5kx+kx^2-3k+10x^2-2x^3-6x
=-2x^3+10x^2+kx^2+5kx-6x-3k(in descending order)
(b)
-2x^3+10x^2+kx^2+5kx-6x-3k
=2662+1210+121k+(-55k)-(-66)-3k
=2662+1210+121k-55k+66-3k
=3938-63k
(c)3938
(d)y=6x33=198

(I think all these are rite ga la~)

2006-12-31 21:02:28 補充：
CONSTANT TERM MUSTN"T HAVE VARIABLE!!!VARIABLE:eg:x,y,a,b,c,d,e,f,................