F.4 math........show步驟.......**附答案ga**

3(c)(iii)

when x<-1 , y=-(x+1)

when x>-1 , y=x+1

And then you can draw the graph easily

4(a)

First you draw the y=x^2 and x+y=2

Then substitute(3,0), then 0<=9 and 3+0>=2

So you can find the suitable half-plane of the equation y<=x^2 and x+y>=2

The required region is the intersection of these two which is your graph provided.

2nPn=8(2n-1)P(n-1)

2n(2n-1)...(2n-n+1)=8(2n-1)(2n-2)...(2n-1-n+1+1)
2n(2n-1)...(n+1)=8(2n-1)(2n-2)...(n+1)
2n=8
n=4

2(a)

3/(x-1)<=2
3(x-1)<=2(x-1)^2
(x-1)[2(x-1)-3]>=0
(x-1)[2x-5]>=0
x<=1 or x>=5/2

But when x=1 (x-1) is undifined

So the solution is

x<1 and x>=5/2

(ii)

49^(-1/2)*27^(2/3)
=1/[49^(1/2)]* 27^(2/3)
=1/[49^(1/2)]* [27^(1/3)]^2
=3^2/7
=9/7

2008-05-18 18:39:18 補充：
因為yahoo故意將> 變成&gt, <變成&gt﹐期望人去手機認證。但是我無手機所以無辦法﹐亦不會因為無<功能而不答問題

2008-05-18 19:44:37 補充：
y=|x+1|

if y=0, then |x+1|=0

So x=-1

|x| should be >=0 for any value of x!!!!

49^-1/2 x 27^2/3
= 27^2/3 / (49^1/2)
= 3^2 / 7
9/7