F.4 Maths !! Polynomials

a = xy / (x - y)b = 1/x - 1/y = (y - x) / (xy)So ab = - 1 or b = - 1/a (a² + ab) / (b² + ab) x (ab - b) / (ab + a)= (a² - 1) / (b² - 1) x (- 1 - b) / (- 1 + a)= (a - 1)(a + 1) / [(b - 1)(b + 1)] x [- (b + 1)] / (a - 1)= (a + 1) / (1 - b)= (a + 1) / (1 + 1/a)= a(1 + 1/a) / (1 + 1/a)= a= xy / (x - y)

2011-04-05 10:58:43 補充：
Alternatively :

You can simplify first :

(a² + ab) / (b² + ab) x (ab - b) / (ab + a)

= a(a + b) / [b(b + a)] x b(a - 1) / [a(b + 1)]

= (a / b) x (b / a) (a - 1)/(b + 1)

= (a - 1) / (b + 1) ......(*)

2011-04-05 10:58:48 補充：
Since
a = xy / (x - y)
b = 1/x - 1/y = (y - x) / (xy)
So ab = - 1 ==> b = - 1/a , sub into (*) :

(a - 1) / (- 1/a + 1)
= (a - 1) / [(- 1 + a)/a]
= a
= xy / (x - y)

Method 1

a = xy/(x - y)
b = 1/x - 1/y = (y - x)/xy

(a^2 + ab)/(b^2 + ab) * (ab - b)/(ab + a)
= a(a + b)/b(a + b) * b(a - 1)/a(b + 1)
= (a - 1)/(b + 1)
= [ xy/(x - y) - 1] / [(y - x)/xy + 1]
= {[xy - (x - y)]/(x - y)} / {[(y - x) + xy]/xy}
= [(xy - x + y)/(x - y)] / [(xy - x + y)/xy]
= xy/(x - y)


Method 2

a = xy/(x - y)
b = 1/x - 1/y = (y - x)/xy

ab = xy/(x - y) * (y - x)/xy
= -1
b = -1/a

(a^2 + ab)/(b^2 + ab) * (ab - b)/(ab + a)
= (a^2 - 1)/(b^2 - 1) * (-1 - b)/(-1 + a)
= [(a+1)(a-1)]/[(b+1)(b-1)] * -(b+1)/(a-1)
= -(a+1)/(b-1)
= -(a+1)/(-1/a - 1)
= - (a+1) / [-(a+1)/a]
= a
= xy/(x-y)

2011-04-13 13:48:58 補充：
我完全無意圖想抄人地答案, 只係想教多一個方法比樓主姐.