Question

求一道分式的答案!

已知一个分式除以a的平方减4份之a的平方减6a加9的商是a加2份之a减3的倒数,求原来的这个分式.

Answer 1

1/[n(n+1)]=1/n-1/(n+1)

2/(x+1)(x+3) =1/(x+1)-1/(x+3)
2/(x+3)(x+5) =1/(x+3)-1/(x+5)
2/(x+5)(x+7) =1/(x+5)-1/(x+7)
……
2/(x+2005)(x+2007) =1/(x+20051)-1/(x+2007)
所以，
2/(x+1)(x+3) + 2/(x+3)(x+5) + 2/(x+5)(x+7) + …+ 2/(x+2005)(X+2007)
＝1/(x+1)-1/(x+3)+1/(x+3)-1/(x+5)+1/(x+5)-1/(x+7)+…+1/(x+20051)-1/(x+2007)
＝1/(x+1)-1/(x+2007)