Question

求解3道初二分式题

1. 1/a(a+1) + 1/（a+1)(a+2) + 1/（a+2)(a+3) + .........+1/（a+2003)(a+2004)
2. 1/x - 1/x-1 （x²-1/x -x+1）
3. 1/a+2 + a-2
第二三题重写一下~~
2. (1/x)- (1/x-1)【（x²-1/x）-x+1】用小括号括的是一个分式前面是分子后面是分母
3. （1/a+2）+a-2

Answer 1

1. 1/a(a+1) + 1/（a+1)(a+2) + 1/（a+2)(a+3) + .........+1/（a+2003)(a+2004)
=1a-1/(a+1) + 1/(a+1)-1/(a+2) +1/(a+2)+......-1/(a+2003)+1/(a+2003)-1/a+2004)
=1/a-1/(a+2004)
=(a+2004-a)/{a(a+2004)
=2004/{a(a+2004)

2. 1/x - 1/x-1 （x²-1/x -x+1）
= 1/x - 1/(x-1) {(x^2-1) -x(x-1)}/x
= 1/x - 1/(x-1) {(x^2-1 -x^2 +x)}/x
= 1/x - 1/(x-1)*(x-1)/x
= 1/x - 1/x
=0

3. 1/(a+2) + ( a-2)
={1+(a+2)(a-2)}/(a+2)
={1+a^2-4)}/(a+2)
=(a^2-3)/(a+2)

Answer 2

1. 1/a(a+1) + 1/（a+1)(a+2) + 1/（a+2)(a+3) + .........+1/（a+2003)(a+2004)
=1/a-1/(a+1)+1/(a+1)-1/(a+2)+1/(a+2)-1/(a+3)+.....+1/(a+2003)-1/(a+2004)
=1/a-1/(a+2004)
=(a+2004)/[a(a+2004)]-a/[a(a+2004)]
=(a+2004-a)/[a(a+2004)]
=2004/[a(a+2004)]

2. 1/x - 1/x-1(x²-1/(x -x+1)?????

3. 1/a+2 + a-2???????

Answer 3

1用分解，1/a(a+1)=1/a-1/(a+1).用这个分解方法分解所有项，然后相消。只余下第一项和最后项相减，答案简单，自己算算吧。这个是思路
你第二三两题写得不明白，麻烦写清楚包括括号一类的。我再回答

Answer 4

1. 1/a(a+1) + 1/（a+1)(a+2) + 1/（a+2)(a+3) + .........+1/（a+2003)(a+2004)
=[1a-1/(a+1)]+[1/(a+1)-1/(a+2)]+[1/(a+2)-1/(a+3)]+.....+[1/(a+2003)-1/(a+2004)]
=1a-1/(a+1)+1/(a+1)-1/(a+2)+1/(a+2)-1/(a+3)+.....+1/(a+2003)-1/(a+2004)
=1/a-1/(a+2004)
=(a+2004)/a(a+20004)-a/a(a+2004)
=(a+2004-a)/a(a+2004)
=2004/a(a+2004)

2. 1/x - 1/(x-1)* （x²-1/x -x+1）
=1/x-1(x-1)*[(x+1)(x-1)/x-(x-1)]
=1/x-[(x+1)/x-1]
=1/x-(1+1/x-1)
=1/x-1-1/x+1
=2

3. 1/(a+2) + a-2
=1/(a+2)+(a-2)*(a+2)/(a+2)
=(1+a^-4)/(a+2)
=(a^-3)/(a+2)