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-22.(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
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
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
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???????
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)
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)
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