1/（x+1）(x+2) + 1/(x+2)(x+3)+.+1/(x+1994)(x+1995)=2x+3987/3x+

1/（x+1）(x+2) + 1/(x+2)(x+3)+.+1/(x+1994)(x+1995)=2x+3987/3x+5985
1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)...+1/(x+1994)-1/(x+1995)=2x+3987/3x+5985 (这一步是计算中常用的方法,可以把项依次销掉)
1/(x+1)-1/(x+1995)=2x+3987/3x+5985 (然后就世界分式方程了)
x+1995) (2x+3987)/(3x+5985)=(2x+3987)/[3(x+1995)]
两边都乘以(x+1995),得
1994/(x+1)(x+1995)=(2x+3987)/[3(x+1995)]
1994/(x+1)=(2x+3987)/3
1994*3=(x+1)(2x+3987)
2x^2+3989x--1995=0
2x-1)(x+1995)=0
x=1/2 (代入检验,成立）
方法最关键!
如果这个过程你不明白
可能就在这里
要把1/（x+1)(x+2)分解成1/(x+1)-1/(x+2)
如果你不相信这个等式成立 可以自己通分以下
这是计算很重要的拆分方法
可以把项都销掉~

1/（x+1）(x+2) + 1/(x+2)(x+3)+....+1/(x+1994)(x+1995)=(2x+3987)/(3x+5985)
[1/(x+1)-1/(x+2)]+……+[1/(x+1994)-1/(x+1995)]=(2x+3987)/(3x+5985)
1/(x+1)-1/(x+1995)=(2x+3987)/(3x+5985)=[2(x+1995)-3]/3(x+1995)
1/(x+1)-1/(x+1995)=2/3-1/(x+1995)
1/(x+1)=2/3
2x+2=3
x=1/2
代入检验，成立

1/（x+1）(x+2) + 1/(x+2)(x+3)+....+1/(x+1994)(x+1995)=2x+3987/3x+5985
1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)...+1/(x+1994)-1/(x+1995)=2x+3987/3x+5985 (这一步是计算中常用的方法，可以把项依次销掉)
1/(x+1)-1/(x+1995)=2x+398...

左边=1/（x+1）(x+2) + 1/(x+2)(x+3)+....+1/(x+1994)(x+1995)=((x+2)-(x+1))/(x+1)(x+2)+((x+3)-(x+2))/(x+2)(x+3)+...+((x+1995)-(x+1994))/(x+1994)(x+1995)=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+...+1/(x+1994)-1/(x+19...

略左式=[1/（x+1)]-[1/(x+2)]+[1/(x+2)]-[1/(x+3)]+...+1/(x+1994)-1/(1+1995)
=1/(x+1)-1/(x+1995)
=1994/(x^2+1996x+1995)=2x+3987/3x+5985，解这个方程即可得x

1/（x+1）(x+2)=1/（x+2）-1/（x+1）
其他同理
上式子=1/（x+1995）-1/（x+1）
通分就能得到结果

1/(x+1)(x+2)= 1/(x+1) - 1/(x+2)
1/(x+2)(x+3)= 1/(x+2) - 1/(x+3)
................
1/(x+1994)(x+1995)= 1/(x+1994) - 1/(x+1995)
所以
1/（x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+1994)(x+1995...

1/(x+1)(x+2)= 1/(x+1) - 1/(x+2)
1/(x+2)(x+3)= 1/(x+2) - 1/(x+3)
................
1/(x+1994)(x+1995)= 1/(x+1994) - 1/(x+1995)
（前一项和后一项的相互抵消掉了，所以最后只剩下，头尾两项了）
所以 (楼主，还不够详细？)
1...

1/（x+1）(x+2) + 1/(x+2)(x+3)+....+1/(x+1994)(x+1995)=2x+3987/3x+5985
1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)...+1/(x+1994)-1/(x+1995)=2x+3987/3x+5985 (这一步是计算中常用的方法，可以把项依次销掉)
1/(x+1)-1/(x+1995)=2x+398...