【急】3道七年级数学题目(特殊分式）① 1/1*2+1/2*3+1/3*4+……91/98*99+1/99*100② 化简下列算是：1/x+1/x(x+1)+1/(x+1)(x+2)+……+1/(x+2004)(x+2005)③ 求下列分式方程 1/x(x+2)+1/(x+2)(x+4)-1/2x=1很急的……明天测验啊!三道当中回答一道也OK啊 可是老师刚刚将要考这个啊~

① 1/1*2+1/2*3+1/3*4+……91/98*99+1/99*100
分析：1/1*2=1-1/2
1/2*3=1/2-1/3
1/3*4=1/3-1/4……
1/98*99=1/98-1/99
1/99*100=1/99-1/100
所以原式=1-1/2+1/2-1/3+1/3-1/4……1/99-1/100=1-1/100=99/100
② 化简下列算是：1/x+1/x(x+1)+1/(x+1)(x+2)+……+1/(x+2004)(x+2005)
分析：与上题方式相同，得：1/x+1/x-1/x+2005=2/x-1/x+2005=x-4010/x^2+2005x
③ 求下列分式方程 1/x(x+2)+1/(x+2)(x+4)-1/2x=1
这题也要用上面方式来转换：
(1/x-1/x+2)*1/2+(1/x+2-1/x+4)*1/2-1/2x=1
1/2x-1/2x+4+1/2x+4-1/2x+8-1/2x=1
-1/2x+8=1
2x+8=-1
2x=-9
x=-9/2

① 1/1*2+1/2*3+1/3*4+……91/98*99+1/99*100
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/99-1/100)
=1-1/100=99/100
② 1/x+1/x(x+1)+1/(x+1)(x+2)+……+1/(x+2004)(x+2005)
=1/x+[1/x-1/(x+1)]+[1/(x+1)-1/(x+2)]+...+[1/(x+2004)-1/(x+2005)]
=2/x -1/(x+2005)=(x+4010)/[x(x+2005)]
③ 1/x(x+2)+1/(x+2)(x+4)-1/2x=1
[1/x-1/(x+2)+1/(x+2)-1/(x+4)]/2-1/2x=1
1/2x-1/2(x+4)-1/2x=1
-1/2(x+4)=1
2(x+4)=-1
x+4=-1/2
x=-4-1/2=-9/2

① 1/1*2+1/2*3+1/3*4+……91/98*99+1/99*100
1/1*2+1/2*3+1/3*4+……91/98*99+1/99*100
=1-1/2+1/2-1/3+1/3-1/4+....+1/99-1/100
=1-1/100
=99/100
② 化简下列算是：1/x+1/x(x+1)+1/(x+1)(x+2)+……+1/(x+2004)(x+2005)
1/x+1/x(x+1)+1/(x+1)(x+2)+……+1/(x+2004)(x+2005)
=1/x+1/x-1/(x+1)+1/(x+1)-.....+1/(x+2004)-1/(x+2005)
=2/x-1/(x+2005)
=(x+4010)/x(x+2005)
③ 求下列分式方程 1/x(x+2)+1/(x+2)(x+4)-1/2x=1
1/x(x+2)+1/(x+2)(x+4)-1/2x=1
1/2x-1/2(x+2)+1/2(x+2)-1/2(x+4)-1/2x=1
2(x+4)=-1
x=-9/2

1. 99/100
2. 2004/x

① 1/1*2+1/2*3+1/3*4+……91/98*99+1/99*100=(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/99-1/100)中间抵消=1-1/100=99/100② 化简下列算是：1/x+1/x(x+1)+1/(x+1)(x+2)+……+1/(x+2004)(x+2005)1/x(x+1)=[(x+1)-x]/x(...

1、原式=（1-1/2）+（1/2-1/3）+（1/3-1/4）+…………+（1/98-1/99）+(1/99-1/100)=1-1/100=99/100

①1/1*2+1/2*3+1/3*4+…+1/9*10=

1/(1*2)=1/1-1/2
1/(2*3)=1/2-1/3
1/(3*4)=1/3-1/4
:
:
1/(99*100)=1/99-1/100
所以原式=(1/1-1/2)+……+（1/99-1/100）=99/100