Question

七年级数学题

1．如何化简x+1分之1 + x(x+1)分之1 + (x+1)(x+2)分之1 +...+（x+2005)(x+2006)分之1？
2．解下分式列方程
x（x＋3）分之1 ＋（x＋3）（x＋6）分之1－ 3x分之1＝1
不会做（应该是有方法的。）请各位老师帮帮忙！谢谢！

Answer 1

解：(1)设k为非负整数
1/（x+k)(x+k+1)=[1/(x+k)]-[1/（x+k+1)]
所以1/x(x+1)+1/(x+1)(x+2)+...+1/(x+2005)(x+2006)
=1/x-1/(+1)+1/(x+1)-1/(x+2)+...+1/(x+2005)-1/(x+2006)
=1/x-1/(x+2006)
(2)1/x(x+3)+1/(x+3)(x+6)-1/3x
=[1/x-1/(x+3)]/3+[1/(x+3)-1/(x+6)]/3-1/3x
=[1/x-1/(x+3)+1/(x+3)-1/(x+6)-1/x]/3
=-1/3(x+6)=1
所以x=-19/3

Answer 2

1/x+1+(1/x-1/x+1)+(1/x+1-1/x+2)+类似
(1/X-1/X+3)/3+(1/X+3-1/X+6)/3

Answer 3

x+1分之1 + x(x+1)分之1 + (x+1)(x+2)分之1 +...+（x+2005)(x+2006)分之1=1/x-1/(x+2006)=2006/(x*x+2006x)
x（x＋3）分之1 ＋（x＋3）（x＋6）分之1－ 3x分之1＝1/3(1/x-1/(x+3))+1/3((x+3)-(x+6))-1/3x=-1/3(x+6)=1
x=-3

Answer 4

用分部分式方法就行了。实在是不好输入啊。