一题初3数学的题目...高手帮忙解答一下

来源：百度知道 编辑：UC知道 时间：2024/05/24 20:42:08
1/(2根号1+1根号2)+1/(3根号2+2根号3)+.......+1/(100根号99+99根号100)
麻烦把解题过程写出来~~谢谢

1/(2根号1+1根号2)
=(2根号1-1根号2)/(4-2)
=根号1-(根号2)/2

1/(3根号2+2根号3)
=(3根号2-2根号3)/(18-12)
=(根号2)/2-(根号3)/3

1/(2根号1+1根号2)+1/(3根号2+2根号3)+.......+1/(100根号99+99根号100)
=根号1-(根号)/2+(根号)/2-(根号3)/3+(根号)/3-...+(根号99)/99-(根号100)/100
=根号1-(根号100)/100
=1-1/10
=9/10

1/(2+根号2)=(2-根号2)/(2+根号2)*(2-根号2)
=(2-根号2)/2
=1-(根号2)/2
1/(3根号2+2根号3)=(3根号2-2根号3)/(3根号2-2根号3)*(3根号2+2根号3) =(3根号2-2根号3)/6
=(根号2)/2 -(根号3)/3
同理,1/(100根号99+99根号100)=(根号99)/99 -(根号100)/100
所以, 1/(2根号1+1根号2)+1/(3根号2+2根号3)+.......+1/(100根号99+99根号100)
=[1-(根号2)/2]+[(根号2)/2-(根号3)/3]+......+[(根号99)/99 -(根号100)/100]
=1-(根号100)/100
=1-1/10=0.9