利用规律计算观察下列各式:1/2=1/(1*2)=1-1/21/6=1/(2*3)=1/2-1/31/12=1/(3*4)=1/3-1/41.利用上述规律计算:1/2+1/6+1/12+……+1/[(n-1)n]+1/[(n+1)n]2.利用上述规律解方程:1/[(x-4)(x-3)]+1/[(x-3)(x-2)]+1/[(x-2)(x-1)]+1/[(x-1)
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/28 10:08:14
![利用规律计算观察下列各式:1/2=1/(1*2)=1-1/21/6=1/(2*3)=1/2-1/31/12=1/(3*4)=1/3-1/41.利用上述规律计算:1/2+1/6+1/12+……+1/[(n-1)n]+1/[(n+1)n]2.利用上述规律解方程:1/[(x-4)(x-3)]+1/[(x-3)(x-2)]+1/[(x-2)(x-1)]+1/[(x-1)](/uploads/image/z/1744601-41-1.jpg?t=%E5%88%A9%E7%94%A8%E8%A7%84%E5%BE%8B%E8%AE%A1%E7%AE%97%E8%A7%82%E5%AF%9F%E4%B8%8B%E5%88%97%E5%90%84%E5%BC%8F%EF%BC%9A1%2F2%3D1%2F%281%2A2%29%3D1-1%2F21%2F6%3D1%2F%282%2A3%29%3D1%2F2-1%2F31%2F12%3D1%2F%283%2A4%29%3D1%2F3-1%2F41.%E5%88%A9%E7%94%A8%E4%B8%8A%E8%BF%B0%E8%A7%84%E5%BE%8B%E8%AE%A1%E7%AE%97%EF%BC%9A1%2F2%2B1%2F6%2B1%2F12%2B%E2%80%A6%E2%80%A6%2B1%2F%5B%28n-1%29n%5D%2B1%2F%5B%28n%2B1%29n%5D2.%E5%88%A9%E7%94%A8%E4%B8%8A%E8%BF%B0%E8%A7%84%E5%BE%8B%E8%A7%A3%E6%96%B9%E7%A8%8B%EF%BC%9A1%2F%5B%28x-4%29%28x-3%29%5D%2B1%2F%5B%28x-3%29%28x-2%29%5D%2B1%2F%5B%28x-2%29%28x-1%29%5D%2B1%2F%5B%28x-1%29)
利用规律计算观察下列各式:1/2=1/(1*2)=1-1/21/6=1/(2*3)=1/2-1/31/12=1/(3*4)=1/3-1/41.利用上述规律计算:1/2+1/6+1/12+……+1/[(n-1)n]+1/[(n+1)n]2.利用上述规律解方程:1/[(x-4)(x-3)]+1/[(x-3)(x-2)]+1/[(x-2)(x-1)]+1/[(x-1)
利用规律计算
观察下列各式:
1/2=1/(1*2)=1-1/2
1/6=1/(2*3)=1/2-1/3
1/12=1/(3*4)=1/3-1/4
1.利用上述规律计算:
1/2+1/6+1/12+……+1/[(n-1)n]+1/[(n+1)n]
2.利用上述规律解方程:
1/[(x-4)(x-3)]+1/[(x-3)(x-2)]+1/[(x-2)(x-1)]+1/[(x-1)x]+1/[(x+1)x]=1/(x+1)
利用规律计算观察下列各式:1/2=1/(1*2)=1-1/21/6=1/(2*3)=1/2-1/31/12=1/(3*4)=1/3-1/41.利用上述规律计算:1/2+1/6+1/12+……+1/[(n-1)n]+1/[(n+1)n]2.利用上述规律解方程:1/[(x-4)(x-3)]+1/[(x-3)(x-2)]+1/[(x-2)(x-1)]+1/[(x-1)
1、
1/2+1/6+1/12+……+1/[(n-1)n]+1/[(n+1)n]
=1-1/2+1/2-1/3+1/3-1/4+...+1/(n-1)-1/n+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)
2、
1/[(x-4)(x-3)]+1/[(x-3)(x-2)]+1/[(x-2)(x-1)]+1/[(x-1)x]+1/[(x+1)x]=1/(x+1)
可化为:
1/(x-4)-1/(x-3)+1/(x-3)-1/(x-2)+1/(x-2)-1/(x-1)+1/(x-1)-1/x+1/x-1/(x+1)=1/(x+1)
所以
1/(x-4)=2/(x+1)
解得x=9
经检验,x=9是原方程的根
所以原方程的根是:x=9
很妙的问题