一道初二数学,找规律
来源:百度知道 编辑:UC知道 时间:2024/05/30 19:50:15
1/6=1/2*3=1/2-1/3, 1/12=1/3*4=1/3-1/4 请用含有字母n的等式表示这种规律,请直接用上述规律计算1/(x-2)(x-3) - 1/(x-1)(x-3) + 1/(x-1)(x-2)
1/n(n+1)=1/n-1/(n+1)
1/(x-3)-1/(x-2)-0.5*[1/(x-3)-1/(x-1)]+1/(x-2)-1/(x-1)=1/(x-3)-1/(x-1)通分即可
等于0
1/(x-2)(x-3) - 1/(x-1)(x-3) + 1/(x-1)(x-2)=(1/x-2)-(1/x-3)-(1/x-1)+(1/x-3)+(1/x-1)-(1/x-2)=0
1/(x-2)(x-3) - 1/(x-1)(x-3) + 1/(x-1)(x-2)
解:原式=1/(x-2)-1/(x-3) -1/(x-1) +1/(x-3)+ 1/(x-1)-1/(x-2)
=0
这不是预初学的吗?
由以上规律,得:
1/(x-2)(x-3)-1/(x-1)(x-3)+1/(x-1)(x-2)
=1/(x-2)-1/(x-3)-【1/(x-1)-1/(x-3)】+1/(x-1)-1/(x-2)
=0
1+1=几