简便计算题1x1⼀2+1⼀2x1⼀3+1⼀3x1⼀4+1⼀4x1⼀5+1⼀5x1⼀6+1⼀6x1⼀7

2024-12-03 04:46:38
推荐回答(5个)
回答1:

将每个乘法看做一项,然后每一项可以拆分。
即1-1/2,1/2-1/3......
以此类推。可以理解为中间的乘法可以换成减法,结果不变。
然后前后正负抵消,可得1-1/7=6/7

回答2:

(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/5-1/6)+(1/6-1/7)=6/7

回答3:

1x1/2+1/2x1/3+1/3x1/4+1/4x1/5+1/5x1/6+1/6x1/7
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7
=1-1/7
=6/7

回答4:

原式=(1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+(1/5-1/6)+(1/6-1/7)
=1-1/7
=6/7

回答5:

1x1/2+1/2x1/3+1/3x1/4+1/4x1/5+1/5x1/6+1/6x1/7
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+(1/5-1/6)+(1/6-1/7)
=1-1/7
=6/7