230问答网 > 计算:(1+1⼀2+1⼀3+1⼀4)(1⼀2+1⼀3+1⼀4+1⼀5)-(1+1⼀2+1⼀3+1⼀4+1⼀5)(1⼀2+1⼀3+1⼀4)

计算:(1+1⼀2+1⼀3+1⼀4)(1⼀2+1⼀3+1⼀4+1⼀5)-(1+1⼀2+1⼀3+1⼀4+1⼀5)(1⼀2+1⼀3+1⼀4)

如果觉得麻烦，把（1/2+1/3+1/4）看作一个整体。

2024-11-27 17:19:36

推荐回答（4个）

回答1：

解：设a=1+1/2+1/3+1/4, b=1+1/2+1/3+1/4+1/5
则原式=a(b-1)-b(a-1)
=ab-a-ab+b
=b-a
=1/5

回答2：

把1/2+1/3+1/4看做一个整体,设1/2+1/3+1/4=a,则原来的式子化为:(1+a)(a+1/5)-(1+a+1/5)a
=a+1/5+a^2+(1/5)a-a-a^2-(1/5)a=1/5,所以:(1+1/2+1/3+1/4)(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)(1/2+1/3+1/4)=1/5

回答3：

(1+1/2+1/3+1/4)×(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)×(1/2+1/3+1/4)
=(1+1/2+1/3+1/4)×(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4)×(1/2+1/3+1/4)-1/5×(1/2+1/3+1/4)
=(1+1/2+1/3+1/4)×[(1/2+1/3+1/4+1/5)-(1/2+1/3+1/4)]-1/5×(1/2+1/3+1/4)
=(1+1/2+1/3+1/4)×[1/2+1/3+1/4+1/5-1/2-1/3-1/4]-1/5×(1/2+1/3+1/4)
=(1+1/2+1/3+1/4)×1/5-1/5×(1/2+1/3+1/4)
=1/5×[(1+1/2+1/3+1/4)-(1/2+1/3+1/4)]
=1/5×(1+1/2+1/3+1/4-1/2-1/3-1/4)
=1/5×1
=1/5 行吗

回答4：