原式=（1+a）*b-（1+b）*a
=b+ab-a-ab
=b-a
=1/5追问
要小学的，中学的太深奥
追答(1+1/2+1/3+1/4)x(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)x(1/2+1/3+1/4)
=(1+1/2+1/3+1/4)x(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4)x(1/2+1/3+1/4)-1/5 x(1/2+1/3+1/4)
=(1+1/2+1/3+1/4) x(1/2+1/3+1/4+1/5-1/2-1/3-1/4)-1/5 x(1/2+1/3+1/4)
=(1+1/2+1/3+1/4) x1/5-1/5 x(1/2+1/3+1/4)
=1/5 x(1+1/2+1/3+1/4-1/2-1/3-1/4)
=1/5 x1
=1/5
简便计算(1+1\/2+1\/3+1\/4)×(1\/2+1\/3+1\/4+1\/5)-(1+1\/2+1\/3+1\/4+1\/5...
设a=（1\/2+1\/3+1\/4），b=（1\/2+1\/3+1\/4+1\/5）原式=（1+a）*b-（1+b）*a =b+ab-a-ab =b-a =1\/5
计算:(1+1\/2+1\/3+1\/4)×(1\/2+1\/3+1\/4+1\/5)-(1+1\/2+1\/3+1\/4+1\/5)×...
令A=1＋1\/2＋1\/3＋1\/4 B=1＋1\/2＋1\/3＋1\/4＋1\/5 所以 原式=A(B-1)-B(A-1)=B-A =1\/5
...1\/2+1\/3+1\/4+1\/5)-(1+1\/2+1\/3+1\/4+1\/5)x(1\/2+1\/3+1\/4)
假设1\/2+1\/3+1\/4=a 那么原式=(1+a)(a+1\/5)-(1+a+1\/5)a =a+1\/5+a*a+a\/5-a-a*a-a\/5 =1\/5
...3+1\/4)-(1+1\/2+1\/3+1\/4+1\/5)×(1\/2+1\/3+1\/4) 简便方法计算
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...1\/2+1\/3+1\/4+1\/5)-(1+1\/2+1\/3+1\/4+1\/5)*(1\/2+1\/3+1\/4)
简便运算题 （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\/5)*[（1+1\/2+1\/3+1\/4)-(1\/2+1\/3+1\/4)]-(1+1\/2+1\/3+1\/4)=(1+1\/2+1\/3+1\/4+1\/5)*1-(1+1\/2+1\/3+1\/4)=1\/5 ...
...乘以(1\/2+1\/3+1\/4+1\/5)-(1+1\/2+1\/3+1\/4+1\/5)乘以(1\/2+1\/3+1\/4...
+ （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\/2+1\/3+1\/4+1\/5）+ （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）- （1\/2+1\/3+1\/4）= 1\/5 ...
...*(1\/2+1\/3+1\/4+1\/5+1\/6)-(1+1\/2+1\/3+1\/4+1\/5+1\/6)*(1\/2+1\/3+1\/...
解：为方便书写，设1\/2+1\/3+1\/4+1\/5=t，则 (1+1\/2+1\/3+1\/4+1\/5)*(1\/2+1\/3+1\/4+1\/5+1\/6)-(1+1\/2+1\/3+1\/4+1\/5+1\/6)*(1\/2+1\/3+1\/4+1\/5)=(t+1)*(t+1\/6)-(1+t+1\/6)*t =(t²+1\/6t+t+1\/6)-(t+7\/6)*t =(t²+7\/6t+1\/...
简便计算(1+1\/2+1\/3+1\/4)*(1\/2+1\/3+1\/4+1\/6)-(1+1\/2+1\/3+
（1+1\/2+1\/3+1\/4）x（1\/2+1\/3+1\/4+1\/6）—（1+1\/2+1\/3+1\/4+1\/5）x（1\/2+1\/3+1\/4）=（1+1\/2+1\/3+1\/4）x（1\/2+1\/3+1\/4)+（1+1\/2+1\/3+1\/4）x1\/6—（1+1\/2+1\/3+1\/4）x（1\/2+1\/3+1\/4） -1\/5x（1\/2+1\/3+1\/4）=（1+1\/2+1\/3+1\/...
(1+1\/2+1\/3+1\/4)*(1\/2+1\/3+1\/4+1\/5)怎么做,求讲解 !
解：令a=1\/2+1\/3+1\/4 则1+1\/2+1\/3+1\/4=1+a 1\/2+1\/3+1\/4+1\/5=a+1\/5 1+1\/2+1\/3+1\/4+1\/5=1+a+1\/5 所以原式=(1+a)(a+1\/5)-a(1+a+1\/5)=a(1+a)+(1+a)*1\/5-a(1+a)-a*1\/5 =(1+a)*1\/5-a*1\/5 =1\/5+a*1\/5-a*1\/5 =1\/5 ...
求高手解题:(1+1\/2+1\/3+1\/4)×(1\/2+1\/3+1\/4+1\/5)—(1+1\/2+1\/3+1\/4...
令a=1\/2+1\/3+1\/4 则原式=(1+a)(a+1\/5)-(1+a+1\/5)a =(1+a)a+1\/5(1+a)-(1+a)-1\/5a =1\/5(1+a)-1\/5a =1\/5+1\/5a-1\/5a =1\/5
