1/2+1/3+1/4+1/5=y
(1+1/2+1/3+1/4)*(1/2+1/3+1/4+1/5ï¼-ï¼1+1/2+1/3+1/4+1/5ï¼*ï¼1/2+1/3+1/4ï¼
=(1+x)y-(1+y)x
=y+xy-x-xy
=y-x
=1/5追é®
ä¸è¦æ¹ç¨ï¼è¦å°å¦ç®å¼
追ç(1+1/2+1/3+1/4)*(1/2+1/3+1/4+1/5ï¼-ï¼1+1/2+1/3+1/4+1/5ï¼*ï¼1/2+1/3+1/4ï¼
=ï¼1/2+1/3+1/4+1/5ï¼+ï¼1/2+1/2+1/4ï¼*ï¼1/2+1/3+1/4+1/5ï¼-ï¼1/2+1/3+1/4ï¼-ï¼1/2+1/3+1/4ï¼*ï¼1/2+1/3+1/4+1/5ï¼
=ï¼1/2+1/3+1/4+1/5ï¼-ï¼1/2+1/3+1/4ï¼
=1/5
å
¶å®åæ¥çæ¹æ³ä¸æ¯æ¹ç¨ï¼æ¯ä»£æ°æ³
=-（1+1/2+1/3+1/4）-（1+1/2+1/3+1/4+1/5）
=-（2+1+2/3+1/2+2/5）
=
...1\/2+1\/3+1\/4+1\/5)-(1+1\/2+1\/3+1\/4+1\/5)*(1\/2+1\/3+1\/4)
设1+1\/2+1\/3+1\/4=x 1\/2+1\/3+1\/4+1\/5=y 则原式=xy-(1+y)(x-1)=xy-xy-x+y+1 =y-x+1 =1\/5
...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+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
...×(1\/2+1\/3+1\/4)-(1+1\/2+1\/3+1\/4+1\/5)×(1\/2+1\/3+1\/4) 简便方法计 ...
？
...×(1\/2+1\/3+1\/4+1\/5)-(1+1\/2+1\/3+1\/4+1\/5)×(1\/2+1\/3+1\/4)_百度...
应该是：令（1+1\/2+1\/3+1\/4)＝a，(1\/2+1\/3+1\/4+1\/5）＝b 原式＝ab－（a＋1\/5)*(b-1\/5)=ab-(ab+1\/5b-1\/5a-1\/25)=ab-ab-1\/5b+1\/5a+1\/25 =1\/5(a+b)+1\/25 =1\/5*4\/5+1\/25 =4\/25+1\/25 =1\/5 ...
...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 ...
...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)
为了简述方便 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
...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+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\/2+1\/3+1\/4+1\/5）+ （1\/...
...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\/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\/2+1\/3+1\/4+1\/5)-(1+1\/2+1\/3+1\/4+1\/5)×(1\/2+1\/3+1\/4)_百度...
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
巧算:(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...
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\/...
