第1个回答 2009-06-19
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本回答被网友采纳
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本回答被网友采纳
第2个回答 2009-06-19
(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)
=1/5(1+1/2+1/3+1/4-1/2+1/3+1/4)=1/5
=1/5(1+1/2+1/3+1/4-1/2+1/3+1/4)=1/5
第3个回答 2009-06-19
=(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)
=1/5本回答被提问者采纳
=1/5本回答被提问者采纳
第4个回答 2009-06-19
(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)(1/5)-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+1/2+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)(1/5)-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
第5个回答 2009-06-19
(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)+(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)x(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(1/5）-(1/5)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)]-[(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)x(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(1/5）-(1/5)x(1/2+1/3+1/4）
=1/5
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