/是分数线1/51=五十一分之一^2=2次方 ×是乘

问题描述:

/是分数线1/51=五十一分之一^2=2次方 ×是乘
第一题 2009+1949×(1/51-1/2000)+51×(1/1949-1/2000)-2000×(1/51+1/1949)
二:(1/3+1/4+1/5+1/6+1/7)^2+(1/3+1/4+1/5+1/6+1/7)×2/3-(1+1/2+1/3+1/4+1/5+1/6+1/7)×(1/4+1/5+1/6+1/7)
三 (531/135+579/357+753/975)×(579/357+753/975+135/531)-(531/135+579/357)+(753/975+135/531)×(579/357+753/975)
四 1+2+3+4+5+6+5+4+3+2+1/666666×666666
1个回答 分类:数学 2014-09-26

问题解答:

我来补答
2009+1949×(1/51-1/2000)+51×(1/1949-1/2000)-2000×(1/51+1/1949)
=2009 + 1949/51 - 1949/2000 + 51/1949 - 51/2000 -2000/51 - 2000/1949
= 2009 + (1949-2000)/51 + (51-2000)/1949 - (1949+51)/2000
=2009 -1 -1 -1
=2006
(1/3+1/4+1/5+1/6+1/7)^2+(1/3+1/4+1/5+1/6+1/7)×2/3-(1+1/2+1/3+1/4+1/5+1/6+1/7)×(1/4+1/5+1/6+1/7)
令1/3+1/4+1/5+1/6+1/7=A
原式 = A^2 + 2A/3 - (3/2 + A)(A-1/3)
= A^2 + 2A/3 - 3A/2 - A^2 + A/3 + 1/2
= -A/2 + 1/2
= (1-A)/2 【可以代入化简一下】
(531/135+579/357+753/975)×(579/357+753/975+135/531)-(531/135+579/357)+(753/975+135/531)×(579/357+753/975)
设:531/135=a,579/357=b,753/975=c
原式=(a+b+c)*(b+c+1/a)-(a+b+c+1/a)*(b+c)
=ab+ac+1+b^+bc+b/a+bc+c^+c/a-ab-ac-b^-bc-bc-c^-b/a-c/a
=1
1+2+3+4+5+6+5+4+3+2+1/666666×666666
= 36/666666X666666
=6X6/666666X666666
=1/111111X111111
=1/12345654321
不理解就追问,理解了请采纳!
 
 
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