1)五十二又二十五分之十一×79.45+159×47.56+七十九又二十分之十一×52.44=

1)五十二又二十五分之十一×79.45+159×47.56+七十九又二十分之十一×52.44
=52.44×79.45+159×47.56+79.55×52.44
=52.44×(79.45+79.55)+159×47.56
=52.44×159+159×47.56
=159×(52.44+47.56)
=159×100
=15900
3)2002+2001-2000-1999+1998+1997-1996-1995+……+2+1
=(2002-2000)+(2001-1999)+(1998-1996)+(1997-1995)+……+(6-4)+(5-3)+2+1
=2+2+2+2+……+2+2(从3-2002共2000个数,所以有1000个2)+2+1
=1000×2+2+1
=2003
4,5两题均用到一个转换式1/（A×B）=1/（B-A）×（1/A-1/B）
如1/15=1/（3×5）=1/（5-3）×（1/3-1/5）=1/2×（2/15）=1/15可验证一下
4)(1×2分之一)+(2×3分之一)+(3×4分之一)+……+(10×11分之一)
=1/(1×2)+1/(2×3)+1/(3×4)+……+1/(10×11)
=(1-1/2)+(1/2 - 1/3)+（1/3 - 1/4）+……+（1/10 - 1/11）
=1-1/11
=10/11
5)三分之一+十五分之一+三十五分之一+六十三分之一+九十九分之一
=1/(1×3)+1/(3×5)+1/(5×7)+1/(7×9)+1/(9×11)
=1/2×(1-1/3)+1/2×(1/3-1/5)+1/2×(1/5-1/7)+1/2×(1/7-1/9)+1/2×(1/9-1/11)
=1/2×(1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11)
=1/2×(1-1/11)
=1/2×10/11
=5/11
6)一又二分之一-六分之五+十二分之七-二十分之九+三十分之十一-四十二分之十三+五十六分之十五
(根据提示,1又1/2=1+1/2,+1/2+1/3=5/6……)
=(1+1/2)-(1/2+1/3)+(1/3+1/4)-(1/4+1/5)+(1/5+1/6)-(1/6+1/7)+(1/7+1/8)
=1+ 1/2 - 1/2 - 1/3 + 1/3 + 1/4 - 1/4 - 1/5 + 1/5 + 1/6 - 1/6 - 1/7 + 1/7 + 1/8
=1+1/8
=9/8

1 4166.358+7562.04+4171.602=15900
3 4*500+2+1=2003
4 1/2+2/3+……10/11=10-(1/2+1/3……1/11)=8又551/27720
5 1/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11)=1/2*10/11=5/11
6 1+1/2-1/2-1/3+1/3+1/4-1/4-1/5+1/5+1/6-1/6-1/7+1/7+1/8=1+1/8=9/8

1)52又11/25×79又9/20+159×47又14/25+79又11/20×52.44
=52.44*79.45+159*47.56+79.55*52.44
=52.44*79.45+79.55*52.44+159*47.56
=52.44*(79.45+79.55)+159*47.56
=52.44*159+159*47.56
=159*(52.44...

1）原式=4166.358+7562.04+4171.602
=15900
3）原式=2002÷4×4
=2002
4）原式=1/2+2/3+……10/11
=10-(1/2+1/3……1/11)
=
5）原式=1/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11) ...

(3)2002*(2002+1)/2=2005003