解:(1)25-

+

-

,
=25-(

+

)+

,
=25-10+

,
=23

;
(2)25×

+3÷

+2.64×12.5,
=25×

+3×

+2.64×12.5,
=265+3×1.25+26.4×1.25,
=265+(3+26.4)×1.25,
=265+36.75,
=301.75;
(3)

×119,
=

×119,
=(1-

)×119,
=119-

,
=119-3

,
=115

;
(4)1-

-

-

-…-

,
=1-(

+

+

+…+

),
=1-[(1-

)+(

-

)+(

-

)+…+(

-

)],
=1-[1-

],
=1-1+

,
=

;
(5)2011+2010+2009-2008-2007-2006+2005+…-8+7+6+5-4-3-2+1,
=(2011-2008)+(2010-2007)+(2009-2006)+(2005-2002)…+(7-4)+(6-3)+(5-2)+1,
=3+3+3+3+…+3+1,
=3×(2010÷2)+1,
=3×1005+1,
=3015+1,
=3016.
分析:(1)運用加法交換律與結(jié)合律簡算;
(2)原式變?yōu)?5×

+3×

+2.64×12.5,即265+3×1.25+26.4×1.25,運用乘法分配律簡算;
(3)把29看作30-1,運用乘法分配律簡算;
(4)原式變?yōu)?-(

+

+

+…+

),把每個分?jǐn)?shù)拆成兩個分?jǐn)?shù)相減的形式,然后通過加減相抵消的方法,求出結(jié)果;
(5)通過觀察,發(fā)現(xiàn)2011-2008=3,2010-2007=3,2009-2006=3,…,共有2010÷2=1050個3,最后加上1即可.
點評:此題考查了運算定律與簡便運算,四則混合運算,靈活運用所學(xué)的運算定律律或運算技巧進行簡便計算.