题目内容
25-![]() ![]() ![]() | 25×![]() ![]() |
![]() | 1-![]() ![]() ![]() ![]() |
2011+2010+2009-2008-2007-2006+2005+…-8+7+6+5-4-3-2+1. |
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)运用加法交换律与结合律简算;
(2)原式变为25×
+3×
+2.64×12.5,即265+3×1.25+26.4×1.25,运用乘法分配律简算;
(3)把29看作30-1,运用乘法分配律简算;
(4)原式变为1-(
+
+
+…+
),把每个分数拆成两个分数相减的形式,然后通过加减相抵消的方法,求出结果;
(5)通过观察,发现2011-2008=3,2010-2007=3,2009-2006=3,…,共有2010÷2=1050个3,最后加上1即可.
点评:此题考查了运算定律与简便运算,四则混合运算,灵活运用所学的运算定律律或运算技巧进行简便计算.



=25-(



=25-10+

=23

(2)25×


=25×


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

=

=(1-

=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)运用加法交换律与结合律简算;
(2)原式变为25×


(3)把29看作30-1,运用乘法分配律简算;
(4)原式变为1-(




(5)通过观察,发现2011-2008=3,2010-2007=3,2009-2006=3,…,共有2010÷2=1050个3,最后加上1即可.
点评:此题考查了运算定律与简便运算,四则混合运算,灵活运用所学的运算定律律或运算技巧进行简便计算.

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