解:(1)2005-999,
=2005-(1000-1),
=2005-1000+1,
=1006;
(2)0.25+
,
=0.25+0.2-0.45,
=0;
(3)
,
=0.72÷0.75×2,
=2;
(4)
,
=(
+
-
)×3,
=
×3+
×3-
×3,
=
+1-
,
=0.3;
(5)
,
=[(99
+
)-(29
+30
)]×
,
=40×
,
=4;
(6)
,
=
×
×
×9,
=
;
(7)11×991+99,
=11×991+11×9,
=11×(991+9),
=11000;
(8)
,
=
,
=
,
=1;
(9)1+3+5+7+9+11+13+15+17+19,
=(1+19)+(3+17)+(5+15)+(7+13)+(9+11),
=20×5,
=100;
(10)
…×
,
=
×
×
×…×
,
=
.
分析:(1)把999看作1000,然后加1;
(2)、(3)把分數(shù)都化為小數(shù),進行計算;
(4)先根據(jù)除以一個分數(shù),等于乘這個分數(shù)的倒數(shù),然后運用乘法分配律進行解答;
(5)先把括號里的運算進行交換,再結(jié)合,進行簡算,然后再和
相乘;
(6)按照乘除法計算方法進行解答;
(7)把99分解成11×9,然后運用乘法分配律進行簡算;
(8)先算1-
=
,再算1÷
=4,繼而算5-4=1,最后算1÷1=1;
(9)首位進行結(jié)合,再運用乘法分配律進行解答;
(10)分別算出括號里的,然后相乘,進行約分,即可.
點評:解答此題的關(guān)鍵:熟練掌握加、減、乘、除中的一些運用定律和性質(zhì),能根據(jù)數(shù)的特點,運用簡便方法,進行簡算.