解:(1)
÷[
-(
-
)],
=
÷[
-
],
=
÷
,
=
×3,
=
;
(2)0.65×6.4-6.5×0.64+65×9%,
=0.65×6.4-6.5×0.64+65×0.09,
=0.65×6.4-0.65×6.4+0.65×9,
=0.65×(6.4-6.4+9),
=0.65×9,
=0.65×(10-1),
=6.5-0.65,
=5.85;
(3)[4
-(2
+0.64)÷3
+
]×4.5,
=[4
-(2.4+0.64)÷3.04+
]×4.5,
=[4
-3.04÷3.04+
]×4.5,
=[4
-1+
]×
,
=[
+
]×
,
=
×
+
×
,
=16+3
,
=19
;
(4)1-2+3-4+5-6+…+2009-2010+2011,
=(2011-2010)+(2009-2008)+…+(7-6)+(5-4)+(3-2)+1,
=1+1+…+1,
=1×1005+1,
=1006;
(5)
+
+
+…+
,
=
×[(
-
)+(
-
)+(
-
)+…+(
-
)],
=
×[
-
],
=
×
,
=
;
(6)1
+2
+3
+…+9
+10
,
=(1+2+3+…+9+10)+(
+
+
+…+
+
),
=(1+10)×10÷2+(1-
+
-
+
-
+…+
-
+
-
),
=55+(1-
),
=55+
,
=55
.
分析:(1)先算小括號(hào)內(nèi)的,再算中括號(hào)內(nèi)的,最后算括號(hào)外的;
(2)根據(jù)數(shù)字特點(diǎn),原式變?yōu)?.65×6.4-0.65×6.4+0.65×9,運(yùn)用乘法分配律簡(jiǎn)算;
(3)先算小括號(hào)內(nèi)的,再算中括號(hào)內(nèi)的除法,然后算中括號(hào)內(nèi)的減法和減法,最后算括號(hào)外的乘法.同時(shí)注意數(shù)字轉(zhuǎn)化;
(4)把此題調(diào)整一下運(yùn)算順序,從后向前,每?jī)蓚(gè)數(shù)為一組,每一組結(jié)果為1,共分成(2011-1)÷2=1005組,最后剩余1,用1005加上1即可;
(5)通過觀察,每個(gè)分?jǐn)?shù)的分母中的兩個(gè)數(shù)相差都是3,于是把原式變?yōu)?img class='latex' src='http://thumb.zyjl.cn/pic5/latex/8.png' />×[(
-
)+(
-
)+(
-
)+…+(
-
)],括號(hào)內(nèi)通過分?jǐn)?shù)加減相互抵消,求得結(jié)果;
(6)把每個(gè)分?jǐn)?shù)拆成整數(shù)+分?jǐn)?shù)的形式,然后整數(shù)與整數(shù)相加,分?jǐn)?shù)與分?jǐn)?shù)相加.整數(shù)部分用高斯求和公式解答.分?jǐn)?shù)部分中的每個(gè)分?jǐn)?shù)又可以拆分成兩個(gè)分?jǐn)?shù)相減的形式,通過分?jǐn)?shù)加減相互抵消,求得結(jié)果.
點(diǎn)評(píng):考查了小數(shù)、分?jǐn)?shù)的四則混合運(yùn)算,注意運(yùn)算順序和運(yùn)算法則,靈活運(yùn)用所學(xué)的運(yùn)算定律進(jìn)行簡(jiǎn)便計(jì)算.