分析:(1)(2)通過數(shù)字變化,運(yùn)用乘法分配律簡(jiǎn)算
(3)166
=164+2
,然后再根據(jù)乘法分配律進(jìn)行簡(jiǎn)算.
(4)把帶分?jǐn)?shù)化為假分?jǐn)?shù),分子不必算出來,通過約分解決問題.
(5)(9)(10)把每個(gè)分?jǐn)?shù)拆成兩個(gè)分?jǐn)?shù)相減的形式,然后通過加減相互抵消,求得結(jié)果.
(6)把200320032003變?yōu)?003×100010001,200220022002變?yōu)?002×100010001,從而發(fā)現(xiàn)減號(hào)前后的算式相同,因此得數(shù)為0
(7)通過觀察,此算式中的數(shù)字有一定特點(diǎn),把原式變?yōu)椋?111+123)+(2222+119)+(3333+79)+(4444-321),計(jì)算即可.
(8)每個(gè)分?jǐn)?shù)的分母中的兩個(gè)因數(shù)相差2,因此提出
,把每個(gè)分?jǐn)?shù)拆成兩個(gè)分?jǐn)?shù)相減的形式,進(jìn)而簡(jiǎn)算即可
(11)先把分母運(yùn)用高斯求和公式表示出來,原式變?yōu)?span id="h0myo5g" class="MathJye" mathtag="math" style="whiteSpace:nowrap;wordSpacing:normal;wordWrap:normal">
+
+
+…+
,再把每個(gè)分?jǐn)?shù)拆成兩個(gè)分?jǐn)?shù)相減的形式,然后通過加減相互抵消,求得結(jié)果.
解答:
解:(1)
×32
=
×32
=(1-
)×32
=31-
=30
(2)27×
=(26+1)×
=26×
+
=15+
=15
(3)166
÷41
=,(164+2
)×
=164×
+2
×
=4+
,
=4
(4)1993÷1993
=1993÷
=1993×
=
(5)
+
+
+
+
+
=1-
+
-+
-
+…+
-
=1-
=
(6)2003×200220022002-2002×200320032003
=2002×2003×100010001-2003×2002×100010001
=0
(7)1234+2341+3412+4123
=(1111+123)+(2222+119)+(3333+79)+(4444-321)
=1111+2222+3333+4444+(123+119+79-321)
=1111+2222+3333+4444
=1111×(1+2+3+4)
=1111×10
=11110
(8)
+
+
+…
=
×(
-
+
-
+
-
+…+
-
)
=
×(
-
)
=
×
=
(9)
+
+
=
×(
-
+
-
+
-
)
=
×(
-
)
=
×
=
(10)
+
+
+…+
=
-
+
-
+
+…+
-
=
-
=
(11)
+
+
+…+
=
+
+
+…+
=
+
+
+…+
=2×(
-
+
-
+
-
+…+
-
)
=2×(
-
)
=2×
=