题目内容
计算下列各题
18
-3.64-7.36+3
9
+99
+999
+9999
+99999
+
×5
(10-
×1)+(9-
×2)+(8-
×3)+…+(1-
×10)
(12345+23451+34512+45123+51234)÷5
(1+
+
+…+
)×(
+
+…+
)-(1+
+
+…+
)×(
+
+…+
).
18
| 8 |
| 13 |
| 5 |
| 13 |
9
| 5 |
| 6 |
| 5 |
| 6 |
| 5 |
| 6 |
| 5 |
| 6 |
| 5 |
| 6 |
| 1 |
| 6 |
(10-
| 2 |
| 55 |
| 2 |
| 55 |
| 2 |
| 55 |
| 2 |
| 55 |
(12345+23451+34512+45123+51234)÷5
(1+
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2010 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2011 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2011 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2010 |
考点:分数的巧算
专题:计算问题(巧算速算)
分析:(1)运用加法交换律与结合律简算.
(2)把每个带分数拆成“整数+分数”的形式,从而达到简算的目的.
(3)原式变为(10+9+8+…+1)-(1+2+3+…+10)×
,进而计算即可.
(4)根据题意,被除数中的五个加数,每个数位上的和都是1+2+3+4+5=15,与除数的和相同,再根据题意解答即可.
(5)此题运用设数法解答,然后代入计算.
(2)把每个带分数拆成“整数+分数”的形式,从而达到简算的目的.
(3)原式变为(10+9+8+…+1)-(1+2+3+…+10)×
| 2 |
| 55 |
(4)根据题意,被除数中的五个加数,每个数位上的和都是1+2+3+4+5=15,与除数的和相同,再根据题意解答即可.
(5)此题运用设数法解答,然后代入计算.
解答:
解:(1)18
-3.64-7.36+3
=(18
+3
)-(3.64+7.36)
=22-11
=11
(2)9
+99
+999
+9999
+99999
+
×5
=(9+99+999+9999+99999)+
×6
=(10-1+100-1+1000-1+10000-1+100000-1)+5
=111110-5+5
=111110
(3)(10-
×1)+(9-
×2)+(8-
×3)+…+(1-
×10)
=(10+9+8+…+1)-(1+2+3+…+10)×
=(10+1)×10÷2-(10+1)×10÷2×
=55-55×
=55-2
=53
(4)(12345+23451+34512+45123+51234)÷5
=(1+2+3+4+5)×(10000+1000+100+10+1)÷5
=15×11111÷5
=33333
(5)设1+
+
+…+
=a,
+
+…+
=b,得
(1+
+
+…+
)×(
+
+…+
)-(1+
+
+…+
)×(
+
+…+
)
=a×(b+
)-(a+
)×b
=ab+
a-ab-
b
=
a-
b
=
×(a-b)
=
×1
=
| 8 |
| 13 |
| 5 |
| 13 |
=(18
| 8 |
| 13 |
| 5 |
| 13 |
=22-11
=11
(2)9
| 5 |
| 6 |
| 5 |
| 6 |
| 5 |
| 6 |
| 5 |
| 6 |
| 5 |
| 6 |
| 1 |
| 6 |
=(9+99+999+9999+99999)+
| 5 |
| 6 |
=(10-1+100-1+1000-1+10000-1+100000-1)+5
=111110-5+5
=111110
(3)(10-
| 2 |
| 55 |
| 2 |
| 55 |
| 2 |
| 55 |
| 2 |
| 55 |
=(10+9+8+…+1)-(1+2+3+…+10)×
| 2 |
| 55 |
=(10+1)×10÷2-(10+1)×10÷2×
| 2 |
| 55 |
=55-55×
| 2 |
| 55 |
=55-2
=53
(4)(12345+23451+34512+45123+51234)÷5
=(1+2+3+4+5)×(10000+1000+100+10+1)÷5
=15×11111÷5
=33333
(5)设1+
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2010 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2010 |
(1+
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2010 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2011 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2011 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 2010 |
=a×(b+
| 1 |
| 2011 |
| 1 |
| 2011 |
=ab+
| 1 |
| 2011 |
| 1 |
| 2011 |
=
| 1 |
| 2011 |
| 1 |
| 2011 |
=
| 1 |
| 2011 |
=
| 1 |
| 2011 |
=
| 1 |
| 2011 |
点评:此题考查了学生运用运算定律或运算技巧进行巧算的能力.
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