题目内容
计算:
(1)64×〔1÷(
-2.09)〕
(2)
×〔
-(
+
)〕
(3)
+
+
+
+
(4)
(5)20×20-19×19+18×18-17×17+…+2×2-1×1.
(1)64×〔1÷(
| 21 |
| 10 |
(2)
| 7 |
| 12 |
| 3 |
| 4 |
| 1 |
| 20 |
| 1 |
| 2 |
(3)
| 1 |
| 3 |
| 1 |
| 15 |
| 1 |
| 35 |
| 1 |
| 63 |
| 1 |
| 99 |
(4)
| 2013+2012×2014 |
| 2013×2014-1 |
(5)20×20-19×19+18×18-17×17+…+2×2-1×1.
考点:四则混合运算中的巧算
专题:计算问题(巧算速算)
分析:(1)(2)先算小括号内的,再算中括号内的,最后算括号外的.
(3)通过观察,每个分数的分母中的两个因数相差2,把每个分数扩大2倍,然后把每个分数拆分为两个分数相减的形式,把最后结果乘
即可.
(4)把分子化成与分母相同或部分相同,通过约分解决问题.
(6)运用平方差公式解答.
(3)通过观察,每个分数的分母中的两个因数相差2,把每个分数扩大2倍,然后把每个分数拆分为两个分数相减的形式,把最后结果乘
| 1 |
| 2 |
(4)把分子化成与分母相同或部分相同,通过约分解决问题.
(6)运用平方差公式解答.
解答:
解:(1)64×[1÷(
-2.09)]
=64×[1÷(2.1-2.09)]
=64×[1÷0.01]
=64×100
=6400
(2)
×[
-(
+
)]
=
×[
-
]
=
×
=
(3)
+
+
+
+
=[(1-
)+(
-
)+(
-
)+(
-
)+(
-
)]×
=[1-
]×
=
×
=
(4)
=
=
=
=1
(5)20×20-19×19+18×18-17×17+…+2×2-1×1
=(20+19)×(20-19)+(18+17)×(18-17)+(16+15)×(16-15)+…+(2+1)×(2-1)
=(20+19)×1+(18+17)×1+(16+15)×1+…+(2+1)×1
=20+19+18+17+16+15+…+2+1
=(20×21)÷2
=420÷2
=210
| 21 |
| 10 |
=64×[1÷(2.1-2.09)]
=64×[1÷0.01]
=64×100
=6400
(2)
| 7 |
| 12 |
| 3 |
| 4 |
| 1 |
| 20 |
| 1 |
| 2 |
=
| 7 |
| 12 |
| 3 |
| 4 |
| 11 |
| 20 |
=
| 7 |
| 12 |
| 1 |
| 5 |
=
| 7 |
| 60 |
(3)
| 1 |
| 3 |
| 1 |
| 15 |
| 1 |
| 35 |
| 1 |
| 63 |
| 1 |
| 99 |
=[(1-
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 5 |
| 1 |
| 7 |
| 1 |
| 7 |
| 1 |
| 9 |
| 1 |
| 9 |
| 1 |
| 11 |
| 1 |
| 2 |
=[1-
| 1 |
| 11 |
| 1 |
| 2 |
=
| 10 |
| 11 |
| 1 |
| 2 |
=
| 5 |
| 11 |
(4)
| 2013+2012×2014 |
| 2013×2014-1 |
=
| 2013+(2013-1)×2014 |
| 2013×2014-1 |
=
| 2013+2013×2014-2014 |
| 2013×2014-1 |
=
| 2013×2014-1 |
| 2013×2014-1 |
=1
(5)20×20-19×19+18×18-17×17+…+2×2-1×1
=(20+19)×(20-19)+(18+17)×(18-17)+(16+15)×(16-15)+…+(2+1)×(2-1)
=(20+19)×1+(18+17)×1+(16+15)×1+…+(2+1)×1
=20+19+18+17+16+15+…+2+1
=(20×21)÷2
=420÷2
=210
点评:注意运算顺序以及运用运算定律或运算技巧,解决问题.
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