题目内容
用合理的方法计算.
÷[
×(
+
)]
5.6×3
-4
×3.75+0.88×37
(947
+94
+9.476+0.9476)÷0.1111
.
1 |
16 |
1 |
25 |
2 |
5 |
1 |
10 |
5.6×3
3 |
4 |
2 |
5 |
1 |
2 |
(947
3 |
5 |
19 |
25 |
2011×(3.4×69+3.5) |
3.5×69-3.4 |
分析:(1)按运算顺序进行计算,先算小括号内的,再算中括号内的,最后算括号外的;
(2)把分数改为小数,根据积不变的规律,把0.88×37.5改为0.88×37.5,然后运用乘法分配律的逆运算简算;
(3)把分数改为小数,根据积不变的规律,原式变为(9476×0.1+9476×0.01+9476×0.001+9476×0.0001)÷0.1111,括号内运用乘法分配律的逆运算简算;
(4)通过观察,此题中的数字很难有特点,然后运用乘法分配律的逆运算,把3.4×69+3.5变为3.4×(70-1)+3.5,把3.5×69-3.4变成(3.4+0.1)×69-3.4,再通过进一步计算,得出结果.
(2)把分数改为小数,根据积不变的规律,把0.88×37.5改为0.88×37.5,然后运用乘法分配律的逆运算简算;
(3)把分数改为小数,根据积不变的规律,原式变为(9476×0.1+9476×0.01+9476×0.001+9476×0.0001)÷0.1111,括号内运用乘法分配律的逆运算简算;
(4)通过观察,此题中的数字很难有特点,然后运用乘法分配律的逆运算,把3.4×69+3.5变为3.4×(70-1)+3.5,把3.5×69-3.4变成(3.4+0.1)×69-3.4,再通过进一步计算,得出结果.
解答:解:(1)
÷[
×(
+
)],
=
÷[
×(
+
)],
=
÷[
×
],
=
÷
,
=
×50,
=
;
(2)5.6×3
-4
×3.75+0.88×37
,
=5.6×3.75-4.4×3.75+0.88×37.5,
=5.6×3.75-4.4×3.75+8.8×3.75,
=(5.6-4.4+8.8)×3.75,
=10×3.75,
=37.5;
(3)(947
+94
+9.476+0.9476)÷0.1111,
=(947.6+94.76+9.476+0.9476)÷0.1111,
=(9476×0.1+9476×0.01+9476×0.001+9476×0.0001)÷0.1111,
=[(0.1+0.01+0.001+0.0001)×9476]÷0.1111,
=0.1111×9476÷0.1111,
=9476;
(4)
,
=
,
=
,
=
,
=2011.
1 |
16 |
1 |
25 |
2 |
5 |
1 |
10 |
=
1 |
16 |
1 |
25 |
4 |
10 |
1 |
10 |
=
1 |
16 |
1 |
25 |
1 |
2 |
=
1 |
16 |
1 |
50 |
=
1 |
16 |
=
25 |
8 |
(2)5.6×3
3 |
4 |
2 |
5 |
1 |
2 |
=5.6×3.75-4.4×3.75+0.88×37.5,
=5.6×3.75-4.4×3.75+8.8×3.75,
=(5.6-4.4+8.8)×3.75,
=10×3.75,
=37.5;
(3)(947
3 |
5 |
19 |
25 |
=(947.6+94.76+9.476+0.9476)÷0.1111,
=(9476×0.1+9476×0.01+9476×0.001+9476×0.0001)÷0.1111,
=[(0.1+0.01+0.001+0.0001)×9476]÷0.1111,
=0.1111×9476÷0.1111,
=9476;
(4)
2011×(3.4×69+3.5) |
3.5×69-3.4 |
=
2011×[3.4×(70-1)+3.5] |
[(3.4+0.1)×69-3.4] |
=
2011×(3.4×70+0.1) |
3.4×69+3.5 |
=
2011×(3.4×70+0.1) |
3.4×70+0.1 |
=2011.
点评:此题除了考查学生对运算顺序的掌握,而且考查了学生运用所学知识,进行灵活巧算的能力.
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