题目内容
用递等式计算,能简算的要简算
(
|
8.1÷〔(4
| ||||||||
1234÷1234
|
9999×2222+3334×3333. |
分析:(1)先把17×25结合在一起,再运用乘法分配律简算;
(2)先算小括号里面的乘法,再算小括号里面的减法,然后算中括号里面的除法,再算中括号里面的加法,最后算括号外的除法;
(3)先根据带分数化成假分数的方法,带分数进行化简,其中把分子运用乘法分配律简算,然后根据分数除法的计算方法求解;
(4)先根据积不变规律把9999×2222变成3333×6666,再运用乘法分配律简算.
(2)先算小括号里面的乘法,再算小括号里面的减法,然后算中括号里面的除法,再算中括号里面的加法,最后算括号外的除法;
(3)先根据带分数化成假分数的方法,带分数进行化简,其中把分子运用乘法分配律简算,然后根据分数除法的计算方法求解;
(4)先根据积不变规律把9999×2222变成3333×6666,再运用乘法分配律简算.
解答:解:(1)(
+
)×17×25,
=(
+
)×(17×25),
=(
×17×25)+(
×25×17),
=(4×25)+(1×17),
=100+17,
=117;
(2)8.1÷[(4
-0.005×700)÷1
+3],
=8.1÷[(4
-3.5)÷1
+3],
=8.1÷[
÷1
+3],
=8.1÷[
+3],
=8.1÷3
,
=
;
(3)1234÷1234
,
=1234÷
,
=1234÷
,
=1234÷
,
=1234×
,
=
;
(4)9999×2222+3334×3333,
=3333×6666+3334×3333,
=3333×(6666+3334),
=3333×10000,
=33330000.
4 |
17 |
1 |
25 |
=(
4 |
17 |
1 |
25 |
=(
4 |
17 |
1 |
25 |
=(4×25)+(1×17),
=100+17,
=117;
(2)8.1÷[(4
1 |
7 |
2 |
7 |
=8.1÷[(4
1 |
7 |
2 |
7 |
=8.1÷[
9 |
14 |
2 |
7 |
=8.1÷[
1 |
2 |
=8.1÷3
1 |
2 |
=
81 |
35 |
(3)1234÷1234
1234 |
1235 |
=1234÷
1234×1235+1234 |
1235 |
=1234÷
1234×(1235+1) |
1235 |
=1234÷
1234×1236 |
1235 |
=1234×
1235 |
1234×1236 |
=
1235 |
1236 |
(4)9999×2222+3334×3333,
=3333×6666+3334×3333,
=3333×(6666+3334),
=3333×10000,
=33330000.
点评:完成本题要注意分析式中数据,运用合适的简便方法计算.
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