题目内容
(2008?高邮市)用递递等式计算.
4÷
-
÷4
105×13-270÷18
÷[(
+
)×
].
4÷
4 |
5 |
4 |
5 |
105×13-270÷18
1 |
3 |
2 |
3 |
1 |
5 |
1 |
13 |
分析:(1)4÷
-
÷4,注意运算顺序,先算除法,再算减法;
(2)105×13-270÷18,注意运算顺序,先算乘、除法,再算减法;
(3)
÷[(
+
)×
],按运算顺序计算,先算小括号里的,再算中括号里的,最后算中括号里的.
4 |
5 |
4 |
5 |
(2)105×13-270÷18,注意运算顺序,先算乘、除法,再算减法;
(3)
1 |
3 |
2 |
3 |
1 |
5 |
1 |
13 |
解答:解:(1)4÷
-
÷4,
=4×
-
×
,
=5-
,
=
;
(2)105×13-270÷18
=1365-15,
=1350;
(3)
÷[(
+
)×
],
=
÷[(
+
)×
],
=
÷[
×
],
=
÷
,
=
×15,
=5.
4 |
5 |
4 |
5 |
=4×
5 |
4 |
4 |
5 |
1 |
4 |
=5-
1 |
5 |
=
24 |
5 |
(2)105×13-270÷18
=1365-15,
=1350;
(3)
1 |
3 |
2 |
3 |
1 |
5 |
1 |
13 |
=
1 |
3 |
10 |
15 |
3 |
15 |
1 |
13 |
=
1 |
3 |
13 |
15 |
1 |
13 |
=
1 |
3 |
1 |
15 |
=
1 |
3 |
=5.
点评:此题重点考查学生对运算顺序的掌握情况,以及用递等式计算的能力.
练习册系列答案
相关题目