题目内容
递等式计算.(能简算的要简算)
2÷
÷
19÷
+3×
10÷[2×(
-
)].
2÷
| 4 |
| 25 |
| 5 |
| 12 |
19÷
| 11 |
| 5 |
| 5 |
| 11 |
10÷[2×(
| 5 |
| 6 |
| 1 |
| 2 |
分析:(1)运用除法性质即可解答,
(2)运用乘法分配律解答,
(3)按照先算小括号里面的,再算中括号里面的,最后算括号外面的顺序即可解答.
(2)运用乘法分配律解答,
(3)按照先算小括号里面的,再算中括号里面的,最后算括号外面的顺序即可解答.
解答:解:(1)2÷
÷
,
=2÷(
×
),
=2÷
,
=30;
(2)19÷
+3×
,
=(19+3)×
,
=22×
,
=10;
(3)10÷[2×(
-
)],
=10÷[2×
],
=10÷
,
=15.
| 4 |
| 25 |
| 5 |
| 12 |
=2÷(
| 4 |
| 25 |
| 5 |
| 12 |
=2÷
| 1 |
| 15 |
=30;
(2)19÷
| 11 |
| 5 |
| 5 |
| 11 |
=(19+3)×
| 5 |
| 11 |
=22×
| 5 |
| 11 |
=10;
(3)10÷[2×(
| 5 |
| 6 |
| 1 |
| 2 |
=10÷[2×
| 1 |
| 3 |
=10÷
| 2 |
| 3 |
=15.
点评:本题主要考查学生依据四则运算顺序正确进行计算,以及正确运用简便方法解决问题的能力.
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