题目内容
计算:
(1)
×
×
;
(2)
÷[3.75×(1.2-
)];
(3)2.42÷
+4.58×
+3÷4;
(4)
×
+
×
;
(5)(
+
)×27÷
;
(6)
×2005.
(1)
| 3 |
| 16 |
| 3 |
| 4 |
| 8 |
| 27 |
(2)
| 3 |
| 4 |
| 1 |
| 5 |
(3)2.42÷
| 4 |
| 3 |
| 3 |
| 4 |
(4)
| 9 |
| 35 |
| 2 |
| 17 |
| 9 |
| 17 |
| 33 |
| 35 |
(5)(
| 2 |
| 39 |
| 5 |
| 27 |
| 1 |
| 39 |
(6)
| 2003 |
| 2004 |
考点:整数、分数、小数、百分数四则混合运算
专题:运算顺序及法则
分析:(1)首先约分,然后分子、分母分别相乘;
(2)首先计算小括号内的减法,然后计算中括号内的乘法,把3.75化成分数是3
=
,最后计算除法;
(3)除以
等于乘
,3÷4=
,然后根据乘法分配律,提取
,计算即可;
(4)根据乘法分配律,提取
,计算即可;
(5)除以
等于乘39,然后根据乘法分配律,首先计算乘,然后求和;
(6)2005=2004+1,然后根据乘法分配律,展开,计算即可.
(2)首先计算小括号内的减法,然后计算中括号内的乘法,把3.75化成分数是3
| 3 |
| 4 |
| 15 |
| 4 |
(3)除以
| 4 |
| 3 |
| 3 |
| 4 |
| 3 |
| 4 |
| 3 |
| 4 |
(4)根据乘法分配律,提取
| 9 |
| 35 |
(5)除以
| 1 |
| 39 |
(6)2005=2004+1,然后根据乘法分配律,展开,计算即可.
解答:
解:(1)
×
×
=
×
×
=
(2)
÷[3.75×(1.2-
)]
=
÷(3.75×1)
=
÷3
=
×
=
(3)2.42÷
+4.58×
+3÷4
=2.42×
+4.58×
+
=
×(2.42+4.58+1)
=
×8
=6
(4)
×
+
×
=
×(
+
)
=
×
=
(5)(
+
)×27÷
=(
+
)×27×39
=
×27×39+
×27×39
=2×27+5×39
=54+195
=249
(6)
×2005
=
×(2004+1)
=
×2004+
×1
=2003+
=2003
| 3 |
| 16 |
| 3 |
| 4 |
| 8 |
| 27 |
=
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 3 |
=
| 1 |
| 24 |
(2)
| 3 |
| 4 |
| 1 |
| 5 |
=
| 3 |
| 4 |
=
| 3 |
| 4 |
| 3 |
| 4 |
=
| 3 |
| 4 |
| 4 |
| 15 |
=
| 1 |
| 5 |
(3)2.42÷
| 4 |
| 3 |
| 3 |
| 4 |
=2.42×
| 3 |
| 4 |
| 3 |
| 4 |
| 3 |
| 4 |
=
| 3 |
| 4 |
=
| 3 |
| 4 |
=6
(4)
| 9 |
| 35 |
| 2 |
| 17 |
| 9 |
| 17 |
| 33 |
| 35 |
=
| 9 |
| 35 |
| 2 |
| 17 |
| 33 |
| 17 |
=
| 9 |
| 35 |
| 35 |
| 17 |
=
| 9 |
| 17 |
(5)(
| 2 |
| 39 |
| 5 |
| 27 |
| 1 |
| 39 |
=(
| 2 |
| 39 |
| 5 |
| 27 |
=
| 2 |
| 39 |
| 5 |
| 27 |
=2×27+5×39
=54+195
=249
(6)
| 2003 |
| 2004 |
=
| 2003 |
| 2004 |
=
| 2003 |
| 2004 |
| 2003 |
| 2004 |
=2003+
| 2003 |
| 2004 |
=2003
| 2003 |
| 2004 |
点评:考查了运算定律与简便运算,四则混合运算.注意运算顺序和运算法则,灵活运用所学的运算律简便计算.
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