题目内容
用简便方法计算
(1)(
÷
+
)×5×7
(2)[
-(
-
)÷
]÷
(3)2002÷
(4)
×(4.85÷
-3.6+6.15×3
)+[5.5-1.75×(1
+
)].
(1)(
| 1 |
| 10 |
| 1 |
| 10 |
| 1998 |
| 1999 |
(2)[
| 3 |
| 5 |
| 2 |
| 3 |
| 1 |
| 4 |
| 5 |
| 6 |
| 1 |
| 10 |
(3)2002÷
| 1998 |
| 1999 |
(4)
| 1 |
| 4 |
| 5 |
| 18 |
| 3 |
| 5 |
| 2 |
| 3 |
| 19 |
| 21 |
分析:(1)、(2)、(4)根据整数、分数、小数四则混合运算的运算顺序:先算乘除,后算加减,有括号的先算括号的,分别进行解答;
(3)把2002拆分为1998+4,然后运用乘法分配律进行简算即可.
(3)把2002拆分为1998+4,然后运用乘法分配律进行简算即可.
解答:解:(1)(
÷
+
)×5×7,
=(1+
)×5×7,
=1×5×7+
×5×7,
=35+34
=69
;
(2)[
-(
-
)÷
]÷
,
=(
-
÷
)÷
,
=(
-
)÷
,
=
÷
,
=1;
(3)2002÷
,
=(1998+4)×
,
=1998×
+4×
,
=1999+4
,
=2003
;
(4)
×(4.85÷
-3.6+6.15×3
)+[5.5-1.75×(1
+
)],
=
×(4.85×
-
+6.15×
)+(5.5-1.75×
),
=
×
(4.58-1+6.15)+(5.5-4.5),
=
×
×10+1,
=9+1,
=10.
| 1 |
| 10 |
| 1 |
| 10 |
| 1998 |
| 1999 |
=(1+
| 1998 |
| 1999 |
=1×5×7+
| 1998 |
| 1999 |
=35+34
| 1964 |
| 1999 |
=69
| 1964 |
| 1999 |
(2)[
| 3 |
| 5 |
| 2 |
| 3 |
| 1 |
| 4 |
| 5 |
| 6 |
| 1 |
| 10 |
=(
| 3 |
| 5 |
| 5 |
| 12 |
| 5 |
| 6 |
| 1 |
| 10 |
=(
| 3 |
| 5 |
| 1 |
| 2 |
| 1 |
| 10 |
=
| 1 |
| 10 |
| 1 |
| 10 |
=1;
(3)2002÷
| 1998 |
| 1999 |
=(1998+4)×
| 1998 |
| 1999 |
=1998×
| 1999 |
| 1998 |
| 1999 |
| 1998 |
=1999+4
| 2 |
| 999 |
=2003
| 2 |
| 999 |
(4)
| 1 |
| 4 |
| 5 |
| 18 |
| 3 |
| 5 |
| 2 |
| 3 |
| 19 |
| 21 |
=
| 1 |
| 4 |
| 18 |
| 5 |
| 18 |
| 5 |
| 18 |
| 5 |
| 54 |
| 21 |
=
| 1 |
| 4 |
| 18 |
| 5 |
=
| 1 |
| 4 |
| 18 |
| 5 |
=9+1,
=10.
点评:此题是考查四则混合运算,要仔细观察算式的特点,灵活运用一些定律进行简便计算.
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