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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