题目内容

用简便方法计算
(1)(
1
10
÷
1
10
+
1998
1999
)×5×7
(2)[
3
5
-(
2
3
-
1
4
5
6
1
10

(3)2002÷
1998
1999

(4)
1
4
×(4.85÷
5
18
-3.6+6.15×3
3
5
)+[5.5-1.75×(1
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.
解答:解:(1)(
1
10
÷
1
10
+
1998
1999
)×5×7,
=(1+
1998
1999
)×5×7,
=1×5×7+
1998
1999
×5×7,
=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
+4×
1999
1998

=1999+4
2
999

=2003
2
999


(4)
1
4
×(4.85÷
5
18
-3.6+6.15×3
3
5
)+[5.5-1.75×(1
2
3
+
19
21
)]

=
1
4
×(4.85×
18
5
-
18
5
+6.15×
18
5
)+(5.5-1.75×
54
21
)

=
1
4
×
18
5
(4.58-1+6.15)+(5.5-4.5)

=
1
4
×
18
5
×10+1,
=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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