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(1)125
7
8
÷(11
3
4
-4
3
20
+2.25-0.35);
(2)1+
1
2
-
1
3
+
1
4
-
1
5
+
1
6
-
1
12

(3)
31
37
÷
25
111
+
36
37
×4
11
25
+4.44÷4
5
8

(4)
1
3
8
+
2
3
×1
1
14
(4
1
12
+3.625)÷20
5
9

(5)(3.25+2.1+5.4)(2.1+5.4+8)-(3.25+2.1+5.4+8)(2.1+5.4);
(6)1-
2
1×(1+2)
-
3
(1+2)×(1+2+3)
-…
9
(1+2+…+8)×(1+2+…+8)
分析:(1)把小数化为分数,括号内运用加法交换律与结合律简算;
(2)除
1
5
外,先通分计算,变为1+
1
2
-
1
5
,再次通分计算;
(3)把除法变为乘法,运用乘法分配律简算;
(4)分子分母同时计算;
(5)运用设数法解答;
(6)根据题目特点,运用拆项的方法,通过加减相互抵消,求出结果.
解答:解:(1)125
7
8
÷(11
3
4
-4
3
20
+2.25-0.35)
=125
7
8
÷(11
3
4
-4
3
20
+2
1
4
-
7
20

=125
7
8
÷[(11
3
4
+2
1
4
)-(4
3
20
+
7
20
)]
=125
7
8
÷[14-4
1
2
]
=
1007
8
÷
19
2

=
1007
8
×
2
19

=
53
4


(2)1+
1
2
-
1
3
+
1
4
-
1
5
+
1
6
-
1
12

=1+
6
12
-
4
12
+
3
12
-
1
5
+
2
12
-
1
12

=1+
1
2
-
1
5

=1+
5
10
-
2
10

=1
3
10


(3)
31
37
÷
25
111
+
36
37
×4
11
25
+4.44÷4
5
8

=
31
37
×
111
25
+
36
37
×
111
25
+
111
25
×
8
37

=(
31
37
+
36
37
+
8
37
)×
111
25

=
75
37
×
111
25

=9

(4)
1
3
8
+
2
3
×1
1
14
(4
1
12
+3.625)÷20
5
9

=
1
3
8
+
2
3
×
15
14
(4
1
12
+3
5
8
9
185

=
1
3
8
+
5
7
185
24
×
9
185

=
117
56
3
8

=
39
7


(5)设3.25+2.1+5.4=a,2.1+5.4=b,得
(3.25+2.1+5.4)(2.1+5.4+8)-(3.25+2.1+5.4+8)(2.1+5.4)
=a×(b+8)-(a+8)×b
=ab+8a-ab-8b
=8a-8b
=8×(a-b)
=8×3.25
=26

(6)1-
2
1×(1+2)
-
3
(1+2)×(1+2+3)
-…
9
(1+2+…+8)×(1+2+…+8)

=2×[
1
1×2
-
1
2×3
+
1
2×3
-
1
3×4
+…+
1
9×10
-
1
10×11
]
=2×[
1
1×2
-
1
10×11
]
=2×[
1
2
-
1
110
]
=2×
54
110

=
54
55
点评:要想算得快、算得巧,就要仔细注意观察题目中数字构成的特点和规律,运用运算定律或运算技巧,进行简便计算.
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