问题 解答题
计算:
(1)(3
5
+4
2
)÷(7
6
-2
3
);
(2)
5
4-
11
-
4
11
-
7
-
2
3+
7

(3)
3
2
2
-
3
-
2
3
+3
3
+2
+
18
3
-
2

(4)(
1
3
-
2
-
2
3
-1
+
1
2
+1
)÷
1
1-
2
答案

(1)原式=

(3
5
+4
2
)(7
6
+2
3
)   
(7
6
-2
3
)(7
6
+2
3
)   
=
21
30
+6
15
+56
3
+8
6
282

(2)原式=

5(4+
11
)
42-(
11
)
2
-
4(
11
+
7
(
11
)
2
-(
7
)
2
-
2(3-
7
)
32-(
7
)
2
=4+
11
-(
11
+
7
)-(3-
7
)=1;

(3)原式=

3
2
2
-
3
-
3
(2+
3
3
+2
+
3
2
3
-
2
=-
3

(4)原式=(

3
+
2
(
3
+
2
)(
3
-
2
)   
-
2(
3
+1)
(
3
-1)(
3
+1) 
+
2
-1
(
2
+1)(
2
-1) 
)×(1-
2
),

=(2

2
-2)(1-
2
),

=4

2
-6.

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