Đề kiểm tra 1 tiết Đại 7 lần 1

TRƯỜNG THCS THỊ TRẤN ĐU

Họ và tên: ................................................................
Lớp: 7……

BÀI KIỂM TRA
Môn: Đại số 7
(Thời gian làm bài: 45 phút)

ĐIỂM

ĐỀ BÀI
Câu 1 (3 điểm): Thực hiện phép tính:
5 7 5 5
a) �  �
3 2 4 3

2

�3 � 10 � 18 �
b) 4 �
 �
�2 � 9 �
�
� �
� 5�

4 8
3
c)  �
4  0,8 �

5 11
11

Câu 2 (2,5 điểm): Tìm x biết:
a)

3
4 17
�
x- =
5
5
50

b) 3x  2 = 4

c) 2x  3  3
2

5 1
� 5 1
�
6 5 �
6
�5 6 �
15
2
6 4  �
A�
3 5

3 2 25 �
�
�
12 �
6  12 �
2
4 3
� 4 6
�
Câu 3 (1,5 điểm): Tính giá trị của biểu thức:

Câu 4 (2 điểm): Hưởng ứng phong trào “Nuôi lợn nhựa tiết kiệm”, cuối học kì I, bốn
chi đội 7A, 7B, 7C, 7D thu được tổng số tiền là 780 nghìn đồng. Biết rằng số tiền của
bốn chi đội lần lượt tỉ lệ với 8, 6, 7, 5. Hãy tính số tiền mà mỗi chi đội thu được?
200
300
5 
3


Câu 5 (1 điểm): So sánh:
và
BÀI LÀM
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Câu

Đáp án

a)

Điểm
1đ

5 7 5 5 5 �7 5 � 5 9 15
� � �
�  � � 
3 2 4 3 3 �2 4 � 3 4 4

1đ

2

1

2

�3 � 10 � 18 � 9 10 18
b) 4 �
 � 4 �  �  9  4  13
�2 � 9 �
�
� �
� 5� 4 9 5
4 8
3
4 8 4 3
4� 8 3 � 4
c)  �

4  0,8 �   �
4  �  �
4   �  �
5  4
5 11
11
5 11 5 11
5 � 11 11 � 5
3
4 17
3
57
57 5
19
a) �
x- =
� �
x=
�x=
��x =
5
5
50
5
50
50 3
10
2
x=
b) 3x  2 = 4 � 3x  2 = 4

3
hoặc 3x  2 =  4 � x = 2 hoặc

c)

1đ
0,75đ
1đ

3
x�
2x  3  3 (ĐK:
2)

0,75đ

� 2x  3  9 � 2x  12 � x  6 (thỏa mãn)
Vậy x = 6
2

3

2
2
�5 �1
1� �
5 1
35
51
� 5 1

�
�
�
5 �
6 6 � �
6 5 �
6
�
�
�5 6 �
�6 �4 3 �
15 �
6 �2
4� 3
2
6
4
�
�
A�

�M

�M�

3 5
3 2 25 �
51 25 5 �
�2 13 �6 5  2 2 � 5 �
�

�
�
�
12 �
6  12 �
2
�
�
� �
�
4 3
4
6
� 4 6
�
�
�
6
6
3
�
�
�
�
2

2

1,5đ

2

�35 6 3 �
�7 3 � �4 � 16
� M  � .  �� M  �  � � �
�6 25 5 �
�5 5 � �5 � 25

4

Gọi số tiền mà các chi đội 7A, 7B, 7C, 7D thu được lần lượt là x, y, z, t (nghìn
đồng) (0 < x, y, z, t < 780)
Vì bốn chi đội 7A, 7B, 7C, 7D đã thu được tổng cộng 780 nghìn đồng nên ta
có: x + y + z + t = 780 (1)
Vì số tiền thu được của bốn chi đội lần lượt tỉ lệ với 8, 6, 7, 5 nên ta có:
x y z t
x y z t x + y+zt
x y z t 780
   �    
�    
 30
8 6 7 5 8 6 7 5 8655
8 6 7 5 26
(theo (1))

2đ

Từ đó, ta tính được: x = 240; y = 180; z = 210; t = 150 (thỏa mãn)
Vậy lớp 7A thu được 240 nghìn đồng, lớp 7B thu được 180 nghìn đồng, lớp
7C thu được 210 nghìn đồng, lớp 7A thu được 150 nghìn đồng.

 3
Ta có:
5

300

 3300  33.100   33 

100
100
Mà 27 > 25 > 0 nên 27  25

Vậy

 3

300

  5 

200

100

 27100

 5
;

200

 5200  52.100   52 

100

 25100

1đ