[2016 - 2017] Đề kiểm tra đại số chương 4 lớp 10: Bất đẳng thức bất phương trình

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Giáo viên: Lê Tuấn Cường – Trường Phổ thông liên cấp Olympia
0988.405.543

BÀI KIỂM TRA ĐẠI SỐ 10 CHƯƠNG III
Năm học: 2016 – 2017

ĐỀ SỐ 01

Thời gian: 60 phút (không kể thời gian phát đề)
Ngày: ……………………..…………. Đề gồm: 06 trang
Họ và tên: ………………………………….……………... Lớp: ………………
(Học sinh không sử dụng tài liệu và làm bài vào đề kiểm tra. Giám thị không giải thích gì thêm.)

I. CÂU HỎI TRẮC NGHIỆM

Khoanh tròn vào chữ cái đứng trước đáp án đúng:
Câu 1. (0,5 điểm) Mệnh đề: “Với mọi số thực 𝑎, |𝑎| là số không âm” được viết dưới dạng kí
hiệu toán học là:
A) ∀𝑥 ∈ 𝑹, |𝑥| ≥ 0
B) ∀𝑎 ∈ 𝑹, |𝑎| < 0
C) ∃ 𝑎 ∈ 𝑹, |𝑎| ≥ 0
Câu 2. (0,5 điểm)

𝑎
𝑏

D) ∀𝑎 ∈ 𝑹, |𝑎| ≥ 0
> 0 khi và chỉ khi:

A) 𝑏 > 0

B) 𝑎 > 0

B) 𝑎 + 𝑏 > 0

D) 𝑎. 𝑏 > 0

Câu 3. (0,5 điểm) Giá trị 𝑥 nào là nghiệm của bất phương trình √𝑥 − 2 < 4?
A) 𝑥 = 2
B) 𝑥 = 1
C) 𝑥 = 0
D) 𝑥 = −1
Câu 4. (0,5 điểm) Nhị thức 𝑦 = −5𝑥 + 1 nhận giá trị âm khi:
A) 𝑥 <

1
5

B) 𝑥 < −

1
5

1
C) 𝑥 > 5

C) 𝑥 > −

1
5

𝑥−1>0
Câu 5. (0,5 điểm) Biểu diễn tập nghiệm của hệ bất phương trình {
là:
3−𝑥 ≥0
A)

B)

C)

D)

Câu 6. (0,5 điểm) Tập nghiệm của bất phương trình (𝑥 + 2)(4 − 𝑥) < 0 là:
A) 𝑆 = [−2; 4]
B) 𝑆 = (−2; 4)
C) 𝑆 = (−∞; −2) ∪ (4; +∞)

D) 𝑆 = (−∞; −2) ∩ (4; +∞)

Câu 7. (0,5 điểm) Tam thức bậc hai 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 có hai nghiệm phân biệt 𝑥1 , 𝑥2 và
bảng xét dấu:
𝑥
−∞
0
𝑥1
𝑥2
+∞
𝑓(𝑥)
+
0

−
0
+
Khi đó dấu của hệ số 𝑎, 𝑏, 𝑐 là:
A) 𝑎 < 0, 𝑏 < 0, 𝑐 < 0
C) 𝑎 > 0, 𝑏 > 0, 𝑐 > 0

B) 𝑎 > 0, 𝑏 < 0, 𝑐 < 0
D) 𝑎 > 0, 𝑏 < 0, 𝑐 > 0

1

Giáo viên: Lê Tuấn Cường – Trường Phổ thông liên cấp Olympia
0988.405.543

Câu 8. (0,5 điểm) Tập nghiệm của bất phương trình −𝑥 2 + 𝑥 + 6 ≥ 0 là:
A) 𝑆 = (−2; 3)
B) 𝑆 = [2; 3]
C) 𝑆 = [−2; 3]

D) 𝑆 = (−∞; −2] ∪ [3; +∞)

Câu 9. (0,5 điểm) Cho bất phương trình 𝑥 + 4𝑦 < 10. Một nghiệm của bất phương trình là:
C) (2; 2)
A) (−2; 3)
B) (−2; 0)
D) (2; 4)
Câu 10. (0,5 điểm) Cho hình vẽ.

𝐵(0; 6)

.

. .

𝑂(0; 0)

𝐴(2; 0)

Miền bị tô đậm (không kể biên) là miền nghiệm của bất phương trình:
A) 3𝑥 + 𝑦 ≤ 6
B) 3𝑥 + 𝑦 < 6
C) 3𝑥 + 𝑦 ≥ 6

D) 3𝑥 + 𝑦 > 6

II. BÀI TẬP TỰ LUẬN
Câu 11. Giải các bất phương trình sau:
a) (1,5 điểm) (𝑥 − 1)(−2𝑥 + 4) < 0

b) (1,5 điểm)

2𝑥 2 +5𝑥−3
−𝑥 2 +𝑥+6

≤0

c) (0,5 điểm) |𝑥 − 6| > −2𝑥 − 12
Câu 12. An thiết kế một khu nuôi gà hình chữ nhật với

chu vi là 20 𝑚.
a) (0,5 điểm) Để diện tích khu nuôi gà không dưới 21 𝑚2
thì chiều rộng của nó phải thỏa mãn điều kiện gì?
b) (0,5 điểm) Để diện tích của khu nuôi gà là lớn nhất thì
chiều rộng phải là bao nhiêu? Xác định diện tích lớn nhất.
Câu 13. (0,5 điểm) Cho 𝑎, 𝑏, 𝑐 > 0. Chứng minh rằng:

1
𝑎2 +𝑏𝑐

+

1
𝑏2 +𝑎𝑐

+

1
𝑐 2 +𝑎𝑏

≤

𝑎+𝑏+𝑐
2𝑎𝑏𝑐

---------*Hết*--------2

.

Giáo viên: Lê Tuấn Cường – Trường Phổ thông liên cấp Olympia
0988.405.543
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Giáo viên: Lê Tuấn Cường – Trường Phổ thông liên cấp Olympia
0988.405.543
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Giáo viên: Lê Tuấn Cường – Trường Phổ thông liên cấp Olympia
0988.405.543
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Giáo viên: Lê Tuấn Cường – Trường Phổ thông liên cấp Olympia
0988.405.543
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D

0,5

2

D

0,5

3

A

0,5

4

C

0,5

5

D

0,5

6

C

0,5

7

D

0,5

8

C

0,5

9

B

0,5

10

B

0,5

11a

 (𝑥 − 1)(−2𝑥 + 4) < 0
 Đặt 𝑓(𝑥) = (𝑥 − 1)(−2𝑥 + 4)
 Ta có bảng xét dấu:
𝑥
−∞
1
𝑓(𝑥)
−
0
+
 Tập nghiệm 𝑆 = (−∞; 1) ∪ (2; +∞)


2𝑥 2 +5𝑥−3
−𝑥 2 +𝑥+6

0,25
2
0

+∞

1,0

−

0,25

≤0

2𝑥2 +5𝑥−3
−𝑥2 +𝑥+6
 Ta có bảng xét dấu:
𝑥
−∞
 Đặt 𝑓(𝑥) =

11b

2𝑥 2 + 5𝑥 − 3

0,25
−3

+
_
_

−𝑥 2 + 𝑥 − 6
𝑓(𝑥)

0
l
0

−2

_
_
+

1
 Tập nghiệm 𝑆 = (−∞; −3] ∪ (−2; 2] ∪ (3; +∞)

12b

0
l
0

3

+
+
+

l
0
ll

+∞

+
_
_

1,0

0,25

|𝑥 − 6| > −2𝑥 − 12
𝑥 − 6 > −2𝑥 − 12
⇔[
𝑥 − 6 < 2𝑥 + 12
𝑥 > −2
⇔[
𝑥 > −18
⇔ 𝑥 > −18
 Tập nghiệm 𝑆 = (−18; +∞)

11c

12a

_
l
0 +
ll _

1
2

0,25

0,25

 Giả sử chiều rộng khu nuôi gà là 𝑥 (𝑥 > 0) Đơn vị: 𝑚
Thì chiều dài khu nuôi gà sẽ là 10 − 𝑥 (Điều kiện: 10 − 𝑥 ≥ 𝑥 ⇔ 𝑥 ≤ 5)
 Theo giả thiết ta có: 𝑥(10 − 𝑥) ≥ 21 ⇔ 3 ≤ 𝑥 ≤ 7
 Kết hợp điều kiện, ta có 3 ≤ 𝑥 ≤ 5 (𝑚)

 Ta có: 𝑥(10 − 𝑥) ≤

[𝑥+(10−𝑥)]

4

2

0,25
0,25
0,25

= 25
7

Giáo viên: Lê Tuấn Cường – Trường Phổ thông liên cấp Olympia
0988.405.543
2
0,25
 Diện tích khu nuôi gà đạt giá trị lớn nhất bằng 25 𝑚 ⇔ 𝑥 = 10 − 𝑥 ⇔ 𝑥 = 5
Hay chiều dài bằng chiều rộng và bằng 5 𝑚.
1
1
1
𝑎+𝑏+𝑐
 Cho 𝑎, 𝑏, 𝑐 > 0. Chứng minh rằng: 2
+ 2
+ 2
≤

(∗)
𝑎 +𝑏𝑐 𝑏 +𝑎𝑐
𝑐 +𝑎𝑏
2𝑎𝑏𝑐
1
1
1
1
1
1
Ta có: 2
+ 2
+ 2
≤
+
+
(1)
𝑎 +𝑏𝑐
𝑏 +𝑎𝑐
𝑐 +𝑎𝑏
2𝑎√𝑏𝑐 2𝑏√𝑎𝑐
2𝑐√𝑎𝑏
13

1

1

1

𝑏+𝑐

𝑎+𝑐

𝑎+𝑏

√𝑎𝑏
√𝑎𝑐
Mà
+
+
=
+
+
≤ 2 + 2 + 2
2𝑎√𝑏𝑐 2𝑏√𝑎𝑐
2𝑐√𝑎𝑏
2𝑎𝑏𝑐 2𝑎𝑏𝑐
2𝑎𝑏𝑐
2𝑎𝑏𝑐
2𝑎𝑏𝑐
2𝑎𝑏𝑐
𝑎+𝑏+𝑐
= 2𝑎𝑏𝑐 (2)
Từ (1) và (2), ta có bất đẳng thức (∗) được chứng minh.
Dấu bằng xảy ra ⇔ 𝑎 = 𝑏 = 𝑐
√𝑏𝑐

0,25
0,25

[2016 - 2017] Đề kiểm tra đại số chương 4 lớp 10: Bất đẳng thức bất phương trình

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