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1. Bài 1 (2điểm):Giải bất phương trình.
a) 3(x – 1) > x - 3 b)
3𝑥−5
5
− 1 ≥
2𝑥−3
3
+ 𝑥
Bài 2 (3điểm):Giải phương trình
a) 2x + 3 = 7 – 3x b) (𝑥 − 20)(3𝑥 + 1) = 0 c) 3
5 8 12 4
3 9 ( 3)
x x
x x x x x
Bài 3 (1,5 điểm) Một ô tô chạy trên quảng đường từ Tp. Hồ Chí Minh đi Bến Tre. Lúc đi
xe chạy với vận tốc 35km/h. Lúc về xe chạy nhanh hơn lúc đi là 5km/h vì vậy thời gian
lúc về ít hơn thời gian lúc đi là 30 phút. Tính quảng đường từ
Tp. Hồ Chí Minh đến Bến Tre?
Bài 4 (1điểm) Bóng của một ống khói nhà
máy trên mặt đất có độ dài 36,9m. Cùng
thời điểm đó 1 thanh sắt cao 2,1m, cắm
vuông với mặt đất có bóng dài 1,62m.
Tính chiều cao của ống khói(làm tròn đến
số thập phân thứ nhất)
Bài 5 (2,5đ) Cho tam giác ABC có 3 góc nhọn, đường cao AH, CK cắt nhau tại I
a)Chứng minh AI. IH = IK.IC
b)Chứng minh: 𝐵𝐾𝐻
̂ = 𝐵𝐶𝐴
̂
c)BI cắt AC tại E. Chứng minh tia KC là tia phân giác 𝐸𝐾𝐻
̂
PHÒNG GD&ĐT QUẬN GÒ VẤP
TRƯỜNG HERMANN GMEINER
ĐỀ CHÍNH THỨC
(Đề chỉ có một trang)
ĐỀ KIỂM TRA HỌC KỲ II
NĂM HỌC 2019 - 2020
Môn: TOÁN 8
Ngày kiểm tra:
Thời gian làm bài: 90 phút
A
B
C
C M
C
D
C
E
2. ĐÁP ÁN – BIỂU ĐIỂM ĐỀ CHÍNH THỨC
Bài 1 Giải bất phương trình
a)(1đ) 3(x – 1) > -5x – 3
3x + 5x > - 3 + 3 0,25
8x > 0 0,25
x > 0 0,25
Vậy bất phương trình có tập hợp nghiệm là: S = {𝑥 𝑥 > 0
⁄ } 0,25
b) (1đ)
3 5 2 3
1
5 3
3(3 5) 15 5(2 3) 15 (0,25)
9 15 15 10 15 15
9 10 15 15 (0,25)
16 15
15
6
x x
x
x x x
x x x
x x x
x
x
(0,25)
Vậy bất phương trình có tập nghiệm là: S =
15
/
6
x x
(0,25)
Bài 2 Giải các phương trình.
a)(1đ) 2x + 3 = 7 – 3x
2x + 3x = 7 – 3 0,25
5x = 4 0,25
x =
4
5
0,25
Vậy phương trình có tập nghiệm là S = {
4
5
} 0,25
b)(1đ) (𝑥 − 20)(3𝑥 + 1) = 0
x – 20 = 0 Hay 3x + 1 = 0 0,25
x = 20 Hay 3x = -1 0,25
x = 20 Hay x =
−1
3
0,25
Vậy phương trình có tập nghiệm S = {20;
−1
3
} 0,25
c)(1đ)
3
5 8 12 4
3 9 ( 3)
5 8 12 4
3 (3 )(3 ) ( 3)
5 8 12 4
(0,25)
3 ( 3)(3 ) ( 3)
: ( 3)( 3) : 0; 3 (0,25)
x x
x x x x x
x x
x x x x x x
x x
x x x x x x
MTC x x x Dk x x
3. Với điều kiện trên, phương trình trở thành:
2 2
2 2
2
5 ( 3) 8 12 ( 4)( 3)
5 15 8 12 3 4 12
5 15 8 3 4 12 12
4 6 0
2 (2 3) 0
0
2 3
0 (loai)
(0,25)
3
(nhan)
2
3
2
x x x x x
x x x x x x
x x x x x x
x x
x x
x
x
x
x
S
(0,25)
Bài 3: 1,5đ
Gọi quảng đường từ TP. Hồ Chí Minh đến Bến Tre là x (km) (x >0) 0,25
Thời gian ôtô lúc đi là
𝑥
35
(h) 0,25
Thời gian ôtô lúc về là
𝑥
40
(h) 0,25
Theo đề ra ta có:
𝑥
35
-
𝑥
40
=
1
2
0,25
Giải phương trình trên tìm ra x = 140 (km) (nhận) 0,25
Vậy quãng đường từ Tp. Hồ Chí Minh đến Bến Tre là 140 (km) 0,25
Bài 4 B
E
2,1
A 36,9 C D 1,62 M
∆ABC đồng dạng ∆DEM (g-g)
=>
𝐴𝐵
𝐷𝐸
=
𝐴𝐶
𝐷𝑀
=>
𝐴𝐵
2,1
=
36,6
1,62
=> 𝐴𝐵 =
36,9 . 2,1
1,62
≈ 47,8𝑚
Vậy ống khói cao gần bằng 47,8 m
4. a) Chứng minh: AI.IH = IK.CI
Xét ∆AIK và ∆CIH có:
𝐴𝐾𝐼
̂ = 𝐶𝐻𝐼
̂ = 900
0,25
𝐴𝐼𝐾
̂ = 𝐶𝐼𝐻
̂ (2 góc đốiđỉnh) 0,25
∆ AIK đồng dạng ∆ CIH (g-g) 0,25
AI.IH = IK.CI 0,25
b) (1đ)Chứng minh: 𝐵𝐾𝐻
̂ = 𝐵𝐶𝐴
̂
Ta chứng minh được:∆BHA và ∆BKC đồng dạng 0,25
BH BA
BK BC
0,25
Xét ∆BHK và ∆BAC có:
BH BA
BK BC
(cmt)
𝐵
̂chung
∆BHK đồng dạng ∆BAC (c-g-c) 0,25
𝐵𝐾𝐻
̂ = 𝐵𝐶𝐴
̂( 2 góc tương ứng bằng nhau) (1) 0,25
c)(0,5đ) Chứng minh: KC là tia phân giác góc 𝐸𝐾𝐻
̂
I là trực tâm của ∆ABC ( do I là giao điểm của đường cao AH, CK)
BE vuông góc với AC tại E
Cm: ∆AKE đồng dạng ∆ACB => 𝐴𝐾𝐸
̂ = 𝐵𝐶𝐴
̂ (2) 0,25
Từ (1)(2) suy ra 𝐴𝐾𝐸
̂ = 𝐵𝐾𝐻
̂
Mà 𝐴𝐾𝐶
̂ = 𝐵𝐾𝐶
̂ = 900
=> 𝐴𝐾𝐶
̂ − 𝐴𝐾𝐸
̂ = 𝐵𝐾𝐶
̂ − 𝐵𝐾𝐻
̂
=> 𝐸𝐾𝐶
̂ = 𝐻𝐾𝐶
̂
=> KC là tia phân giác của góc EKH 0,25