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1. ĐỀ
Bài 1: (3,5 điểm) Giải các phương trình sau:
a) 5 x + 6 = 18 8x
b) (2x 3)(x + 1) = 0
c) 2
2 1 2
2 2
x
x x x x
Bài 2: (1,5 điểm) Giải bất phương trình sau:
2 3 5 9 7
7 3 21
x x x
Bài 3: (1.5 điểm) Một miếng đất hình chữ nhật có chu vi 26m, chiều dài hơn
chiều rộng 5m. Tính diện tích miếng đất đó.
Bài 4: (3,5 điểm) Cho tam giác ABC vuông tại A có AB = 6cm, AC = 8cm.
a) Tính độ dài cạnh BC
b) Gọi AH là đường cao (H thuộc cạnh BC). Chứng minh: 𝛥 𝐴𝐵𝐶 ∽ 𝛥 𝐻𝐴𝐶
và 2
AC =BC.HC.
c) CD là đường phân giác của ACB (D thuộc cạnh AB). GọiE là giao điểm của
AH và CD. Chứng minh: AE.AD=HE.BD
- - - Hết - - -
UBND HUYỆN BÌNH CHÁNH
TRƯỜNG THCS BÌNH CHÁNH
ĐỀ CHÍNH THỨC
(Đề có 01 trang)
ĐỀ KIỂM TRA HỌC KỲ 2
NĂM HỌC 2019-2020
MÔN: TOÁN LỚP 8
NGÀY KIỂM TRA:20/6/2020
Thời gian làm bài 90 phút (Không kể thời gian phát đề)
2. UBND HUYỆN BÌNH CHÁNH
TRƯỜNG THCS BÌNH CHÁNH HƯỚNG DẪN CHẤM KIỂM TRA
HỌC KỲ II- NĂM HỌC 2019-2020
MÔN: TOÁN 8
Ngày kiểm tra: 20/6/2020
ĐÁP ÁN ĐỀ CHÍNH THỨC
Bài 1: (3,5 điểm) Giải các phương trình sau:
a) 5 x + 6 = 18 8x
x + 8x = 1865 0,25
7x = 7 0,25
x = 1 0,25
Vậy S = { 1 } 0,25
b) (2x 3)(x + 1) = 0
2x 3 = 0 hoặc x + 1= 0 0,25
2x = 3 hoặc x = -1 0,25
x = 3/2 hoặc x = -1 0,25
Vậy S = 3/2; -1 0,25
c) 2
2 1 2
2 2
x
x x x x
ĐKXĐ: 0; 2
x x
2 1 2
2 ( 2)
x
x x x x
0,25
2 2 2
( 2) ( 2) ( 2)
x x x
x x x x x x
0,25
2 ( 2) 2
x x x
2
2 2 2 0
x x x
2
0
x x 0,25
( 1) 0
x x 0,25
0 0
1 0 1
x x
x x
0,25+0,25
3. Vậy:
0; 1
S
Bài 2: (1,5 điểm) Giải bất phương trình sau:
2 3 5 9 7
7 3 21
x x x
6 3 7( 5) 9 7
21 21 21
x x x
0,5
6 3 7( 5) 9 7
x x x
0,25
6 18 7 35 9 7 0
x x x
0,25
4 24
x
0,25
6
x
Vậy S = x/ 6
x
0,25
Bài 3: (1,5 điểm)
Gọi chiều rộng của miếng đất hình chữ nhật là x (m) , 0 < x < 13 0,25
Chiều dài của miếng đất hình chữ nhật là x + 5 (m) 0,25
Chu vi của hình chữ nhật là 26 m
Ta có phương trình:
2( 5) 26
x x 0,25
2(2 5) 26
4 10 26
x
x
4 16 4
x x (tmđk) 0,25
Suy ra chiều rộng của miếng đất hình chữ nhật là 4m
Chiều dài của miếng đất hình chữ nhật là 4 + 5 = 9 (m) 0,25
Vậy diện tích của miếng đất hình chữ nhật là 4 . 9 = 36 (m2) 0,25
Bài 4: ( 3,5 điểm)
4. Ghi GT-KL đúng: 0,25; Vẽ hình đúng: 0,25 ; Vẽ hình sai không chấm bài 4.
a) Xét ABCvuông tại A có: BC2 = AC2 + AB2( định lý Pytago)
BC2 = 62 + 82 = 102 BC = 10 cm ( 0,5đ+0,5 đ)
b) Chứng minh:
ABC HAC
∽ và 2
AC =BC.HC
Xét ABC và HAC , có
0
BAC=AHC=90 (ABC vuông tại A và AH là đường cao)
C chung
Vậy 𝛥 ABC ∽ 𝛥 HAC (g-g) (0,5đ)
Suy ra
AC BC
=
HC AC
hay
2
AC =BC.HC (đpcm) (0,5đ)
c) Chứng minh: AE.AD=HE.BD
Xét ABC, có CD là phân giác của góc ACB (D thuộc AB)
Nên
BD BC
=
AD AC
(1) (0,25đ)
Xét HAC, có CE là phân giác của góc ACH (E thuộc AH)
Nên
AE AC
=
HE HC
(0,25)
Mà
AC BC
=
HC AC
(cmt) (2) (0,25đ)
Từ (1) và (2) suy ra
AE BD
=
HE AD
hay AE.AD=HE.BD (đpcm) (0,25đ)