Chuyên đề và bộ đề ôn thi học kỳ 1 Toán 10 năm học 2018 – 2019 – Lê Văn Đoàn

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Trung taõm Hoaứng Gia
56 Phoỏ Chụù P. Taõn Thaứnh Q. Taõn Phuự

Môn Toán Lớp 10
Năm học 2018 2019

MỤC LỤC
Trang
Chuyên đề 1. Parabol & một số vấn đề liên quan ..................................................................... 1
Chuyên đề 2. Giải và biện luận phương trình bậc nhất ........................................................... 5
Chuyên đề 3. Bài toán chứa tham số trong phương trình bậc hai .......................................... 7
Chuyên đề 4. Phương trình quy về phương trình bậc hai ....................................................... 13
Chuyên đề 5. Bất đẳng thức và GTLN, GTNN ......................................................................... 23
Chuyên đề 6. Hệ trục tọa độ ........................................................................................................ 29
Chuyên đề 7. Tích vô hướng và hệ thức lượng ......................................................................... 42
Đề số 01. THPT Bình Hưng Hòa (2017 – 2018) .................................................................. 49
Đề số 02. THPT Trần Phú (2017 – 2018) .............................................................................. 51
Đề số 03. THPT Lê Trọng Tấn (2017 – 2018) ....................................................................... 53
Đề số 04. THPT Bình Tân (2017 – 2018) ............................................................................... 56
Đề số 05. THPT Nguyễn Hữu Cảnh (2017 – 2018) ............................................................ 58
Đề số 06. THPT Trần Quang Khải (2017 – 2018) ................................................................ 61
Đề số 07. THPT Nguyễn Thượng Hiền (2017 – 2018) ....................................................... 63
Đề số 08. THPT Hàn Thuyên (2017 – 2018) ........................................................................ 66
Đề số 09. THPT Nguyễn Chí Thanh (2017 – 2018) ............................................................ 69
Đề số 10. THPT Tây Thạnh (2017 – 2018) ............................................................................ 72
Đề số 11. THPT Chuyên Lê Hồng Phong (2017 – 2018) .................................................... 74
Đề số 12. THPT Nguyễn Thị Minh Khai (2017 – 2018) ..................................................... 77
Đề số 13. THPT Bùi Thị Xuân (2017 – 2018) ....................................................................... 79

Chóc c¸c trß rÌn luyÖn tèt vµ ®¹t kÕt qu¶ cao trong kú thi s¾p ®Õn !

Ôn tập thi học kì 1 lớp 10 năm học 2018 2019

Chuyên đề 1. Parabol & một số bài toán liên quan
b

b
Cn nh: Parabol (P ) : y ax 2 bx c cú nh I ; v trc i xng x
2a
4a
2a
(honh nh). Khi a 0 : th cú dng v a 0 : th cú dng .
1. Tỡm parabol (P ) : y ax 2 4x c, bit
rng (P ) i qua A(1; 2) v B(2; 3).

2. Tỡm parabol (P) : y ax 2 bx 2, bit rng
(P ) i qua A(1;5), B(2; 8).

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3. Tỡm parabol (P ) : y ax 2 bx 3, bit 4. Tỡm parabol (P ) : y ax 2 4x c, bit
(P ) i qua im A(3; 0) v cú trc i
(P ) cú honh nh bng 3 v i qua
xng l x 1.
im A(2;1).
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Biên soạn & giảng dạy: Ths. Lê Văn Đoàn 0933.755.607

Trang - 1 -

¤n tËp thi häc k× 1 líp 10 n¨m häc 2018 – 2019
5. Tìm parabol (P ) : y  ax 2  bx  c, biết 6. Tìm parabol (P ) : y  ax 2  bx  c, biết (P )
(P ) đi qua A(1; 0), B(2;8), C (0; 6).
đi qua điểm A(0;5) và có đỉnh I (3; 4).
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1



1



3




4

0

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Biªn so¹n & gi¶ng d¹y: Ths. Lª V¨n §oµn – 0933.755.607

Trang - 2 -

¤n tËp thi häc k× 1 líp 10 n¨m häc 2018 – 2019
9. Tìm parabol (P) : y  ax 2  bx  c khi biết 10. Tìm parabol (P) : y  ax 2  bx  c khi biết đồ
đồ thị của nó là
thị của nó là

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11. Vẽ parabol (P ) : y  x 2  2x  2. Dựa 12. Vẽ parabol (P ) : y  x 2  4x  5. Dựa vào

vào đồ thị biện luận nghiệm phương
đồ thị biện luận nghiệm phương trình:
trình: 2x 2  4x  m  3  0.

x 2  4x  5  m  0.

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Biªn so¹n & gi¶ng d¹y: Ths. Lª V¨n §oµn – 0933.755.607

Trang - 3 -

¤n tËp thi häc k× 1 líp 10 n¨m häc 2018 – 2019
13. Vẽ parabol (P ) : y  x 2  4x  3. Tìm 14. Vẽ parabol (P ) : y  x 2  4x  5. Dựa
vào đồ thị biện luận nghiệm phương trình:
m để phương trình x 2  4x  m  0 có

2 nghiệm thỏa 0  x1  2  x2.

x 2  4x  5  m  0.

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15. Vẽ parabol (P ) : y  x 2  2x. Suy ra đồ 16. Vẽ (P ) : y  x 2  6x  5. Hãy biên luận
thị hàm số (P ) : y  x 2  2x .

2
nghiệm x  6 x  4  m trên (1; 4].

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Gii. Phng trỡnh m 2x m 9x 3

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m 2x 9x m 3

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(m 2 9)x m 3

()

Vi m 2 9 0 m 3.
Khi m 3 thỡ () tr thnh 0x 6,
suy ra phng trỡnh vụ nghim.
Khi m 3 thỡ () tr thnh 0x 0
phng trỡnh nghim ỳng x .

Vi m 2 9 0 m 3

() x

m3
1

2
m 9 m 3

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Kt lun:

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m 3 : Phng trỡnh vụ nghim.

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m 3 : Phng trỡnh nghim ỳng x . .................................................................................
1
m 3 : Phng trỡnh cú nghim x

m3
3.

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Gii v bin lun: (m2 2m 8)x 4 m. 4.

Gii v bin lun: (4m2 2)x 1 2m x.

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Biên soạn & giảng dạy: Ths. Lê Văn Đoàn 0933.755.607

Trang - 5 -

¤n tËp thi häc k× 1 líp 10 n¨m häc 2018 – 2019
5.

Tìm m để phương trình có nghiệm:

3x  m
x 1

 x 1 

2x  5m  3
x 1

6.

Tìm m để phương trình có nghiệm:

2mx  1



x 1

 2 x 1 

m 1
x 1



Giải. Điều kiện: x  1  0  x  1.

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Quy đồng và bỏ mẫu, phương trình đã cho

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 3x  m  x  1  2x  5m  3

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 2x  6m  2

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 x  3m  1.

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Vì x  1 nên phương trình có nghiệm
2
khi x  3m  1  1  m   
3

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7.

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Tìm tham số m để phương trình sau có 8.
nghiệm nguyên: (m  2)x  m  1.

Giải. Với m  2  0  m  2 thì phương
m  1 (m  2)  3
trình  x 

m 2
m 2

3
 x 1
m 2

Tìm tham số m để phương trình sau có
nghiệm nguyên: m(x  3)  x  m.

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Vì x   nên 3 (m  2)

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m  2  3

m  5


m  2  3
m  1


 
.
m

2

1

m  3


m  2  1
m  1

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9.

Tìm tham số m để phương trình

(m 2  m )x  2x  m 2  1 vô nghiệm.

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10. Tìm tham số m để phương trình

m 2x  4x  m 2  m  2 có nghiệm.

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a 0

khi:

0
(2m 3)2 4(m 2 4) 0

Th x 7 vo phng trỡnh, ta c:

(7)2 7(2m 3) m 2 4 0
m 2
m 2 14m 24 0

m 12

Vi m 12 thỡ phng trỡnh tr thnh

12
()
25
b
Theo Viột: S x 1 x 2 2m 3 v
a
c
P x 1x 2 m 2 4.
a

x 2 27x 140 0 x 7 x 20.

2
2
2
Theo : x 1 x 2 17 S 2P 17

Kt lun:

(2m 3)2 2(m 2 4) 17

Vi m 2 thỡ phng trỡnh tr thnh:

x 2 7x 0 x 0 hoc x 7.

12m 25 0 m

Vi m 2 thỡ nghim cũn li l x 0.

2m 2 12m 0 m 0 hoc m 6.
Vi m 12 thỡ nghim cũn li l x 20. So vi (), giỏ tr cn tỡm l m 0.
2. Cho phng trỡnh x 2 (2m 3)x m 2 4 0. Tỡm tham s m phng trỡnh:
a) Cú 1 nghim 7. Tỡm nghim cũn li.

b) Cú 2 nghim pb x 1, x 2 tha x 12 x 22 17.

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Biên soạn & giảng dạy: Ths. Lê Văn Đoàn 0933.755.607

Trang - 7 -

¤n tËp thi häc k× 1 líp 10 n¨m häc 2018 – 2019
3. Cho phương trình x 2  2mx  m 2  3m  0. Tìm tham số m để phương trình:

a) Có nghiệm kép. Tính nghiệm kép đó.

b) Có 2 nghiệm pb x 1, x 2 thỏa x 12  x 22  8.

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4. Cho phương trình (m  1)x 2  3x  1  0. Tìm tham số m để phương trình:
a) Có 1 nghiệm bằng 3. Tìm nghiệm còn lại. b) Có 2 nghiệm pb x 1, x 2 thỏa x 1  1  x 2 .
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Biªn so¹n & gi¶ng d¹y: Ths. Lª V¨n §oµn – 0933.755.607

Trang - 8 -

¤n tËp thi häc k× 1 líp 10 n¨m häc 2018 – 2019
5. Cho phương trình (2m  3)x 2  2(2m  3)x  1  2m  0. Tìm m để phương trình:
a) Có 1 nghiệm bằng 1. Tìm nghiệm còn b) Có 2 nghiệm phân biệt x 1, x 2 thỏa mãn
lại của phương trình.
(5x 1  1)(5x 2  1)  13x 1x 2  1.

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6. Cho phương trình x 2  4x  m  1  0. Tìm tham số m để phương trình:
a) Có hai nghiệm trái dấu ?
Có hai nghiệm dương phân biệt.

b) Có 2 nghiệm phân biệt x 1, x 2 thỏa mãn

x 1x 2  6  2 x 1x 2 .

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Biªn so¹n & gi¶ng d¹y: Ths. Lª V¨n §oµn – 0933.755.607

Trang - 9 -

¤n tËp thi häc k× 1 líp 10 n¨m häc 2018 – 2019
7. Cho phương trình mx 2  2(m  3)x  m  6  0. Tìm tham số m để phương trình:
a) Có 2 nghiệm phân biệt thỏa

b) Có hai nghiệm trái dấu và có giá trị tuyệt
1 1
 1.
đối bằng nhau.
x1 x2

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8. Cho phương trình mx 2  2x  1  0. Tìm tham số m để phương trình:
a) Có hai nghiệm trái dấu ? Có hai nghiệm b) Có hai nghiệm là độ dài của hai cạnh góc
phân biệt cùng dương ? Có hai nghiệm
vuông trong một tam giác vuông có độ dài
đối nhau ?
cạnh huyền bằng 2.
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Biªn so¹n & gi¶ng d¹y: Ths. Lª V¨n §oµn – 0933.755.607

Trang - 10 -

¤n tËp thi häc k× 1 líp 10 n¨m häc 2018 – 2019
9. Cho phương trình x 2  (m  5)x  m  0. Tìm tham số m để phương trình:
a) Chứng minh phương trình luôn có hai b) Có hai nghiệm phân biệt x1, x 2 thỏa mãn
nghiệm phân biệt. Tìm m để phương
điều kiện x 1  2x 2  5.
trình có hai nghiệm dương phân biệt ?
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10. Cho phương trình x 2  (2m  2)x  m 2  4  0. Tìm tham số m để phương trình:
a) Có nghiệm ? Có hai nghiệm pb dương ?

b) Có hai nghiệm pb x1, x 2 thỏa x 1  2x 2 .

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12. Cho phương trình x 3  2mx 2  2mx  1  0. Tìm tham số m để phương trình:
a) Có ba nghiệm phân biệt ?

b) Có hai nghiệm ?

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Biªn so¹n & gi¶ng d¹y: Ths. Lª V¨n §oµn – 0933.755.607

Trang - 12 -

Ôn tập thi học kì 1 lớp 10 năm học 2018 2019

Chuyên đề 4. Phương trình quy về phương trình bậc hai
1. Gii: 4(2x 2 3x 1)(2x 2 4x 1) 3x 2.

2.

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2
6.
2x 5x 3 2x x 3
2

4.

Gii: (x 1)(x 2)(x 6)(x 12) 6x 2.

Gii:

4x
5x
10
2

9
x 2x 3 x 4x 3
2

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5.

x 2 2x 3 x 2 5x 3 3

Gii: 2
x 4x 3 x 2 6x 3 4

6.

x 2 2x 15
3x
2

Gii: 2
x 4x 15 x 6x 15

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Trang - 13 -

¤n tËp thi häc k× 1 líp 10 n¨m häc 2018 – 2019

7.

Giải: 4x  1  x 2  2x  4.

8.

B  0

Lời giải. Áp dụng A  B  
A  B

A  B


Giải: 3x  5  2x 2  x  3.

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Điều kiện: x 2  2x  4  0.
 4x  1  x 2  2x  4
Phương trình  
2
 4x  1  (x  2x  4)

.................................................................................

x


x
x 2  2x  3  0


 2

x
x  6x  3  0

x


.................................................................................

 1
3
 3  2 3

.

 3  2 3

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Thế các nghiệm vào điều kiện, các nghiệm

.................................................................................

thỏa mãn là x  3 và x  3  2 3.

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9.

Giải: x  2  x 2  4x  2.

10. Giải: x 2  2x  2  x 2  7x  9.

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11. Giải: x 2  4x  2  x  2.

12. Giải: 2x 2  3x  1  1  2x .

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Biªn so¹n & gi¶ng d¹y: Ths. Lª V¨n §oµn – 0933.755.607

Trang - 14 -

¤n tËp thi häc k× 1 líp 10 n¨m häc 2018 – 2019
13. Giải: 3x 2  2x  6  x 2 .

14. Giải: 3x  4  x  2 .

.................................................................................
A  B
Lời giải. Áp dụng A  B  
thì
.................................................................................
A  B
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 3x 2  2x  6  x 2
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3x 2  2x  6  x 2   2
2
3
x

2
x

x

6


.................................................................................
x   1
4x 2  2x  6  0
.................................................................................

  2


x  3
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2x  2x  6  0 : VN o

2
15. Giải: 5x 2  3x  2  x 2  1 .

16. Giải: x 2  2x  2x 2  x  2 .

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2

4x 2  12x  9  3x  2 .

A  B  A  B2

Lời giải. Áp dụng
thì phương trình

18. Giải:

2

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.................................................................................

 x  6x  9  4x  4x  1

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2
 3x 2  10x  8  0  x  4, x   
3

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19. Giải: 2 3  2x  x  1 .

20. Giải:

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Biªn so¹n & gi¶ng d¹y: Ths. Lª V¨n §oµn – 0933.755.607

3x 2  9x  1  x  2 .

Trang - 15 -

¤n tËp thi häc k× 1 líp 10 n¨m häc 2018 – 2019
21. Giải: (x  3). x  1  4x

()

 TH1: Nếu x  1  0  x  1.
() trở thành (x  3)(x  1)  4x

 x 2  2x  3  0  x  1, x  3.
So với x  1, nhận nghiệm x  3.
 TH2: Nếu x  1  0  x  1.
() trở thành (x  3)(1  x )  4x
2

 x  6x  3  0  x  3  2 3.
So với x  1 nhận nghiệm x  3  2 3.
Kết luận: Tập nghiệm S  {3  2 3;3}.
23. Giải:

4x 2  2x  2x  1
4x  3

 2x  1.

22. Giải: (x  1). x  3  4(x  2).
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24. Giải:

x 1
1
2x  1

 2

x
x 1 x  x

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26. Giải: x 2 

1
1
 10  2 x   (ẩn phụ)
2
x
x

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Trang - 16 -

¤n tËp thi häc k× 1 líp 10 n¨m häc 2018 – 2019
27. Giải:

x 2  3x  2  x  3.

B  0 (hay A  0)
Lời giải. A  B  
.

A  B

x  3  0
Phương trình  
 2
x  3x  2  x  3

x  3
x  3


 2

x  4x  1  0
x  2  3



 x  2  3 là nghiệm cần tìm.
29. Giải:

x  1  2 2x  5.

28. Giải:

6x 2  4x  3  x  4  0.

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30. Giải: 3 x  1  x 2  8x  11.

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 2
x  x  1  (3  x )2

x  3
x  3


 2


x  x  1  9  6x  x 2
x  8


7

8
 x  là nghiệm cần tìm.
7

32. Giải:

5x 2  21x  8  x  2.

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33. Giải: 2 3x 2  2x  1  1  3x .

34. Giải: 2x  12x 2  18x  1  2.

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Biªn so¹n & gi¶ng d¹y: Ths. Lª V¨n §oµn – 0933.755.607

Trang - 17 -

¤n tËp thi häc k× 1 líp 10 n¨m häc 2018 – 2019
35. Giải: x 2  5x  4  5 x 2  5x  28  0.

36. Giải: 5 x 2  2x  7  x 2  2x  3.

Lời giải. Đặt t  x 2  5x thì phương trình trở

.............................................................................

thành t  4  5 t  28  0

.............................................................................

t  4  0
 5 t  28  t  4  
25(t  28)  (t  4)2

t  4

t  4

 2
  t  36  t  36
t  17t  684  0



 t  19

x  4
Với t  36  x 2  5x  36  
.
x  9

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 Cách khác: Đặt t  x 2  5x  28  0  t 2.
37. Giải: x 2  3x  3 3x 2  9x  7  1  0.

38. Giải: 2x  x 2  6x 2  12x  7  0.

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39. Giải: (x  3)(1  x )  5 x 2  2x  7.

40. (x  2)(x  3)  x 4  2x 3  x 2  2  2.

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2x  1  0
Lời giải. Điều kiện 
 x  3.

x  3  0


.............................................................................

Phương trình  2x  1  2  x  3

.............................................................................

 ( 2x  1)2  (2  x  3)2

.............................................................................

 2x  1  4  4 x  3  x  3
x  4
x  0

 4 x  3  x  

x  12 .
16(x  3)  x 2


So với điều kiện và thử lại, suy ra S  {4;12}.

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45. Giải:

x  4  1  x  1  2x .

46. Giải: 2 3x  1  x  1  2 2x  1.

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Trang - 19 -

¤n tËp thi häc k× 1 líp 10 n¨m häc 2018 – 2019
47. Giải: (x  3) x 2  5x  4  2x  6.

Lời giải. Điều kiện x 2  5x  4  0

48. Giải: (x  3) x 2  4  x 2  9.

()

.............................................................................

PT  (x  3) x 2  5x  4  2(x  3)  0

.............................................................................

 (x  3) x 2  5x  4  2(x  3)  0

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 (x  3)( x 2  5x  4  2)  0
x  3  0
x  3
  2
  2
 x  5x  4  2
x  5x  4  4
x  3
 
. Thế các nghiệm vào điều
x  0, x  5
kiện (), nghiệm cần tìm là x  0, x  5.

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49. Giải: (x  1) 2x  3  x 2  4x  3.

50. Giải: (2x  1) x  1  2x 2  7x  3.

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Lời giải. Điều kiện 
 x  2.

x  1  0


.............................................................................

PT  (x  2)(x  1)  2 x  2  ( x  1  2)

.............................................................................

 x  2.( x  1  2)  ( x  1  2)  0
 ( x  1  2).( x  2  1)  0
 x 1  2
x  1  4


 
 x  3.
x

2

1
 x 2  1


So với điều kiện, nghiệm cần tìm là x  3.

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Trang - 20 -

¤n tËp thi häc k× 1 líp 10 n¨m häc 2018 – 2019
53.

3  x  6  x  3  (3  x)(6  x ).

54.

x  2  x  2  2 x 2  4  2x  2.

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55. Giải:

2x 2  x  9  2x 2  x  1  x  4.

56. Giải:

x 2  15  3x  2  x 2  8.

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Trang - 21 -

¤n tËp thi häc k× 1 líp 10 n¨m häc 2018 – 2019
2x  y  7  0

57. Giải hệ:  2

y  x 2  2x  2y  4  0


x 2  y 2  6x  2y  0

58. Giải hệ: 

x  y  8  0


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
 2
x y  y 2x  2


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Trang - 22 -

Ôn tập thi học kì 1 lớp 10 năm học 2018 2019

Chuyên đề 5. Bất đẳng thức Giá trị lớn nhất & giá trị nhỏ nhất
1.

Chng minh rng vi mi x , ta luụn 2.

Chng minh rng vi mi a, b, c

cú x 4 4x 3 0.

thỡ a 2 b 2 4 ab 2a 2b.

Gii. Thờm bt a v hng ng thc, tc

.............................................................................

x 4 4x 3 (x 2 )2 2x 2 1 2(x 2 2x 1)

.............................................................................

(x 2 1)2 2(x 1)2 0, x .
x 2 1 0

Du " " xy ra khi
x 1.
x 1 0

Nhn xột: i vi bi toỏn x , ta nờn
s dng hng ng thc a v dng:
2

2

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2

A B C 0 : luụn ỳng v du " "
xy ra khi A B C 0.
3.

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.............................................................................

Chng minh rng vi mi x , y thỡ ta 4.
2

2

Chng minh rng vi mi a, b, c
thỡ cú a 2 b 2 c 2 12 4(a b c ).

luụn cú x y xy 3x 3y 3 0.
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Chuyên đề và bộ đề ôn thi học kỳ 1 Toán 10 năm học 2018 – 2019 – Lê Văn Đoàn

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