1. Trang chủ
  2. » Khoa Học Tự Nhiên

Hướng dẫn giải bài tập phương trình mũ

12 638 0

Đang tải... (xem toàn văn)

Tài liệu hạn chế xem trước, để xem đầy đủ mời bạn chọn Tải xuống

THÔNG TIN TÀI LIỆU

Thông tin cơ bản

Định dạng
Số trang 12
Dung lượng 1,2 MB

Nội dung

2 Traàn Quang Lost time is never found again30... 3 Traàn Quang Time goes, you say?. Alas, time stays, we goPhương pháp4: Đưa về phương trình tích và Đặt ẩn phụ không toàn phần... 4 T

Trang 1

1 Traàn Quang Time goes, you say? Ah, no! Alas, time stays, we go

PHÖÔNG TRÌNH MUÕ

Phương pháp 1: Đưa về cùng cơ số

1 x2  3x 2  x 1

2 x2  6x 5/2 

3 3  x 2 4x  1 / 243

4

32 0 25 128,

5 5x34 1252x8 3/

6 4x 82x1

7 2 3x x 1.5x 2  12

8 2 9 27

    

   

   

x x

9   x

x x

2

1 1

3 2

10 2 2 5 2 1

3 x x  27 x

11 4.9x1  3 22x1

12 (2 3)3x 1 (2 3)5x 8

13

x 1

( 5 2) ( 5 2)

14

1 3

3 1

( 10 3) ( 10 3)

x x

x x

15

3

1 2 1 3

2x 4 x 8x 2 2 0 125 ,

16 2x3 4x x3 0 125,  4 23

17

4

5 0,2 125.0, 04

x

18

4

x

19

2( 1)

1

2 x 3.2x 7

20 2 33x x 23x1.3x1 192

21

2

x

x  x  

22 2x21 3x2 3x21 2x22

23

2 2

 

 

 

x

x

24 (x2 x 1)x2 1 1

25

2 3

3 3

1

3

x

 

 

xx  x  x

27 x 4  x 3  x x 2

28 x 1  x 2  x 4  x 3

29 2x  2x 1  2x 2  3x 3x 1  3x 2

30  2  9 3 2

2 2 2

2

2

1 2 cos

2

x

Phương pháp 2: Đặt ẩn phụ

1 2 16 15 4 8 0 xx 

2 9x8.3x 7 0

3 32x 8 4.3x 5 27 0 

4 4x1 2x4 2x2 6

4x 6.2x  8 0

6

2

7 6.0, 7 7 100

x

x

7

3

64x 2 x 12 0

4xx 5.2x  x   6 0

9 4 x2 1610.2 x2

25 x 5 4.x

11 2 log ( 3 2 16) log ( 3 2 16) 1

12 9x2 x 110.3x2 x 2  1 0

13 22x 6  2x 7  17  0

14     10

53 x  103 x 84

.4 21 13.4 2

x   x

16 3 2cos 1 cos

4 x7.4 x 2 0

17 4xx25 12 2 x 1 x25  8 0

18 32x32x 30

5 5 26

20 D032x2x 22  x x2 3

21 9sin2x  9cos2x  10

5.2 x 3.2 x 7 0

23 5x1 5.0,2x2  26

4x4 x 3.2xx

25

sin cos

26 cos 2 cos2

4 x4 x 3

28 9  x2 2x 1 34.152x x 2  252x x 2 1 0

29

Trang 2

2 Traàn Quang Lost time is never found again

30

6.9x13.6x 6.4x 0

25x  9x  15x 0

32 A063.8x 4.12x 18x 2.27x 0

33 2.22x 9.14x 7.72x 0

34 6.9 13.6xx6.4x 0

35 27x12x 2.8x

36 4.3x 9.2x  5.6x/2

37

15.25x 34.15x 15.9x 0

38 25x12.2x6, 25.0,16x 0

39 125x  50x  23x1

40 8x18x 2.27x

41 2  3 x  2 3x  14

42  2  3 x 2  3x 2

(2  3)  (2  3)   4 0

45 B07( 2 1)  x ( 2 1)  x 2 2  0

46 ( 3  2)x  (7 4 3)(2  3)x  4(2  3)

47 3 5 x  3 5x 7.2x 0

48 (3  5)x  16(3  5)x  2x 3

50 26 15 3 2 7 4 3 2 2 3 1

2

52 7 3 5  x 7 3 5x 14.2x

53 ( 2  3) x  ( 2  3) x  4

54

55 5 24 x  5 24x 10

56 7 5  21 x 5  21x  2x 3

58 52 6tanx 52 6tanx 10

x

61 3

16

64 8x 1 8.0,53x 6.2x 2 125 24.0,5 x

65 53x 9.5x 27.(125x 5x)64

67 4.33x 3x1  19x

68 5.32x17.3x1 16.3x 9x1 0

69 4.23x 3.2x  122x2 24x2

10(2 3)

72 23x17.22x7.2x 2 0

Phương pháp 3:

Sử dụng tính đơn điệu của hàm số

1 4x9x 25x

2 3x 4x  5x

3 3x    x 4 0

x

4 1

5 3x2 2x 2  2 2xx2

6 2x 3x/2 1

7 3.16x  2.8x  5.36x

8

x x

x x

20 2 4 5

9 ►

1/

2, 9

x

11  3  2x  ( 3  2 )x  ( 10 )x

5 2 2 3 5

x

2

4 2   2 

6 2 1 7

9  

3x   ( x  4)3x   1 0

17 ►3 2x x  3x  2 x  1

x

         

x

         

20 2013x 2015x 2.2014x

Trang 3

3 Traàn Quang Time goes, you say? Ah, no! Alas, time stays, we go

Phương pháp4:

Đưa về phương trình tích và Đặt ẩn phụ

không toàn phần

6 1 3

2   

2 8.3x + 3.2x = 24 + 6x

5x 7x 175x350

4 D062x2x 4.2x2x22x 40

x   xx   x

7 12.3x 3.15x 5x120

8 4x2 3 2x 4x2 6 5x 42x2 3 7x 1

9 32x 2x 9.3x 9.2x 0

10 2 32x 2.12x0

x x

11 9x 2.x2.3x 2x50

12 3.25 2 3 10.5 2 3 0

x

x

13 4x2x 21x2 2 x12 1

14 22x21 9.2x2x 22x2 0

15 x2.2x  8 2x2 2x2

17 x2.2x12x 3 2x2.2x 3 42x1

18 9x2x2 3 x2x 5 0

x2  3 2 x2 1 2 0

21 3.16x2 (3 x  10)4x2    3 x 0

4 x  2 x  2x  16  0

23 ►54x 1 25.53x 26.52x 5x 1 5 0

24 ►81x 4.27x 10.9x 4.3x 1  3 0

Phương pháp5: Lôgarit hóa

Phương pháp 6:

Hàm đặc trưng và PP đánh giá

1 22x1  32x  52x1  2x  3x1  5x2

2

2

1 1 2

2

2

x

3 2x1 4x   x 1

1 2

4   2  

x

x x

5 2x23.cosx 2x24.cos3x 7.cos3x

6 2 3 x1  74 3xx1

7 2x  x 1

8 3x 2x 1

9

x

x

 

 

 

1

1

10 32 22 2 3 1 2 1  1

x

x x x x x

x

cos

2

2 2

12 3x2 cos2x

13 3cos2x  3 x2

4

x

 

 

 

2 x x

x x

17 4x  6x  25 x  2

18 3x 5x 6 2

x

19 3x 2x 3x2

20 3x 2x  3x2 2

21 2x2 3x2 4x2 3

22 2x2 3x2 7x2 8x2  41x2

23 2x2 4.10x2  7 3x2

24 8.3x 3.2x 24 6 x

PHÖÔNG TRÌNH LOGARIT

Phương pháp 1: Đưa về cùng cơ số

1 log (52 x 1) 4

5 log x(x 2x65)2

3 logx2 x 2x1

log (x  1) log (x 1)

9

4 1 3 2

xx

10 23 32

11

2 5 6 3

12

2

5 3x x 1

13 2x4.3x2 22x1.33x2

14

3

2

3 2 6

x

x x 

15

1

x

x x

2x 3x x 3x x 2x

1 5 x x18x  100

2

2

x

x x

4 4.9x13 22x1

5

2 2

2xx.3x 1,5

6

2 1 1

5 2 50

x

x x

 

7 34x 43x

8 57x 75x

Trang 4

4 Traàn Quang Lost time is never found again

5 log (5 x 3) log ( 25 x6)

6 log 3 2 1

2

x   xx

7 log2 6 x log 32  x1

2

2 log 36

log 4

x  x

 

9 log3xlog9xlog27x11

10 log3xlog (3 x2) 1

11 log (9 x 8) log (3 x26) 2 0

log (x  3) log (6x10) 1 0

log( 1) log( 2 1) log

2

log xlog xlog x 5

log (1 x 1) 3log x400

5 5

log (4x 6) log (2 x 2) 2

3

log x log x log 3x  3

log (x   1) log 5  log (x  2) 2log (x 2)

3

log logxlog(logx 2)0

log 1 x 3log 1  x 2 log 1x

22 log (4 x 3) log (2 x  7) 2 0

8

log (x 2) 6log 3x 5 2

2

2

log x 1 log (3x)log (x1)

2

2 log 2x2 log 9x 1 1

26 log (4 x2).log 2 1x

27 2

9 log 27.logx 4

log (x ) log (x ) 3

log (x  3x 2)  log (x  7x 12)   3 log 3

2log xlog x.log ( 2x 1 1)

31 log5xlog3xlog 3.log 2255 9

log (x1)  2 log 4 x log (x4)

log  ( x   1 x)  log  ( x   1 x)  0

x

2 1

log x 4 2

5 log log ( 25) 0

5

x

x x

38 log55x 4 1 x

2

3

x

x

log x 1   2 log 4  x log 4 x

1 1

7 2

x

x x

3

1 log 3 1 2

2

x   xx

2log ( x   2) log ( x  4)  0

44 xlog 9  9logx  6

45 5logx  50  log5

x

46

2

log

1

x

47

3

log x x x x

log 10 1 log 4

log 2 log (3 2) 2 log 5

x

  

49 log29 2 

1 3

x

x

50 log (4x4) x log (2x 13)

2

2

1

52 D07

1 log (4 15.2 27) 2log 0

4.2 3

x

2

log 3 2 2 log16 log 4

x x   x

55

1

1

2 log 2 1 log 3 log 3 27 0

2

x

x

log 4x 1 log 2x 6

Trang 5

5 Traàn Quang Time goes, you say? Ah, no! Alas, time stays, we go

1 4 log 1 7 2.

1

2 1

2 1

x x

x x

 

1 log 2 log 1 log 1 3 log

2

x

log x x 1 log x x 1

log x x 1 log x x 1

2

log ( x   3) log (6 x  10) 1 0  

2

log ( x  2)  log x  4 x   4 9

3

4

1 log

x

x

x

3

1

66 log (22 x  4)   x log2 2x  12   3

Phương pháp 2: Đặt ẩn phụ

1 log2x  log x3   2 0

2 log22x2log2 x 2 0

3 3 log2xlog (8 ) 32 x  0

1

4

5 log (33 x1).log (33 x1 3) 6

6 log (55 x1).log (525 x1 5) 1

4

3

8 log (4 x1)2log (2 x1)3 25

9 4 log9xlog 3x 3

10 log(x1)16  log (2 x  1)

11 1 log ( 2 x 1) logx14

12 log 16x2 log2x643

13 log (33 x2).log 3 12x

14 log (22 ) log 2 x 2

x  xx

5 log x log ( )x 1

x

16 log 2 2log 4x  2x log 2x8

2 3 27

16log x x3log x x 0

8

x

19 4 logx/2 x  2 log4x x2  3 log2x x3

20 log3/x2 log  23 x  1

21 3log 16x  4log16x  2log2x

22 logx/2 x2  40 log4x x  14 log16x x3  0

log x(6x  5x  1) log x(4x  4x   1) 2 0

log x (9 12  x 4x )  log x (6x  23x 21)  4

log x(2x   x 1) logx (2x1) 4

26 4log(10 )x 6logx 2.3log(100x2)

27 log 2 2 log 6 2 log 4 2 2

4 xx 2.3 x

28 log2/x2 log 4 2 x3

29 log 4 logx x2 22 x  12

30 log2 x  4 log4 x   5 0

31 log23 x log23 x  1 5 0

32 log 3 logx  3xlog x3 log 3 x 1 / 2

2

34 log 3 x log 3 3 logx  33 3  6

35 log log4 2 x  log log2 4 x  2

38 1 log(2 1) 2

2

1 log( 1)

1 log ( 1)

x

x x

39 1 log 2x  4 log4 x 2 4

40 CĐ08log (22 x 1) 6log2 x  1 2 0

Phương pháp 3.

Sử dụng tính đơn điệu của hàm số

1 log (5 x  3) 4 x

2 log (3 x  3)   8 x

3 0,5

log

log (x x 6) x log (x 2) 4

Trang 6

6 Traàn Quang Lost time is never found again

log(x  x 12) x log(x 3) 5

6 log 2

10 log22x 1 log34x22

11 log3x 2 log2x 3 2

12 log (3 x 1) log (25 x 1) 2

13 4(x 2) log  3x  2 log 2x 3 15(x 1)

x

2

2 log ( 2) 6

2 1

Chứng minh :  2 

nghiệm thực phân biệt

Phương pháp 4:

Phương trình tích và đặt ẩn phụ không

toàn phần

2

2 log x (x 1)log2 x 6 2x

2

3 2log log log  2 1 1

3 3 2

9 xx x 

4

1

5   2     

6 log2x.log3x 1 log2xlog3x

7 log2x.log3xlog2 xlog3x

8 log ( 4 xx2 1).log ( 5 xx2  1) log ( 20 xx2 1)

9

2

log x(x3).log x  x 2 0

10

2

(log 3) 4 log 0

11 log .log log log2 3

3 3 3

2 x xxx

12 x 2log23x 1  4 x 1log3x 1 16  0

13 log2 1  5log3 1 2 6 0

3 x  xx  x 

14  2  2 2   2   2 

15 log2 x2log7x2log2 x.log7 x

16

2

log x(x12) log x  11 x 0

17

2

log 2( 1) log 4 0

18 2 3 5

log log log

log log log log log log

19

3

log log log log

2 3

x

20 (2 2)log2xx(2 2)log2x 1 x2

x xx x xx xx

22 log log2x 3x log3x3 log2x 3

23  2    2

3 log 1 2 logx 2 3 logx 2 2 log 1

xx    x   x

log xxlog x3 x/ 2 2 log x3 log  x

y

Phương pháp 5: Mũ hóa

4 18

5

7 2 log log  

3

21 log log  

4

log ( x  2 x   1) log ( x  2 ) x

5 2log tan3 xlog sin2 x

6 log2xlog3x1

7 log5xlog3xlog15

8 log log2 2 x log log3 3 x

9 log log2 3 x log log3 2 x log log3 3 x

10 log log log2 3 4 x log log log4 3 2 x

11 log 92 2.3log2x log 32

12

3 2

3 log log 3

13

x

x

x

log 5

5 log

3 10

14

2

2

15

2 log 3log 4,5 2log

10

16 log 2 log 3 2 3log 2

2

2 log

2

Phương pháp 6:

PP đánh giá và dùng hàm đặc trưng

1

1

2

2

3 2

log 1

3 2 xx2  log (5 x2    x 4) log5x

4 ln(x2  x 1) ln(2x2 1) x2x

Trang 7

7 Traàn Quang Time goes, you say? Ah, no! Alas, time stays, we go

5

2

2

2 2

1

6

1

x x

x

7

2

2

2015 2

3

x x

x x

x x

8

2

2

3 2

3

9

2x .log ( x   1) 4x.(log x   1 1)

10

sin( )

x

3

8

log (4 4 4)

   

1

1

x

x

x

x

x x

Biện luận

A02: Cho phương trình

log x log x 1 2m 1 0 (1) (m là

tham số)

a) Giải phương trình (1) khi m = 2

b) Tìm m để phương trình (1) có ít nhất

một nghiệm thuộc đoạn 1 ; 3 3

Trang 8

8 Traàn Quang Lost time is never found again

BAÁT PHÖÔNG TRÌNH MUÕ

1.3x22x27

2.5 1

2

log3

x

x

3

1

( 5 2) ( 5 2)

x

4

2 2 16

( ) ( )

xx x

5

1 2

1 1

2

16

x

x   

   

6.2x2x12x2 3x 3x13x2

7

2 3 2 2 3 3 2 3 4

2x x 3x x 5x x 12

8

( 10 3) ( 10 3)

9

2

1 3 xx 9

10

2 5 6 2

3

3 x x x

11  2 2 7

2 x x 1

12 2x2x1 3x 3x1

13 ( 2 1) 1 ( 2 1) 1

x

14

2

1

3

3

x x

x x

 

   

15 9x2.3x 3 0

16 22x62x7170

17 2 x23 x 9

18 2.49x7.4x9.14x

19 5.2x7 10x 2.5x

20 4x3.2xx4 x1

21

6.9 xx13.6 xx6.4 xx0

22 4x2x.3 x31 x 2 3x2 x2x6

23

2

1

x x

x x

   

24 2x27x 9

25

12 3

1 3 3

1 2









26 16loga x 43.xloga4

27  xxxx  xx

1 5 3 2

1

28 32x8.3xx4 9.9 x4 0

29 3x24 24.3x2 1

x

30 4 16 2.log48

1 

x

31    

52 8

2

1 2 1

xx

x

32

0 2 2

1 21 2

3 2 1

2     

x x

33

9.4x 5.6x 4.9x

34 8.3 x x 9 x 1 9 x

4 4

35  2 1x 1

x x

36

6.9 xx13.6 xx6.4 xx 0

37 25x3x2 2x2x.3x 25x3x2 4x2.3x

38 4 2 3 . 31 2.3 . 22 6

x x x

39     1

1 1

1 5 2

x x

40

2 2

2

15 34 9

25 xx   xx   xx

41 5  5 10

2

5 log log xx

x

42

 log log 2 1

1 1

3

3 5 12

, 0

x x

x x

43 3x24 24.3x21

x

44

1 5

9 log 3 3 log log 2 log 3 3 2

2 1

x x

2 log log xx

x

12 5

3

2log2x log2 x1 log2x2 

47 9 2 1 7.3 2 1 2

2 2

x

48 log2x4 32

x

49 4 .2 3.2 .2 8 12

2 2

x x

x

50 3x122x1122x 0

BAÁT PHÖÔNG TRÌNH LOGARIT

1

2 1

1 8 log

2

x

x x

2

 3 2 1 log 2

2

1 xx 

3 log93x24x21log33x24x2

1 6

5 log 3

x

x

x

1 1

1 2 log4 

x x

6

2 4

1

 x

x

7

x x

x

2 1 2

2

3 2 2 1 4

2 9log 32 4.log

8 log





Trang 9

9 Traàn Quang Time goes, you say? Ah, no! Alas, time stays, we go

8

2

2 1 2

2

9 logx24x51

10 log2xlog2x84

11   1 8 2 6 1 0

5 log 1 3

5

x

x x

x

12 4x2 16x7log3x30

13

log 2

1 2 log

6 5 log

3 1 3

1 2

3 xx  x  x

14

1 1

3 2

log3 

x x

15 log2xlog3x1log2x.log3x

16

 1

log

1 1

3 2

log

1

3 1 2

3

1 xx  x

17 log2x2 3x2

18 log5x2 11x432

19

 4 6 2

log 2

2

1 xx 

20

x1log 2x

2

1

21   log  1

1 2

9 6

2

2

x x

x x

22

8

2 18 log 2 18

 

23 log log93x 9 1

x

24

 1 log  1 log 5  1

3 1 3

1 x  x  x

25 log3xx23x1

26 logxx2 x21

x x

x x

x

x 7 12 2 1 14 2 24 2 logx2

 

28 logx5x28x32

29

log ( x  2 x   1) log ( x  2 ) x

30

4

3 16

1 3 log 1 3

log

4 1



x

31 log0,3 x5x10

32

0 3

5 2

11 4 log

11 4 log

2

3 2

11

2 2

x x

x x

x

33

9

4

1 log

log x  x 

1 2

5 4 log 2 

x

x

x

35

2 1 2

1 1 log 1 2 log

2

1

x

36

1 log 2

1 log

2

3

2 3

4 xx

1 4 log

5

2

x x

38 log 2 1 log2 2 2

2 x    x

39

 

x

x x

2 log 1 1 2

40

 

0 8 2

1 log 2 2

1

x x x

41

x x

8 1 2

8

1 1 4log log

9

42 log log24x6 1

x

43 log3xlog5xlog3 x.log5x

44

3 log

8 9 log

2

2

x

x x

45

2 log

1 log

2 2 log

2 2

x

46

1 1

1 2

x

x

x

47

1 log 1

log 1 3

2

3 

x x

48

1 log

2 log

4

3

4 3

2 xx

49

5  3

log

35 log 5

3

x x

1 1 2 2 logx2x1 x2  x 

51

2 log

2

1 log7 x 7 x

52

0 4

3

1 log 1 log

2

3 3 2

x x

x x

53

2 2 lg lg

2 3

lg 2

x

x x

3 logx 5x 18x16 2

55 log2x64logx2163

56

0 log

4 1 2

2

1 xx

Trang 10

10 Traàn Quang Lost time is never found again

57 log x2 x2log x12

58

1 1 2

1 log

1 log

.

2

5 1 5

2



x x

59 log42x2 3x21log22x2 3x2

60

x x

x

2 1 2

2

3 2 2 1 4

8 log





61

log 3 5

3 log

2 1 2

2 xx   x

62

9 1 2 log 3 7

2 1 1

2

1 x    x 

63 log x3 log x 3 03  

64

3

log log x 5 0

65

 2     

5

log x 6x 8 2 log x 4 0

66

 

3

5 log x log 3

2

67 log 2.logx 2x2.log 4x2 1

68 log2x 3   1 log2x 1 

69

8

2

2 log (x 2) log (x 3)

3

70

2 log log x 0

71 log5 3x 4.log 5 1 x 

72

 

2

3 2

x 4x 3

x x 5

73

2

log x log x 1

74  2  

2x log x 5x 6 1

75 

2

2 3x

x 1

5

2

76

    

3

x 1

x 2

77 2  

log x log x 0

78

2 16

1 log 2.log 2

log x 6

log x 4 log x 9 2 log x 3

80

2 log x 4 log x 2 4 log x

81 log22x1log34x 22

Trang 11

11 Traàn Quang Time goes, you say? Ah, no! Alas, time stays, we go

HEÄ PHÖÔNG TRÌNH MUÕ - LOGARIT

I Hệ phương trình mũ

1 

 

1

3

3 5 4

y

x

y

x

x y x

y

y x y

x

x

y

y

x

3

log

32

4

3 



4 2 3

9

9

3

1

y

x x

y

x

y x y

4

 

y y y

x

x

y y

x

1

3 5

2

3

3

3

2

2

2

5 

2 lg

lg

1

2 2

y x

xy

6 

7 2

3

77 2

3

2

2

y

x

y

x

7 

19 3

2

6 3

2

2

3

1

1 y

x

y x

8

3 3

3

3 5 5

5

y x y x

y x

y x

9 

y

y y

x

x x x

2 2

2 4

4 5 2

1

2 3

10



0 6

8

1 3

4 4

4 4

y x

x y

y x

y x

11    

3 lg 4

lg

lg lg

3 4

4 3

y x

y x

12

 



1 log

3 log 4

2

5 log

x y y x

y

x x y

13 

1 1

3

2 3 2 2

2

3 2

1 3

x xy

x

x y y

x

14



4 2 1 2 2 3

4 2 1 2 2 3

x y

y x

15

 



x y 3x 2y 3

16

 

x y (x y) 1

17



2x y

3 2 77

18

  

 



2 2 12

x y 5

19

1

4 2

2 2

x

x x x

y

 

1

3 9 18

y y

x x

20

21

1

2x y 2x

x y



22

y x



II.Hệ phương trình lôgarit

1)

16

2 log log

3

3

2 2

y

x

xy x

y y

x

2)    

3 lg 4

lg

lg lg

3 4

4

3

y x

y x

3)    

1 log

3 log 2 log log

7

2 2

2

y x

y x

4)

8

5 log

log

2

xy

y

y

5) 

1 log log

4

4 4

log log8 8

y x

y

6) 

1 log log

4

4 4

log log 8 8

y x

y

7)



1 log

4 3 3 1 1

3x y

x

x

8)

x x

y x

4 2 2

4

2 4 4

2

log log log

log

log log log

log

9)

    



3

8 1 log

2 log

14 2

2

y xy x

y x

10)

2 2 3 log

2 2 3 log

x y

y x

y x

11)



4 5 3 log 5 3 log

4 5 3 log 5

3 log

x y y

x

x y y

x

y x

y x

Ngày đăng: 20/04/2015, 21:38

Xem thêm

TỪ KHÓA LIÊN QUAN

TÀI LIỆU CÙNG NGƯỜI DÙNG

TÀI LIỆU LIÊN QUAN

w