Bài tập toán cao cấp I - GVHD Phạm Thị Ngũ

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Bài tập toán cao cấp I - GVHD Phạm Thị Ngũ

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Tài liệu tham khảo về bài tập môn toán cao cấp A1 dành cho sinh viên hệ cao đẳng - đại học tham khảo học tập củng cố kiến thức môn học. Tài liệu hay và bổ ích. GVHD: Phạm Thị Ngũ.

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  1. ̀ ̣ ́ ́ Bai tâp toan cao câp I GVHD: Phan Thị Ngũ Chương I: ĐAI SỐ TUYÊN TINH ̣ ́ ́ Bai tâp 1: Cho 2 ma trân A và B ̀ ̣ ̣ 2 1 − 1 2 1 1 A = 3 0 − 1   B = 1 2 1    1 1 0    1 0 0    ́ Tinh: a) At – 2BA + 3Bt b) 2AB - 3BA + 2ABt ́ c) Cho f(x) = x3 + 3x – 2 Tinh f(A) , f(B) ́ Ta co: 2 3 1 2 1 1 8 3 − 3 1 0 1 B = 1 2 0 BA = 9 2 − 3 A = t  ; t   ;    − 1 − 1 0   1 1 0   2 1 − 1   − 16 − 6 6 6 3 3  4 4 3 − 2BA =  − 18 − 4 6   ; 3B = 3 6 0 t   ; AB = 5 3 3    − 4 − 2 2   3 3 0    3 3 2    8 8 6 − 2 0 0  2AB = 10 6 6   ;  0 −2 0  −2=     6 6 4  0  0 − 2   − 24 − 9 9  4 3 2 6 3 3 − 3BA = − 27 − 6 9   ; AB = 5 2 3  t  ; 3B = 3 6 3    − 6 − 9 3   3 3 1     3 0 0    8 6 4 6 1 − 3  12 3 − 7  2 AB = 10 4 6 t   ; AA = 5 2 − 3   ; A = 13 2 − 7  3    6 6 2   5 1 − 2    11 3 − 6   6 3 − 3 6 4 3 19 14 10 3A = 9 0 − 3   ; BB = 5 5 3   ; B = 18 15 10 3   3 3 0    2 1 1   6 4 3    − 8 0 10  − 8 5 19   A − 2 BA + 3B = − 14 2 7  t t   ; 2 AB − 3BA + 2 AB = − 7 4 21 t   −2 0 2    6 9 9   ̃ Nguyên Phan Thanh Lâm MSV: 071250510319 Trang 1/18
  2. ̀ ̣ ́ ́ Bai tâp toan cao câp I GVHD: Phan Thị Ngũ 16 6 − 10 23 17 13 f ( A) = A + 3 A − 2 = 22 0 − 10 3   ; f ( B ) = B + 3B − 2 =  21 19 13 3   14 6 − 8    9 4 1   ̀ ̣ ́ Bai tâp 2: Tinh A-1B + ABt + At +2 khi 1 0 − 1  1 1 0 a) A = 3 1  2  ; B =  0 1 1   0 − 1 1     − 1 1 0   0 1 2 2 0 1  b) A = 3 − 1 1    ; B = 1 − 1 2    1 2 1   1 1 − 1   − 1 0 2 0 2 1  2 1 A= 3 B = 3 − 1 1  c)  ;    − 1 − 1 − 2   0 − 2 0    CÂU A: 1 0 − 1  1 1 0 A = 3 1  2 ; B =  0 1 1   0 − 1 1     − 1 1 0   Vì A = 6 ⇒ ∃A −1 ̀ ́ Tim A-1 theo 2 cach: ́  Cach 1:  1 2 2 3 1+ 3 3 1 C11 = (−1)1+1  =3 ; C12 = (−1)1+ 2   = −3 ; C13 = ( −1) 0 = −3 − 1 1  1 0  − 1   0 − 1 − 1 1 2 + 3 1 0 C 21 = (−1) 2+1  ; =1 C 22 = ( −1) 2 + 2   = 1 ; C 23 = (−1) 0 =1 − 1 1  1 0  − 1  0 − 1 − 1 1 3+ 3 1 0 C 31 = (−1) 3+1   =1 ; C 32 = (−1) 3+ 2   = −5 ; C 33 = (−1) 3 =1 1 2  2 3  1   3 1 1    3 − 3 − 3 3 1 1  6 6 6  1 3 1 5 ⇒ C = 1 1  1  → A −1 = − 3 1 − 5 = −     6 6 − 6  6 1 − 5 1    − 3 1 1   3 1 1    − 6 6 6    ́  Cach 2 ̃ Nguyên Phan Thanh Lâm MSV: 071250510319 Trang 2/18
  3. ̀ ̣ ́ ́ Bai tâp toan cao câp I GVHD: Phan Thị Ngũ 1 0 − 1 1 0 0 h1 → h1 1 0 − 1 1 0 0 h1 → h1 1 0 − 1 1 0 0 3 1  − 3h + h → h 0 1 2 0 1 0 h → h 5 − 3 1 0 2 0 1 5 − 3 1 0   1 2 2 2   0 − 1 1 0 0 1 h3 → h3   0 − 1 1  0 0 1 h2 + h3 → h3 0 0 6 − 3 1 1    1  3 1 1  h3 + h1 → h1 1 0 0 3 1 1 1 0 0 6 6   h → h1 6 6 6 6 6 1  5 h2 → h2 0 1 5 − 3 1 0  − 5h + h → h 0 1 0 − 3 1 −   3 1 1 3 2 2  6 6 6 1 0 0 1 −  h3 → h3 0 0 1 − 3 1 1  h3 → h3   6 6 6 6   6 6 6   ́ Ta co:  2 5 1  6 6 1 3 0 1 0 − 1  1 − 1 − 1  2 6 7 1 A t =  0 1 − 1 ; B t = 1 1 1  ; AB t =  4      3 − 2 ; A −1 B =   −   6 6 6 − 1 2 1    0 1 0    − 1 0 − 1    − 4 1 1  6 −  6 6   26 17 5  6 −  6 6  26 29 17  Vây A B + AB + A + 2 =  6 −  −1 t t ̣ 6 6  − 16 11 13   6  6 6   CÂU B: 0 1 2 2 0 1  A = 3 − 1 1    ; B = 1 − 1 2    1 2 1   1 1 − 1   Vì A = 12 ⇒ ∃A −1 ̀ ́ Tim A-1 theo 2 cach: ́  Cach 1: − 1 1 3 1 3 − 1 C11 = (−1)1+1  = −3 ; C12 = (−1)1+ 2  = −2 ; C13 = ( −1)1+3  =7 2 1 1 1  1 2 1 2 0 2 0 1 C 21 = (−1) 2+1  =3 ; C 22 = ( −1) 2 + 2  = −2 ; C 23 = (−1) 2+3  =1 2 1 0 1 1 2 1 2 0 2 0 1 C 31 = (−1) 3+1  =3 ; C 32 = (−1) 3+ 2  =6 ; C 33 = (−1) 3+3  = −3 − 1 1  3 2  3 − 1  3 3 3  − − 3 − 2 7  − 3 3 3   12 12 12  1  2 2 6  ⇒ C =  3 − 2 1  → A −1 = − 2 − 2 6  = −   −  12    12 12 12  3  6 − 3  7  1 − 3  7  1 3 −   12  12 12   ́  Cach 2 ̃ Nguyên Phan Thanh Lâm MSV: 071250510319 Trang 3/18
  4. ̀ ̣ ́ ́ Bai tâp toan cao câp I GVHD: Phan Thị Ngũ 0 1 2 1 0 0 h1 ↔ h3 1 2 1 0 0 1 h1 → h1 1 2 1 0 0 1 3 − 1 1 0 1 0 h → h 3 − 1 1 0 1 0 − 3h + h → h 0 − 7 − 2 0 1 − 3   2 2  1 2 2  1 2 1 0 0 1 h1 ↔ h3 0 1 2 1 0 0 h3 → h3     0 1  2 1 0 0    h1 → h1 1 2 1 0 0 1 2 1 0 0 1  h1 → h1 1  1  2 1 3  2 1 3  − h2 → h2 0 1 0 −  h2 → h2 0 1 0 −  7 0 1 7 7 7  7 7 7   2 1 0 0  h3 − h2 → h3   12 1 3 h →h 3 3 0 0 1 −    7 7 7    3 2 1  1 0 0 h1 → h1 1 2 1 0 0 1  − 2h + h → h  7 7 7  1  3  3  2 1 2 1 2 1 h2 → h2 0 1 0 −  h2 → h2 0 1 0 −   7 7 7   7 7 7  7 7 1 3 h3 → h3 0 0 1 7 1 3 h3 → h3 0 0 1 −  −  12   12 12 12     12 12 12    3 2 1  3  3 3 3  h1 → h1 1 0 0  − 7 h + h1 → h1 1 0 0 − − 7 7 7 12 12 12  2  2 2 6  3  2 2 6  − h3 + h2 → h2 0 1 0 − −  h2 → h2 0 1 0 − −  7  12 12 12   12 12 12  0 0 7 1 3  h3 → h3 0 0 7 1 3 h3 → h3 1 − 1 −    12 12 12    12 7 7 ́ Ta co:    0 3 1  2 1 1  2 3 − 1 0 0 0  8  A t = 1 − 1 2 ; B t = 0 − 1 1  ; AB t = 7 6 1  ; A −1 B = 0       − 1  12  2 1 1    1 2 − 1   3 1 2    1 − 4 1   12     4 6 0  92  Vây ̣ A −1 B + AB t + A t + 2 = 8 2  12  6 20 6  12    CÂU C: − 1 0 2 0 2 1  2 1 A= 3 B = 3 − 1 1   ;    − 1 − 1 − 2   0 − 2 0    Vì A = −3 ⇒ ∃A −1 ̀ ́ Tim A-1 theo 2 cach: ́  Cach 1: 1 3 2 3  2 1 C11 = (−1) 1+1   =1 ; C12 = (−1)1+ 2   =1 ; C13 = (−1) 1+3   = −1  − 1 − 2  − 1 − 2 − 1 − 1 ̃ Nguyên Phan Thanh Lâm MSV: 071250510319 Trang 4/18
  5. ̀ ̣ ́ ́ Bai tâp toan cao câp I GVHD: Phan Thị Ngũ 0 2  − 1 2  − 1 0  C 21 = (−1) 2 +1   = −2 ; C 22 = (−1) 2 + 2  =4 ; C 23 = (−1) 2 + 3   = −1  − 1 − 2 − 1 − 2 − 1 − 1 0 2  − 1 2  − 1 0 C 31 = (−1) 3+1   = −2 ; C 32 = (−1) 3+ 2  =7 ; C 33 = (−1) 3+3   = −1 1 3  2 3   2 1  1 2 2  − 3   1 1 − 1  1 − 2 − 2  3 3 1 1 4 7 ⇒ C = − 2 4 − 1 → A −1 = −  1    4 7  = −   3 −3 −  3 3 − 2 7 − 1   − 1 − 1 − 1   1   1 1   3  3 3   ́  Cach 2 − 1 0 2 1 0 0 − h1 → h1 1 0 − 2 − 1 0 0 2 1  2h + h → h 0 1 3 0 1 0 1 7 2 1 0  2 2   − 1 − 1 − 2 0 0 1 2h3 + h2 ↔ h3 0 − 1 − 1 0 1 2       h1 → h1 1 0 − 2 − 1 0 0 h1 → h1 1 0 − 2 − 1 0 0 h2 → h2 0 1 7 2 1 0 h2 → h2 0 1 7 2 1 0    1 1 1  h2 + h3 → h3 0 0 6 2 2 2 1  h → h 0 0 1  3 3 3 3 3 6    1 2 2  h1 → h1 1 0 − 2 − 1 0 0  2h + h → h 1 0 0 − 3 3 3  1  7 7 3 1 1 4 1 4 − 7h3 + h2 → h2 0 1 0 − − −  h 2 → h2 0 1 0 − − −   3 3 3  3 3 3 h3 → h3 0 0 1 1 1 1  h3 → h3 0 0 1 1 1 1    3 3 3     3 3 3   ́ Ta co:  6 8 1   3 − − 1 2 − 1  0 3 0   2 −1 0  3 3   4 16 5 A t =  0 1 − 1 ; B t = 2 − 1 − 2 ; AB t =  5      8 − 2 ; A −1 B = −  −   3 3 3  2 3 − 2   1 1  0   − 4 − 4 2     1 1 2   3 −  3 3    5 2  5 − − 3 3  11 49 14  ̣ Vây A −1 B + AB t + A t + 2 =  −   3 3 3 − 5 − 4 8   3  3 3  ̃ Nguyên Phan Thanh Lâm MSV: 071250510319 Trang 5/18
  6. ̀ ̣ ́ ́ Bai tâp toan cao câp I GVHD: Phan Thị Ngũ Bai tâp 3: Giai cac hệ phương trinh sau ̀ ̣ ̉ ́ ̀  4 x1 + 2 x2 − x3 =8  3x1 + x2 + 3x4 =5  1.   x1 − x2 + 4 x3 − 2 x4 = −1  2 x1 + 3x2 + x3 + x4  =8  − 2 x1 + 2 x2 + 4 x3 =0  x1 + 3 x2 − x4 =2  2.  4 x1 + 3x2 − 6 x3 + 2 x4 =1   x1 − x2 − 2 x3 =0  − x1 + x3 − 2 x4 + x5 =7  2 x1 − 2 x3 + 3 x4 =8  3.  3 x1 − 3 x2 + 6 x4 − 3 x5 = 10  x1 + x2 − x3 + x4 − x5  =0 ̃ Nguyên Phan Thanh Lâm MSV: 071250510319 Trang 6/18
  7. ̀ ̣ ́ ́ Bai tâp toan cao câp I GVHD: Phan Thị Ngũ CÂU 1  4 x1 + 2 x2 − x3 =8  3x1 + x2 + 3x4 =5    x1 − x2 + 4 x3 − 2 x4 = −1  2 x1 + 3x2 + x3 + x4  =8 Ta có 4 2 − 1 0 8  h3 → h1 1 − 1 4 − 2 − 1 3 1 0 3 5  h4 → h2 2 3 1 1 8 A=     1 − 1 4 − 2 − 1 h2 → h3 3 1 0 3 5     2 3 1 1 8  h1 → h4 4 2 − 1 0 8 h1 → h1 1 − 1 4 −2 − 1 h1 → h1 1 − 1 4 − 2 − 1 0 5 0 5 − 7 h2 → h2 −7 5 10  − 2h1 + h2 → h2  5 10  4   − h + h → h 0 0 − 22  5 0 − 3h1 + h3 → h3 0 4 − 12 9 8 5 2 3 3  5    6  11  − 4h1 + h4 → h4 0 6 − 17 8 12  − h + h → h 0 0 1 − 0 4  4 3 4  2  h1 → h1 1 − 1 4 − 2 − 1 0 5 −7 5 10  h2 → h2   22 h3 → h3 0 0 − 5 0  5  5  192  h3 + h4 → h4 0 0 0 − 0 22   44  ⇒ r(A) = r( A ) = 4 vây hệ phương trinh có nghiêm duy nhât. ̣ ̀ ̣ ́ Tư đó ta có hệ phương trinh đã cho tương đương vơi hê: ̀ ̣  x1 − x 2 + 4 x3 − 2 x 4 = −1  5x − 7 x + 5x = 10 x1 = 1  2 3 4 x = 2  22  2 (1) ⇔  − x3 + 5 x 4 =0 ⇔  5  x3 = 0  192 x4 = 0   − x4 =0  44 ̃ Nguyên Phan Thanh Lâm MSV: 071250510319 Trang 7/18
  8. ̀ ̣ ́ ́ Bai tâp toan cao câp I GVHD: Phan Thị Ngũ CÂU 2  − 2 x1 + 2 x2 + 4 x3 =0  x1 + 3 x2 − x4 =2   4 x1 + 3x2 − 6 x3 + 2 x4 =1   x1 − x2 − 2 x3 =0 ́ Ta co: − 2 2 4 0 0 h4 → h1  1 − 1 − 2 0 0 1 3 0 −1 2 h2 → h2  1 3 0 −1 2 A=    4 3 −6 2 1  h3 → h3  4 3 −6 2 1      1 −1 − 2 0 0 h1 → h4 − 2 2 4 0 0 h1 → h1 1 − 1 − 2 0 0 h1 → h1 1 −1 − 2 0 0  0 2 −1 2  − h1 + h2 → h2 0 4 2 − 1 2 h → h2  4   7 2 3 15 5 − 4h1 + h3 → h3 0 7 2 2 1  − h2 + h3 → h3 0 0 − −    4  2 4 2 2h1 + h4 → h4 0 0 0 0 0 h4 → h4 0  0 0 0 0   ⇒ r(A) = r( A ) = 3 vây hệ phương trinh có vô số nghiêm. ̣ ̀ ̣ Tư đó ta có hệ phương trinh đã cho tương đương vơi hê: ̀ ̣   x1 − x 2 − 2 x3 = 0  (2) ⇔ 4 x 2 + 2 x3 − x 4 = 2 (*)  3 15 5  − x3 + x 4 = −  2 4 2 ̣ ̀ ́ ̣ ̀ ́ ́ Chon x4 lam biên phu; x1, x2, x3 lam biên chinh. Cho x4 = α vơi α là tham số tuỳ ́ y.  x1 − x 2 − 2 x3 =0  4 x + 2 x − x  x1 = α + 2  2 3 4 =2    4 (*) ⇔  3 15 5 ⇔  x 2 = −4α −  − 2 x3 + 4 x 4 =− 2  3   5 5   x4 =α  x3 = 2 α + 3  Vây hệ phương trinh có vô số nghiêm có nghiêm tông quat la: ̣ ̀ ̣ ̣ ̉ ́ ̀  4 5 5   α + 2;−4α − ; α + ; α  vơi α tuỳ ý  3 2 3  ̃ Nguyên Phan Thanh Lâm MSV: 071250510319 Trang 8/18
  9. ̀ ̣ ́ ́ Bai tâp toan cao câp I GVHD: Phan Thị Ngũ CÂU 3:  − x1 + x3 − 2 x4 + x5 =7  2 x1 − 2 x3 + 3 x4 =8   3 x1 − 3 x2 + 6 x4 − 3 x5 = 10  x1 + x2 − x3 + x4 − x5  =0 ́ Ta co: − 1 0 1 − 2 1 7  h1 → h1 − 1 0 1 −2 1 7 2 0 −2 3 0 8  2h1 + h2 → h2  0 0 0 −1 2 22 A=      3 −3 0 6 − 3 10 3h1 + h3 → h3  0 − 3 3 0 0 31     1 1 − 1 1 − 1 0  h1 + h4 → h4  0 1 0 −1 0 7 h1 → h1 − 1 0 1 − 2 1 7  − 1 1 0 − 2 1 7 h3 → h2  0 − 3 3 0 0 31  31   c 2 ↔ c3  0 3 − 3 0 0  h2 → h3  0 0 0 − 1 2 22  0 0 0 −1 2 22     h4 → h4  0 1 0 −1 0 7   0 0 1 −1 0 7 h1 → h1 − 1 1 0 −2 1 7  h2 → h2  0 3 − 3 0 0 31   h4 → h3  0 0 1 −1 0 7    h3 → h4  0 0 0 − 1 2 22 ⇒ r(A) = r( A ) = 4 vây hệ phương trinh có vô số nghiêm. ̣ ̀ ̣ Tư đó ta có hệ phương trinh đã cho tương đương vơi hê: ̀ ̣ − x1 + x3 − 2 x 4 + x5 =0   − 3 x 2 + 3 x3 = 31 (3) ⇔  (**)  x2 − x4 =7   − x 4 + 2 x5 = 22 ̣ ̀ ́ ̣ ̀ ́ ́ Chon x5 lam biên phu; x1, x2, x3, x4 lam biên chinh. Cho x5 = β vơi β là tham số tuỳ ý − x1 + x3 − 2 x 4 + x5 =0  97   x1 = 3 − 2 β  − 3 x 2 + 3 x3 = 31    x = 2β − 15 (**) ⇔  x2 − x4 =7 ⇔ 2  − x 4 + 2 x5 = 22  x = 2 β − 14   3 3   x5 =β  x = 2β − 22  4 Vây hệ phương trinh có vô số nghiêm có nghiêm tông quat la: ̣ ̀ ̣ ̣ ̉ ́ ̀  97 14   − 2 β ;2 β − 15;2 β − ;2 β − 22; β  Vơi β là tham số tuỳ y. ́  3 3  ̃ Nguyên Phan Thanh Lâm MSV: 071250510319 Trang 9/18
  10. ̀ ̣ ́ ́ Bai tâp toan cao câp I GVHD: Phan Thị Ngũ Bai tâp 4: Biên luân số nghiêm cua hệ phương trinh theo a ̀ ̣ ̣ ̣ ̣ ̉ ̀ (a + 1)x + y + z = 1  x + (a + 1)y + z = a + 1  x + y + (a + 1)z = (a + 1) 2 ́ Ta co: a + 1 1 1 1  h2 → h1  1 a +1 1 a +1   1 A= a +1 1 a +1  1  h → h a + 1 1 1 1  2    1  1 a + 1 (a + 1 ) 2  h3 → h3  1   1 a + 1 (a + 1 )2   h1 → h1 1 a +1 1 a +1  1 1 a +1 a +1  0 − a 2 − 2a − a − a 2 − 2a C ↔ C 0 − a − a 2 − 2a − a 2 − 2a  − (a + 1 )h1 + h2 → h2   2 3  − h1 + h3 → h3 0  −a a a +a  2  0 a  −a a2 + a   h1 → h1 1 1 a +1 a +1  h2 → h2  0 − a − a 2 − 2a − a 2 − 2a   h2 + h3 → h3 0 0 − a 2 − 3a  −a   ̣ ̣ Biên luân:  Nêu (−a − 3a) = 0 ⇔ (a = 0) ∨ (a = −3) 2 ́ ̀  Khi a = 0 thi: 1 1 1 1 A = 0 0 0 0 ⇒ r ( A) = r ( A) = 1 ⇒ Hệ phương trinh đã cho có vô số nghiêm.   ̀ ̣ 0 0 0 0    Vơi nghiêm tông quat có dang: ( 1 − α − β ; α ; β ) vơi α , β là cac tham số tuỳ y. ̣ ̉ ́ ̣ ́ ́  Khi a = -3 1 1 − 2 − 2 A = 0 3 − 3 − 3  ⇒ r ( A) = 2 ≠ r ( A) = 3 ⇒ Hệ phương trinh vô nghiêm.   ̀ ̣ 0 0 0  3    Nêu (−a − 3a) ≠ 0 ⇔ (a ≠ 0) và (a ≠ −3) , khi đó ta có 2 ́ h1 → h1   1 1 a +1 a +1  1 1 a + 1 a +1  0 − a − a 2 − 2 a − a 2 − 2 a  1 A=  − h2 → h2 0 1 a + 2 a+2  a  a  0 0 − a 2 − 3a  −a   1 0 0 1  − 2 h3 → h3  a + 3a  2 a + 3a Do đó hệ phương trinh đã cho tương đương vơi hệ phương trinh sau: ̀ ̀  3a + 7  x = − a + 3  x + (a + 1) y + z = a + 1    1  ( a + 2) y + z = a + 2 ⇔  y =   a+3 a  y= 2 z = a + 3  a + 3a   ́ ̣ Kêt luân:  Khi a = 0 : Hệ có vô số nghiêm có dang (1 − α − β ; α ; β ) vơi α , β là cac tham số ̣ ̣ ́ tuỳ y. ́  Khi a = -3 : Hệ vô nghiêm. ̣ ̃ Nguyên Phan Thanh Lâm MSV: 071250510319 Trang 10/18
  11. ̀ ̣ ́ ́ Bai tâp toan cao câp I GVHD: Phan Thị Ngũ 3a + 7 1  Khi a ≠ 0 và a ≠ −3 : Hệ có nghiêm duy nhât ( − ̣ ́ , ,a + 3) a+3 a + 3 ̃ Nguyên Phan Thanh Lâm MSV: 071250510319 Trang 11/18
  12. ̀ ̣ ́ ́ Bai tâp toan cao câp I GVHD: Phan Thị Ngũ Chương II: HAM MÔT BIÊN THỰC ̀ ̣ ́ Bai tâp 1:Tinh cac giơi han sau: ̀ ̣ ́ ́ ̣  1   2  1. lim   tan 2 x + Cos − tan x   2. lim  Sin2 x − Cotx  π   π x→   x→ 2 2 π tan x − Sinx Cos x 3. lim 4. 2 x→0 x3 lim x→1 1− x 1 + xSinx − Cos 2 x 1+ x −1 lim 5. lim 1 − x→0 3 1+ x 6. x →0 tan 2 x 2 Cosx − 3 Cosx Cos 2 x − 1 7. lim Sin 2 x 8. lim x→0 1 − x2 −1 9. x→0 2 6 x +12 1  2x2 + 3  10. lim  1 + tan x  Sinx lim  2 x 2 + 10   x→+∞      x→0  1 + Sinx  11. x −1  x2 + 4x − 1  2 lim  x 2 + 7 x − 1   x→+∞    ̀ Bai 1: 1 1   ( tan 2 x + − tan x).( tan 2 x + + tan x) 1 Cosx Cosx lim  tan 2 x +  = lim − tan x  x→  π Cosx  x→π 1 2 2 ( tan 2 x + + tan x) Cosx 1 tan 2 x + − tan 2 x = lim Cosx π 1 2 ( tan x + + tan x) x→ 2 Cosx 1 = lim π 1 2 Cosx.( tan x + + tan x) x→ 2 Cosx 1 = lim π Sin 2 x 1 Sinx x→ 2 Cosx.( 2 + + ) Cos x Cosx Cosx 1 1 lim π 2 Sin x 1 Sinx = lim π Sin x + Cosx Sinx 2 x→ 2 Cosx.( 2 + + ) x→ 2 Cosx.( + ) Cos x Cosx Cosx Cos 2 x Cosx 1 = lim x→ π ( Sin 2 x + Cosx + Sinx) 2 1 1 = = 1+ 0 +1 2 ̃ Nguyên Phan Thanh Lâm MSV: 071250510319 Trang 12/18
  13. ̀ ̣ ́ ́ Bai tâp toan cao câp I GVHD: Phan Thị Ngũ  1  1 ̣ Vây lim   tan 2 x + Cosx − tan x  =  2 x→ π   2 ̀ Bai 2:  2   2 Cosx  lim  Sin2 x − Cotx  = lim  2Sinx.Cosx − Sinx  π   π   x→ x→ 2 2  1 Cosx  = lim  −  π  Sinx.Cosx Sinx  x→ 2 1 − Cos 2 x = lim x→ π Sinx.Cosx 2 Sin 2 x Sinx = lim = lim x→ π Sinx.Cosx x → π Cosx 2 2 = lim tgx = ∞ π x→ 2 ̀ Bai 3: Sinx − Sinx tan x − Sinx Sinx − Sinx.Cosx lim x3 = lim Cosx 3 = lim x →0 x→0 x x →0 x 3 .Cosx x2 x. Sinx(1 − Cosx ) = lim 3 = lim 3 2 x→0 x .Cosx x → 0 x .Cosx 1 1 = lim = x → 0 2.Cosx 2 ̀ Bai 4: π π π Cos x − Sin x π 2 = lim x →1 1− x lim 2 − 1 2 = 2 x →1 ̃ Nguyên Phan Thanh Lâm MSV: 071250510319 Trang 13/18
  14. ̀ ̣ ́ ́ Bai tâp toan cao câp I GVHD: Phan Thị Ngũ ̀ Bai 5: 1 + x −1 ( 1 + x − 1).( 1 + x + 1).(1 + 3 1 + x + 3 (1 + x) 2 ) lim 1 − 3 1 + x X →0 .( 1 + x + 1).(1 − 3 1 + x ).(1 + 3 1 + x + 3 (1 + x)2 ) X →0 = lim (1 + x − 1).(1 + 3 1 + x + 3 (1 + x) 2 ) = lim X →0 (1 − 1 − x).( 1 + x + 1) x.(1 + 3 1 + x + 3 (1 + x) 2 ) = lim X →0 − x.( 1 + x + 1) (1 + 3 1 + x + 3 (1 + x ) 2 ) = lim − X →0 ( 1 + x + 1) 3 =− 2 ̀ Bai 6: 1 + xSinx − Cos 2 x ( 1 + xSinx − Cos 2 x ).( 1 + xSinx + Cos 2 x ) lim = lim X →0 tan 2 x 2 X →0 x 2 ( tan 2 . 1 + xSinx + Cos 2 x ) 1 + xSinx − Cos 2 x = lim X →0 x ( tan 2 . 1 + xSinx + Cos 2 x 2 ) xSinx + 2Sin 2 x = lim X →0 x ( tan 2 . 1 + xSinx + Cos 2 x 2 ) 1 ( xSinx + 2 Sin 2 x) = lim .lim X →0 ( 1 + xSinx + Cos 2 x X → 0 ) x tan 2 . 2 x ( xSinx + 2 Sin 2 x).Cos 2 1 2 = lim ( 1 + xSinx + Cos 2 x lim ) . x X →0 X →0 Sin 2 2   1 x xSinx Sin 2 x  = lim Cos 2  lim − 2 lim  2 X →0 2  X → 0 Sin 2 x X →0 2 x  Sin    2 2   1  x2 x2  = .1. lim − 2 lim  2  X →0 ( x )2 X →0 x 2 ( )    2 2  1 1 1 1 = .( − ) = − 2 4 2 8 ̃ Nguyên Phan Thanh Lâm MSV: 071250510319 Trang 14/18
  15. ̀ ̣ ́ ́ Bai tâp toan cao câp I GVHD: Phan Thị Ngũ ̀ Bai 7 : Cosx − 3 Cosx Cosx − 3 Cos 2 x 1 Cosx − 3 Cos 2 x lim = lim = lim x→0 Sin 2 x x→0 Sin x.( Cosx + Cosx ) 2 3 2 x→0 Sin 2 x 1 Cos 3 x − Cos 2 x 2 lim Sin 2 x.(Cos 2 x + 2.Cosx.3 Cos 2 x + 3 Cos 4 x = x→0 1 1 − Cosx − Cos 2 x = lim ( . ) 2 x→0 Sin 2 x. Cos 2 x + 2.Cosx.3 Cos 2 x + 3 Cos 4 x x2 1 1 1 = .( − ). lim 22 = − 2 4 x→0 x 16 Bài 8: Cos 2 x − 1 (Cos 2 x − 1).( 1 − x 2 + 1) (1 − Cos 2 x).( 1 − x 2 + 1) lim x→0 1 − x2 −1 = lim x→0 − x2 = lim x→0 x2 2Sin 2 x.( 1 − x 2 + 1) Sin 2 x = lim = 2 lim 2 .( 1 − x 2 + 1) x→0 x2 x→0 x x2 = 2 lim .( 1 − x 2 + 1) = 4 x→0 x2 ̀ Bai 9: −7 .( 6 x 2 +12 ) 6 x 2 +12 2 6 x +12  − 2 x 2 +10  2 x 2 +10  2x2 + 3   7   7  7  lim  2 x 2 + 10   x→+∞    = lim 1 − 2 x→+∞   2 x + 10  = lim 1 − 2 x→+∞   2 x + 10       −7.( 6 x 2 +12 ) lim 2 x 2 +10 =e x→+∞ − 7.(6 x + 12) 2  18  lim x→+∞ 2 x + 10 2 = 7. lim  − 3 + 2 x→+∞   = −21 2 x + 10  6 x 2 +12  2x2 + 3  Vây lim  2 ̣    = e −21 x→+∞  2 x + 10  ̃ Nguyên Phan Thanh Lâm MSV: 071250510319 Trang 15/18
  16. ̀ ̣ ́ ́ Bai tâp toan cao câp I GVHD: Phan Thị Ngũ ̀ Bai 10: 1 1  1 + tan x  Sinx  1 + tan x + Sinx − Sinx  Sinx lim  1 + Sinx  = lim  X →0   X →0  1 + Sinx   tanx − Sinx 1 .  1 + Sinx  1 + sinx Sinx  tan x − Sinx  tanx − Sinx  = lim 1 +  X → 0  1 + Sinx     tanx - Sinx 1 . 1+sinx Sinx =e Sinx − Sinx  tanx − Sinx 1  Cosx lim 1 + sinx . Sinx  = X →0   lim Sinx.(1 + Sinx) X →0 Sinx − SinxCosx = lim Sinx.Cosx.(1 + Sinx) X →0 1 − Cosx = lim Cosx.(1 + Sinx) = 0 X →0 1 Vây lim  1 + tan x  = e0 = 1 Sinx ̣   X → 0  1 + Sinx  ̀ Bai 11: − . 3x x−1 x −1 x −1 − x +7 x−1  x +7 x−1 2 2 2  x2 + 4x −1  2  3x  2  3x  3x  lim  x 2 + 7 x − 1   x→+∞    = lim 1 − 2 x→+∞   = lim 1 − 2 x + 7x −1 x→+∞   x + 7x −1       3x x −1  − .  lim  2    =e x → +∞  x + 7 x −1 2  ́ Tinh :  1   3( x − x)  2 3 x −x 3  1− 2  lim  −  = − lim 2 = − lim  x =−3 x →+∞ 2( x 2 + 7 x − 1)  2 x→+∞ x + 7 x − 1 2 x→+∞ 1 + 7 − 1  2      x x2  x −1 Vây lim  x 2 + 4 x − 1  2 ̣ 2 1     = x→+∞  x + 7 x − 1  e3 ̃ Nguyên Phan Thanh Lâm MSV: 071250510319 Trang 16/18
  17. ̀ ̣ ́ ́ Bai tâp toan cao câp I GVHD: Phan Thị Ngũ ̀ ̣ ́ ́ ́ Bai tâp 2: Tinh cac tich phân sau: +∞ e dx dx 1. ∫ 2. ∫ x. ln x 1x. 1 + x 2 1 2 −1 dx 1 + x2 3. ∫ 3 4. ∫ x3 dx 1 x −1 −∞ ̉ Giai: ̀ Bai 1. +∞ b dx dx ∫ x. 1 1 + x2 = lim ∫ b→+∞ 1 x. 1 + x 2 ̣ Đăt t = 1 + x 2 ⇒ t 2 = 1 + x 2 ⇔ 2tdt = 2 xdx ⇔ tdt = xdx x = b → t = 1 + b2  ̉ ̣ Đôi cân  x = 1 → t = 2  b b 1+ b 2 1+ b 2 dx xdx tdt dt lim ∫ x. b → +∞ 1 1+ x 2 = lim ∫ b → +∞ 1 x . 1+ x 2 2 = lim b → +∞ ∫ t.(t 2 − 1) lim = b → +∞ ∫ t −1 2 2 2 2 2 1+ b 1+ b 1 dt 1 dt = lim 2 b → +∞ ∫ − lim t − 1 2 b → +∞ ∫ t +1 2 2 1+ b 2 1 1 t −1 = lim ( ln(t − 1) − ln(t + 1) ) 1+ b 2 = lim ln 2 b → +∞ 2 2 b → +∞ t + 1 2    ln 1 + b − 1 − ln 2 − 1  = +∞ 2 1 2 lim  = b → +∞  1 + b2 + 1 2 + 1 +∞ dx ⇒ ∫ x. 1 1+ x 2 Hôi tụ ̣ ̀ Bai 2. e e dx dx ∫ 1 = lim ∫ x. ln x ε → 0 + 1+ ε x. ln x 1 1 dx 1 Đăt t = ̣ ⇒ dt = − 2 dx ⇒ = − 2 dt ln x x. ln x x t  x = e → t =1  ̉ ̣ Đôi cân:  1  x = 1 + ε → t = ln(1 + ε )  ̃ Nguyên Phan Thanh Lâm MSV: 071250510319 Trang 17/18
  18. ̀ ̣ ́ ́ Bai tâp toan cao câp I GVHD: Phan Thị Ngũ e 1 dx 1 lim ∫ ∫ − t dt = lim ( − ln t ) 1 = x. ln x lim 1 ε→0+ 1+ε ε →0+ 1 ε →0+ 1+ε 1+ε  1  = lim −  ln 1 − ln =0 ε →0+  1+ ε  e dx ̣ Vây ∫ x. ln x 1 Hôi tụ ̣ Bài 3: 2 2 2 2 dx dx dx dx ∫ 1 = lim ∫ 3 = lim ∫ × lim ∫ 2 x 3 − 1 ε → 0+ 1+ ε x − 1 ε → −∞ 1+ ε x − 1 ε → 0+ 1+ ε x + 2 x + 1 2 2 dx dx ∫ = lim (ln 1 − ln ε ) × lim ∫ 2 2 = lim ln( x + 1) 1+ ε × lim ε → 0+ ε → 0+ 1+ ε x 2 + 2 x + 1 ε → 0+ ε → 0+ 1+ ε x + 2 x + 1 =∞ ̀ Bai 4. −1 1 + x2 −1 1 + x2  −1 1 −1 2 x  ∫ x3 −∞ dx = lim ∫ 3 dx = lim  ∫ 3 dx + ∫ 3 dx  b→− b ∞ x ∞  b→−  b x b x    −1 1 −1 1   1 −1 −1  = lim  ∫ 3 dx + ∫ dx  = lim  − 2  + ln b  b→−  b x ∞ b x  b → −∞  2 x   b    1 1  1 = lim  − + 2 + ln1 − ln b  = − b→−  ∞ 2 2b  2 −1 1+ x2 ̣ Vây − ∫∞ x 3 dx Hôi tụ ̣ ̃ Nguyên Phan Thanh Lâm MSV: 071250510319 Trang 18/18

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