Tài liệu Bài tập đâị số tuyên tính 2

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