Toán ôn thi đại học - chuyên đề 5 hình học không gian

... dẫn giải CDBT từ các ĐTQG Toán học – 1 65 Bài 5: ĐẠI HỌC KHỐI D NĂM 2010 Cho hình chóp S.ABCD có đáy ABCD là hình vuông cạnh a, cạnh bên SA = a; hình chiếu vuông góc của đỉnh S trên mặt ... 17 17(cm) A B C S I H M N K Hướng dẫn giải CDBT từ các ĐTQG Toán học – 157  Chuyên đề 5: HÌNH HỌC KHÔNG GIAN KIẾN THỨC CĂN BẢN 1. QUAN HỆ SONG SONG I. ĐƯỜNG THẲNG SONG SONG ... Hình chóp đều là hình chóp có đáy là đa giác đều và các cạnh bên bằng nhau. Đỉnh của hình chóp đều có hình chiếu là tâm của đáy. Hình chóp tam giác còn gọi là tứ diện hình tứ diện. Hình...

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AC = 22A C A A a 5 ,   22BC AC AB 2a B’ C’ A’ A D G C B Hướng dẫn giải CDBT từ các ĐTQG Toán học – 157  Chuyên đề 5: HÌNH HỌC KHÔNG GIAN KIẾN THỨC CĂN BẢN 1. ... dẫn giải CDBT từ các ĐTQG Toán học – 1 65 Bài 5: ĐẠI HỌC KHỐI D NĂM 2010 Cho hình chóp S.ABCD có đáy ABCD là hình vuông cạnh a, cạnh bên SA = a; hình chiếu vuông góc của đỉnh S trên mặt ... src="data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAucAAARACAIAAADCrGrzAAAACXBIWXMAABYlAAAWJQFJUiTwAAAgAElEQVR42uzdfXBT 153 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... 5 8H 4 5 8; 1 5 y 1 5          . Phương trình : ( 5 1)x 4 5 8 y 0    Cách 2:    Oy  H  A: không thỏa AH = d(H, Ox)    Ox  H  O: không thỏa ... b2 = 4 5 (1); (C1) tiếp xúc 1, 2  a b a 7b2 5 2 (2). (1) và (2), cho ta:  22 5 a 2 5b 4 5 a b a 7b        22 5 a 2 5b 4I 5 a b a 7b ... 12 5 25 2 5 25 2b ; b77  Thay 3a 1b2 vào (1) ta có: 23a 15a 5 a 12    51 a2 – 14a + 99 = 0 phương trình vô nghiệm  Vậy ta có 2 tiếp tuyến là: 1: x + 7y – 5...

... Giả thi t suy ra:   211t 2029t 88t 683  35t2 – 88t + 53 = 0  t = 1 hoặc t = 53 35 Ta có   53 18 53 3t 1 M 0; 1; 3 ; t M ; ; 35 35 35 35      Bài 8: ĐẠI HỌC ... CDBT từ các ĐTQG Toán học – 2 65  A(2; 0; 1), B( 5; 1; 3). Phương trình của d là: x 2 y z 17 1 4 Bài 3: ĐỀ DỰ BỊ 1 ĐẠI HỌC KHỐI D NĂM 2006 Trong không gian với hệ tọa độ ... trình (P): 2x + 3z – 5 = 0 Vậy (P): 4x + 2y + 7z – 15 = 0 hoặc (P): 2x + 3z – 5 = 0 Bài 9: ĐẠI HỌC KHỐI A NĂM 2008 Trong không gian với hệ tọa độ Oxyz, cho điểm A(2; 5; 3) và đường thẳng...

... tơ: Chun đề LTĐH Huỳnh Chí Hào – boxmath.vn 102 Chuyên đề 11: HÌNH HỌC GIẢI TÍCH TRONG KHÔNG GIAN A. KIẾN THỨC CƠ BẢN PHƯƠNG PHÁP TOẠ ĐỘ TRONG KHÔNG GIAN TỌA ĐỘ ĐIỂM - TỌA ĐỘ ... A(3;1;0), B (-1 ;2 ;-1 ), C(2 ;-1 ;3) Tìm điểm D sao cho tứ giác ABCD là hình bình hành Ví dụ 2: Trong Kg(Oxyz) cho ba điểm A(2 ;-1 ;6), B (-3 ;-1 ;-4 ), C (5 ;-1 ;0) a. Chứng minh rằng tam giác ABC vuông . ... thể tích tứ diện ABCD biết A (-1 ;-2 ;0), B(2 ;-6 ;3), C(3 ;-3 ;-1 ), D (-1 ; -5 ;3) Bài 3: Bài 4: Bài 5: Bài 6: Bài 7: Cho tứ diện ABCD với A(2; 1;6),B( 3; 1; 4),C (5; 1;0),D(1;2;1)− − − − −....

...  OA ( 5; 0; 5) , OB (12; 0; 10) OA,OB (0; 10; 0).         OAB11S OA,OB .10 5 22 (đvdt). Bài 5: ĐỀ DỰ BỊ 1 ĐẠI HỌC KHỐI B NĂM 20 05 Trong không gian với hệ ... Hướng dẫn giải CDBT từ các ĐTQG Toán học – 241 B. ĐỀ THI Bài 1: ĐẠI HỌC KHỐI B NĂM 2009 Trong không gian với hệ toạ độ Oxyz, cho mặt phẳng (P): x – 2y + 2z – 5 = 0 và hai điểm A(3; 0;1), ... Phương trình của  là:   x 1 y 2 z 31 3 5 Bài 3: ĐỀ DỰ BỊ 1 - ĐẠI HỌC KHỐI A NĂM 2006 Trong không gian với hệ trục tọa độ Oxyz cho hình lăng trụ đứng ABC.A'B'C' có...

... 0 < x < 5 4     2 2 224 5x 5 5 3x44Sx 5 4x x 5 4x 5 S 0 x 1 x3        Bảng biến thi n x  1 5 4 S'  0 + S 5 TTLT ĐH VĨNH ... 2233u v 5 u v 5 u v u v uv 3(u v) 15m 10u v 3(u v) 15m 10          2u v 5 u v u v 3uv 3(u v) 15m 10     2u v 5 5 5 3uv ... (t1, t2 không nhất thi t phân biệt).  Xét hàm số   2f(t) t 5t 8 với t 2: Suy ra f'(t) = 2t – 5 và f'(t) = 0  t = 5 2 Bảng biến thi n t  2 2 5/ 2 + f'(t)...

... =2 + k (k  ) hay sin5x + sinx = sinx  x =2 + k hay x =k 5  (k  )  Vấn đề 2: GIẢI PHƯƠNG TRÌNH LƯNG GIÁC TRÊN MỘT MIỀN ĐỀ THI TT Luyện Thi Đại Học VĨNH VIỄN 79 ... Bài 11: ĐẠI HỌC KHỐI D NĂM 2009 Giải phương trình: 3 cos5x 2sin3xcos2x sinx 0   Giải Phương trình đã cho tương đương:  3 cos5x sin5x sinx sin x 0     31cos5x sin5x sin ... x22  sin 5x sinx3           5x x k2 hay 5x x k233 (k  ) Vậy: x =         k hay x k k18 3 6 2 TT Luyện Thi Đại Học VĨNH VIỄN ...

... trình:     225x 10x 1 7 2x x. (1) Giải     225x 10x 1 7 2x x Điều kiện để căn bậc hai có nghóa là: 5x2 + 10x + 1  0      5 2 5 5 2 5 x hoặc x 55 (*) Với điều ... 24v 5   2v 5  4u 5  Suy ra: 42x 5 22x 5   162x 5 42x 5  6x 5  ª Với u = 3 + 2v thế vào (2) ta được (3 + 2v)2 + v2 = 4  5v2 ... Hướng dẫn giải CDBT từ các ĐTQG Toán học – 1 05         22 5 5x 10x 1 36 5x 10x 1 (*) Đặt    2t 5x 10x 1, t 0 (*) trở thành t2 + 5t – 36  0  t  4 (nhận)  t ...

... Luyện Thi Đại Học VĨNH VIỄN 151  Vấn đề 5: ỨNG DỤNG CỦA TÍCH PHÂN A. PHƯƠNG PHÁP GIẢI TÍNH DIỆN TÍCH Bài toán 1: Cho hàm số y = f(x) liên tục và không âm trên đoạn [a, b]. Diện tích hình ... x asin tax 3. b22adxđặt x atan tax B. ĐỀ THI Bài 1: ĐẠI HỌC KHỐI A NĂM 2011 TT Luyện Thi Đại Học VĨNH VIỄN 1 25 Đặc biệt: u(x) = ax + b;       1f(x)dx ...  = 5 3232t6t 21t 10ln t3   = 34 310ln 35 . Bài 3: ĐẠI HỌC KHỐI B NĂM 2010 Hướng dẫn giải CDBT từ các ĐTQG Toán học – 136 Bài 16: ĐỀ DỰ BỊ 2 Tính...