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7 months ago

Bộ đề thi thử THPT QG 2018 Các môn TOÁN - LÍ - HÓA Các trường THPT Cả nước CÓ HƯỚNG DẪN GIẢI (Lần 7) [DC09042018]

https://app.box.com/s/ndw8wijyug5fr425ktyaixesjd9264hw

https://twitter.com/daykemquynhon plus.google.com/+DạyKèmQuyNhơn www.facebook.com/daykem.quynhon https://daykemquynhon.blogspot.com http://daykemquynhon.ucoz.com Nơi bồi dưỡng kiến thức Toán - Lý - Hóa cho học sinh cấp 2+3 / Diễn Đàn Toán - Lý - Hóa 1000B Trần Hưng Đạo Tp.Quy Nhơn Tỉnh Bình Định Ta có: AB là hình chiếu của SB lên (ABC) nên SBA = 60° SA SA a 3 tan SBA = ⇒ AB = = = a = BC AB tan SBO 3 2 2 2 2 AC AB BC a a a = + = + = ( ) 2 ( ) 2 2 2 SB SA AB a 3 a 2a = + = + = Do đó ta có Stp = SSAB + SSBC + SSAC + S ABC Vậy S tp = 1 2 + + + 2 ( SA. AB SB. BC SA. AC AB. BC ) ( 3. 2 . 3. 2 . ) = 1 2 a a + a a + a a + a a 3 + 3 + 6 = . a 2 3 + 3 + 6 = . a 2 Câu 43: Đáp án D 3 Theo đề bài ta có V = 18000 cm , h = 40 cm. 2 2 1 2 3V 3.18000 Do đó, ta có: V = π r h r r 20,72 cm. 3 ⇒ = π h = 40π ⇒ ≈ Vậy bán kính của hình tròn là r = 21 cm. Câu 44: Đáp án A 3 a a a Ta có V1 = a. . = và V 4 4 16 Do đó V > V . 1 2 Câu 45: Đáp án B 2 3 1 3 3 a a a = a. . . . = . 2 3 2 3 36 Đường thẳng d có vectơ chỉ phương u = ( 2; −3;1) , qua H ( −2;4; − 1) Mặt phẳng (P) có vectơ pháp tuyến n = ( A; B; C );( A 2 + B 2 + C 2 > 0) Ta có d ( P) ⎧⎪ u. n = 0 ⎧2A − 3B + C = 0 ⎧C = − 2A + 3B / / ⇔ ⎨ ⇔ ⎨ ⇔ ⎨ * ⎪⎩ H ( −2;4; −1) ∉ ( P) ⎩− 3A + 4B − C ≠ 0 ⎩C ≠ 3A − 4B DIỄN ĐÀN TOÁN - - HÓA 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN Mặt khác (P) qua K ( 1;0;0 ) suy ra ( ) ( ) P : Ax + By + 3B − 2 A . z − A = 0 . ( ) MỌI YÊU CẦU GỬI VỀ HỘP MAIL DAYKEMQUYNHONBUSINESS@GMAIL.COM HOTLINE : +84905779594 (MOBILE/ZALO) https://daykemquynhonofficial.wordpress.com/blog/ DẠY KÈM QUY NHƠN OFFICIAL ST> : Đ/C 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN Trang 21 Đóng góp PDF bởi GV. Nguyễn Thanh Tú www.facebook.com/daykemquynhonofficial www.facebook.com/boiduonghoahocquynhonofficial

https://twitter.com/daykemquynhon plus.google.com/+DạyKèmQuyNhơn www.facebook.com/daykem.quynhon https://daykemquynhon.blogspot.com http://daykemquynhon.ucoz.com Nơi bồi dưỡng kiến thức Toán - Lý - Hóa cho học sinh cấp 2+3 / Diễn Đàn Toán - Lý - Hóa 1000B Trần Hưng Đạo Tp.Quy Nhơn Tỉnh Bình Định Ngoài ra d ( M ( P) ) − 5A + 8B ; = = 3 ( 3 2 ) 2 2 2 A + B + B − A 2 2 ⎡A = B ⇔ 5A − 22AB + 17B = 0 ⇔ ⎢ ⎣5A = 17B Với A = B ⇒ C = B không thỏa mãn (*) Với 5A = 17B , chọn A = 17 , suy ra B = 5 , do đó C = − 19 (nhận) Vậy ( ) :17 + 5 −19 − 17 = 0. P x y z Câu 46: Đáp án B b = x ; y ;z Do ( ) cùng phương với a = ( 1; −2;4) nên b = ( k; −2 k;4k ) Mà 0 0 0 b = 21 = k + 4k + 16k = 21k 2 2 2 2 nên suy ra k = ± 1. Ở bài toán này cần chú ý vectơ b tạo với tia Oy một góc nhọn. Khi đó y 0 > 0 , suy ra k = − 1. Do đó x0 = − 1, y0 = 2, z0 = − 4. Vậy x0 + y0 + z0 = − 3. Câu 47: Đáp án A Kiểm tra thấy A và B nằm khác phía so với mặt phẳng (P) Ta tìm được điểm đối xứng với B qua (P) là B '( −1; − 3;4) Lại có MA − MB = MA − MB ' ≤ AB ' = const . Vậy MA − MB đạt giá trị lớn nhất khi M, A, B’ thẳng hàng hay M là giao điểm của đường thẳng AB’ với mặt phẳng (P). ⎧x = 1+ t ⎪ ⎨ = − ∈ R . ⎪ ⎩z = − 2t Đường thẳng AB’ có phương trình tham số là y 3 ( t ) Tọa độ điểm M ứng với tham số t là nghiệm của phương trình Suy ra a = − 2, b = − 3, c = 6 Vậy a + b + c = 1. Câu 48: Đáp án C ⎧1 + mt ' = t + 5 ⎪ Ta có hệ giao điểm như sau ⎨3 + t ' = 2t + 3 ⎪ ⎩ − 5 + mt ' = − t + 3 ( 1+ t) + ( − 3) + ( −2t ) − 1 = 0 ⇔ t = −3 ⇒ M ( −2; − 3;6) DIỄN ĐÀN TOÁN - - HÓA 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN MỌI YÊU CẦU GỬI VỀ HỘP MAIL DAYKEMQUYNHONBUSINESS@GMAIL.COM HOTLINE : +84905779594 (MOBILE/ZALO) https://daykemquynhonofficial.wordpress.com/blog/ DẠY KÈM QUY NHƠN OFFICIAL ST> : Đ/C 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN Trang 22 Đóng góp PDF bởi GV. Nguyễn Thanh Tú www.facebook.com/daykemquynhonofficial www.facebook.com/boiduonghoahocquynhonofficial

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