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4 months ago

Bộ đề thi thử THPT QG 2018 Các môn TOÁN - LÍ - HÓA Các trường THPT Cả nước CÓ HƯỚNG DẪN GIẢI (Lần 8) [DC11042018]

https://app.box.com/s/nhxs7vnw9na94twmj7hc0inbdjzwv940

https://twitter.com/daykemquynhon plus.google.com/+DạyKèmQuyNhơn www.facebook.com/daykem.quynhon http://daykemquynhon.blogspot.com Giải hệ trên ta thu được ⎧a = 2 ⎨ (thỏa mãn) hoặc ⎩b = 1 ⎧ 22 a = − ⎪ 13 ⎨ (loại) ⎪ 19 b = − ⎪⎩ 13 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 Do đó z = 2 + i và P = 898 Câu 40: Đáp án D C ' C ⊥ ABCD , BD ⊥ OC ⇒ BD ⊥ OC ' ⇒ COC ' = 45° Ta có ( ) a 2 ∆OCC ' vuông cân tại C ⇒ CC ' = OC = 2 3 2 a 2 a 2 Vậy V = a . = 2 2 Câu 41: Đáp án D Kẻ AM ⊥ BC và SH ⊥ AM , khi đó ∆ SHM vuông cân tại H Suy ra HM = HS = h, AM = 3h Vậy 9 3 S = h 4 Câu 42: Đáp án A Kẻ AI ⊥ BC và AH SI Ta có 2 ⊥ . Khi đó ( ) ( ,( )) a 3 AI = (do ∆ ABC đều cạnh a) 2 AH ⊥ SBC ⇒ d A SBC = AH và ( SB ( ABC )) = SBA = 60 ° ⇒ SA = AB.tan 60° = a 3 SA. AI a 15 = = = SA + AI 5 Vậy d ( A( SBC) ) AH 2 2 Câu 43: Đáp án B BC ∆ ABC vuông cân tại A ⇒ AB = = 1 2 V = π AB 2 . AA' = π.1.2 = 2π Câu 44: Đáp án D Bán kính mặt cầu Diện tích mặt cầu 2 2 2 ⎛ 2 a 2 ⎞ ⎛ a ⎞ 3a R = ⎜ + = 2 ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ 2 ⎠ 4 DIỄN ĐÀN TOÁN - - HÓA 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN 3a S = 4π R = 4 π. = 3π a 4 2 2 2 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 23 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 http://daykemquynhon.blogspot.com Câu 45: Đáp án 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 Đặt cạnh hình vuông là x, x > 24 cm 4800 = x − 24 .12 ⇔ x = 44 cm Theo đề ta có ( ) 2 Vậy độ dài cạnh của tấm bìa hình vuông là 44cm Câu 46: Đáp án D (S) có tâm ( 1;0; 1) I − , bán kính R = 2 AB = − − − = − − ( 1; 3; 4 ), AC ( 1; 1; 4) ⎣ ⎦ Gọi ( α ) là mặt phẳng chứa 3 điểm A, B, C nhận n = ⎡ AB, AC⎤ = ( 8; −8;4) phương trình: ( ) ( ) ( ) 8 x + 1 − 8 y + 2 + 4 x + 3 = 0 ⇔ 2x − 2y + z + 1 = 0 d ( I 2 − 0 − 1+ 1 2 , α ) = = < 2 = ( ) ( ) 2 2 2 2 2 1 3 R ⇒ S ∩ α = ∅ + − + Ta có V Gọi 1 2 ( ) 1 = h . S ∆ nên V ABCD lớn nhất ⇔ hD lớn nhất 3 ABCD D ABC D D là đường kính của (S) vuông góc với mặt phẳng ( α ) Vì D là điểm bất kì thuộc (S) nên d ( D, α ) ≤ max { d ( D1 , α ), d ( D2 , α )} Dấu bằng xảy ra khi và chỉ khi D trùng với một trong hai điểm D 1 hoặc D 2 D D qua I nhận vectơ pháp tuyến của ( ) 1 2 ⎧x = 1+ 2t ⎪ D1D 2 : ⎨y = − 2 t , t ∈ R ⎪ ⎩z = − 1 + t Gọi ( 1 2 ; 2 ; 1 ) 0 0 0 1 2 làm vectơ pháp tuyến nên có α làm vectơ chỉ phương nên có phương trình tham số D + d − d − + d ∈ D D là điểm cần tìm. Khi đó D là nghiệm của phương trình: DIỄN ĐÀN TOÁN - - HÓA 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN 2 1+ 2d0 + 4d0 + − 1+ d0 − 2 1+ 2d0 + 2 − 1+ d0 − 2 = 0 ⇔ d0 = ± 3 2 ( ) 2 2 ( ) ( ) ( ) 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 24 Đóng góp PDF bởi GV. Nguyễn Thanh Tú www.facebook.com/daykemquynhonofficial www.facebook.com/boiduonghoahocquynhonofficial

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Bộ đề luyện thi THPT Quốc Gia Môn Ngữ Văn (2016)
CÁC TRẠNG THÁI VẬT LÝ CỦA POLYMER
ĐỘC HỌC MÔI TRƯỜNG VÀ SỨC KHỎE CỘNG ĐỒNG