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8 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 Ta có d ( D α ) 9d + 2 = Vì 3 0 , . 2 ⎛ 2 ⎞ 9. + 2 9. ⎜ − ⎟ + 2 3 ⎝ 3 ⎠ 2 > nên D phải ứng với d 0 = 3 3 3 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 7 4 1 Vậy D ⎛ ⎜ ; − ; − ⎞ ⎟ là điểm cần tìm ⎝ 3 3 3 ⎠ Câu 47: Đáp án B Giả sử ( P) ∩ Ox = A( a;0;0 ),( P) ∩ Oy = A( 0; b;0 ),( P) ∩ Oz = A( 0;0; c) Khi đó (P) có phương trình x + y + z = 1 a b c 4 5 6 Ta có: H ( 4;5;6 ) ∈( P) ⇔ + + = 1 a b c AH = 4 − a;5;6 , BH = 4;5 − b;6 BC = 0; − b; c , AC = −a;0; c ⎪⎧ AH. BC = 0 ⎧− 5b + 6c = 0 Vì H là trực tâm tam giác ABC nên ⎨ ⇔ ⎨ ⎪⎩ BH. AC = 0 ⎩ − 4b + 6c = 0 ( ) ( ) ( ) ( ) Giải hệ phương trình ⎧ 77 ⎧ 4 5 6 1 ⎪ a = ⎪ + + = 4 a b c ⎪ ⎪ 77 ⎨− 5b + 6c = 0 ⇔ ⎨b = ⎪ 5 − 4b + 6c = 0 ⎪ ⎪ ⎪ 77 ⎩ ⎪c = ⎩ 6 Vậy phương trình mặt phẳng (P) là x y z + + = 1 ⇔ 4x + 5y + 6z − 77 = 0 77 77 77 4 5 6 Câu 48: Đáp án B Do ( β ) / / ( α ) nên ( β ) : 2x + 2y − z + D = 0( D ≠ 17) Mặt cầu (S) có tâm ( 1; 2;3) I − , bán kính R = 5 Đường tròn có chu vi là 6π nên bán kính của đường tròn này là r = 3 Ta có ( ( β )) ( ) 2.1+ 2. −2 − 3 + D ⎡D = −7 = − ⇔ = 4 ⇔ − 5 = 12 ⇔ ⎢ 2 + 2 + −1 ⎣D = 17 2 2 d I R r D 2 2 2 Nhận giá trị D = − 7 . Vậy ( ) Câu 49: Đáp án C ( ) DIỄN ĐÀN TOÁN - - HÓA 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN β có phương trình là 2x + 2y − z − 7 = 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 25 Đóng góp PDF bởi GV. Nguyễn Thanh Tú www.facebook.com/daykemquynhonofficial www.facebook.com/boiduonghoahocquynhonofficial

https://twitter.com/daykemquynhon plus.google.com/+DạyKèmQuyNhơn 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 I là trung điểm AB ⇒ I ( 2;2;0) Mặt phẳng (P) có vectơ pháp tuyến là: n = ( 3;2; −1) Vì KI ⊥ ( P) nên đường thẳng KI qua I nhận n = ( 3;2; −1) ⎧x = 2 + 3t ⎪ ⎨y = 2 + 2t ⎪ ⎩z = − t ( 2 3 ;2 2 ; ) K ∈ KI ⇒ K = + t + t − t Theo đề ta có: 6 + 9t + 4 + 4t + t + 4 d ( K ( P) ) = KO ⇔ = ( 2 + 3t ) + ( 2 + 2t ) + t 14 làm vectơ chỉ phương nên có phương trình 2 2 2 2 2 2 14 + 20 + 4 = 14 + 1 ⇔ 14 + 20 + 4 = 14 + 28 + 14 3 t = − 4 Vậy t t t t t t t ⎛ 1 1 3 ⎞ K ⎜ − ; ; ⎟ thỏa mãn yêu cầu bài toán ⎝ 4 2 4 ⎠ Câu 50: Đáp án A Giả sử ( ) S có phương trình là + + − 2 − 2 − 2 + = 0 + + − > 0 2 2 2 2 2 2 x y z ax by cz d . (Điều kiện: a b c d ) ( ) 0 O ∈ S ⇔ d = ( ) ( ) A 0;0;4 ∈ S ⇔ 16 − 8c + d = 0 . Mà d = 0 nên suy ra c = 2 ( ) ( ) A 2;0;0 ∈ S ⇔ 4 − 4a + d = 0 . Mà d = 0 nên suy ra a = 1 Với I ( 1; b ;2) , ta có d ( I; ( P) ) Vậy phương trình mặt cầu (S) cần tìm là: 5 b + 5 5 ⎡b = 0 = ⇔ = ⇔ ⎢ 6 6 6 ⎣b = − 5 + + − 2 − 4 = 0; + + − 2 + 20 − 4 = 0 2 2 2 2 2 2 x y z x z x y z x y z DIỄN ĐÀN TOÁN - - HÓA 1000B TRẦN HƯNG ĐẠO TP.QUY NHƠN ----- HẾT ----- www.facebook.com/daykem.quynhon http://daykemquynhon.blogspot.com 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 26 Đóng góp PDF bởi GV. Nguyễn Thanh Tú www.facebook.com/daykemquynhonofficial www.facebook.com/boiduonghoahocquynhonofficial

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