đề thi thử khối a 2014 chuyên nguyên tất thành yên bái lần 2

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... đều ABD. Ta có 2 3 2 2ABCD ABD a S S∆= = . 0 ,25 Trong tam giác vuông SGA, ta có: 2 2 2 263 3 a aSG SA GA a= − = − = . Do đó: 3.1 2 .3 6S ABCD ABCD a V S SG= = . 0 ,25 Gọi O ... = . 0 ,25 Gọi ();n a b= với + > 2 20 a b là vectơ pháp tuyến c a đường thẳng AH. Ta có ( ) 2 2 2 271 15cos , 35 74 35 0537 37. 2 7b a a bBD AH a ab bb a b a =−= ... −. 0 ,25 (S) tiếp xúc với ( ) : 2 2 7 0P x y z− + − =( ) 2 2 2 2 7,( ) 43 a bd T P TA a b− −⇔ = ⇔ = + + (*) 0 ,25 Thay 1 a b= − vào (*), ta được: ( ) 2 2 2 1* ( 1) 4 3 2 0 2 bb...

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... ( ;7 ) A A a a∈ ∆⇒−. Tam giác ADE cân tại A nên 2 25( 5) ( 3) 643 2 a DEAE a a a == ⇔ − + + = ⇔= − Đỉnh A có hoành độ dương nên ta chọn 5 a =(5 ;2) A ⇒ 0 ,25 Đường ... 357 51574 a AM a KMA BMC AK BC a AH a CM a ∆ ∆ ⇒ = = = ⇒ =∼ Vậy 2 34( ,( ))51d A SCM a =. 0 ,25 Ta có 2 21 1 2 P xy xyxyx y= + + ≥ + 0 ,25 Đặt t xy= ta có 2 10 2 4x yt xy ... 3BC a =, 0.cot30 3. 3 3SB BC a a= = =, 2 2SA a = Suy ra: 31 1 1 6. . . . . . . 3 .2 23 6 6 3MCD a V S SA CD BC SA a a a= = = = . 0 ,25 Trong ( )ABCD, kẻ AK CM⊥. Suy ra (...

... xảy ra ? A. Al+NaOH+3H 2 O → NaAl(OH)4+ 3/2H 2 B. SiO 2 +2NaOHnóng chảy→ Na 2 SiO3+H 2 OC. NaAlO 2 +CO 2 +H 2 O → Al(OH)3+NaHCO3D. Al 2 O3+3CO → 2Al+3CO 2 21. Hai bình ch a các ... ra ? A. Al+NaOH+H 2 O → NaAlO+H 2 B. SiO 2 +2NaOHnóng chảy→ Na 2 SiO3+H 2 OC. NaAlO 2 +CO 2 +H 2 O → Al(OH)3+NaHCO3D. Al 2 O3+3CO → 2Al+3CO 2 45. Hai bình ch a các dung dịch Ca(OH) 2 ... xảy ra ? A. Al+NaOH+3H 2 O → NaAl(OH)4+ 3/2H 2 B. SiO 2 +2NaOHnóng chảy→ Na 2 SiO3+H 2 OC. NaAl(OH)4+CO 2 → Al(OH)3+NaHCO3-D. Al 2 O3+3CO → 2Al+3CO 2 21. Hai bình ch a các...

... A A C A H CA H= = =  .0 ,25 • 2 01 3. .sin 120 2 2ABC a S AC BC∆= =0 ,25 • 2 2 2 0 2 2 . .cos 120 7 7AB AC BC AC BC a AB a= + − = ⇒ =• 2. 21 7ABCS a CHAB∆= =Suy ra: 0 2 21'sin30 ... a A C = =.0 ,25 Xét tam giác vuông AA’C ta được: 2 235' '7 a AA A C AC= − =. Suy ra: 3105. '14ABC a V S AA∆= =.0 ,25 Do ( )'/ / ' '/ / ' 'CC ... AA CC ABB A ,Suy ra: ( ) ( )( )( )( ) 21 ' , ' ', ' ' , ' '7 a d A B CC d CC ABB A d C ABB A CH= = = =.0,5V(1,5điểm)Ta có: ) (21 )()(3) (21 )(4 22 22 yxyxyxyxxyyx++≤−++⇔++≤++ 2 )(3) (21 ...

... WWW.VIETMATHS.COM TRƯỜNG THPT CHUYÊN NGUYỄN TẤT THÀNH ĐỀ KIỂM TRA ĐẠI HỌC NĂM 20 13 TÔ: TOÁN Môn thi: TOÁN – Khối A Thời gian: 120 phút, (không kể thời gian giao) ĐỀ SỐ 5:Câu I (2, 5 điểm). ... tích phân: 1 2 ln = + ÷ ∫eI x xdxx.Câu IV(1,0 điểm). Cho tứ diện SABC có SA vuông góc với mp(ABC), SA = a và diện tích tam giác SBC gấp hai lần diện tích tam giác ABC. Tính khoảng ... 0 120 .Câu II (2 điểm). 1) Tìm các giá trị c a tham số m để phương trình: 2 2 2 2m x x x− + = + có 2 nghiệm phân biệt. 2) Giải phương trình: 31 8 2 2log 1 log (3 ) log ( 1) 0+ − − − − =x x...

... c a C trên 'AC )BC'(CK A ⊥⇒aCK=⇒0 .25 ACCACABAC '(ABC))),((ABC'AB,'∠=∠⇒⊥⊥060'=∠⇒ACC0 .25 3 2 60sin0aCKAC==;aACCC 26 0tan.'0==0 .25 aABACABAH 2 111 22 2=⇒+=34'.3'''. a CCSVABCCBAABC==∆.0 .25 V(1.0 ... ca C cú dng ); 32( aaC+0 .25 Phng trỡnh ng thng 053:=+yxAB0 .25 2 =ABCS 2) ,(. 2 1=ABCdAB 2 10 2 .10 2 1= a 6 2 a a== 0 .25 Vi 6= a ta cú )6;9(C; vi 2 = a ta cú )2; 7(C0 .25 2. (1.0 ... 26 0tan.'0==0 .25 aABACABAH 2 111 22 2=⇒+=34'.3'''. a CCSVABCCBAABC==∆.0 .25 V(1.0 imĐặt zyxt++= 2 3) (23 2 2=+++++=tzxyzxyzxyzxyt.0 .25 Ta có 30 22 2=++++zyxzxyzxy...