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Đề thi thử THPT Quốc gia 2020 lần 1 môn Toán sở GD&ĐT Ninh Bình (Mã đề 001)

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111111111111111111111111111111111111111111111111 DACB12341516718910121SỞ GIÁO DỤDC VÀDC 9101B111410DACB1ÁĐ OCỤ OCT ẠÀ OCỤ OCNO HBÌP GỤỀ ỬQD OCU Ố A DLẦ CỨP 1-ỐĂ M 1-1- ẦỌD O20D (Đáp npgồ m03 t0pr a) Đ Đồn Đáp npgồ Đ 1 111111111111111111111111C9 –ôn –:o áân:u hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhỏ Si ãđề t3n:u hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh Ầ4 56 –:â --Ố P78 Ốu1C10D11001C1A10D1101C1CB11C1CB11 101A110D10CB1CB1CB1Ềh 1Vh 1Ph 1Ih 11P78 1u1C1CB18C1á3â1Ềh 11C 1CB1110D10DC11Vh 11C 118101CC10D10CB118CB101C18CB1Ph 11C 1C18CB1C10DC11Ih 11C 11B10D1C1C010DC11CB11P78 u110A1B11CB10111 g t 1A10A1B11AC10D110A10181C1CC11C101CB1AC1CB1Ềh gt 1Vh gt 1Ph g t t 1Ih 2gt 1P78 u1 12181C1010DCB1CB1BAC1101010C1A1C1011011Ềh 21Vh 21Ph 21Ih 21P78 u1CB1CB10DCB1C1118101A111C11811Ềh 21Vh 2   1Ph 1Ih B 1P78 u1110111101011C10181C101CB1g11g10191A1A1C1CB1Ềh g1Vh g1Ph g1Ih 4g1P78 u1111 11CB1C10C1C1A1 1 CB118CB101C1A1B101C18CB1101C1CBACB1A18101111Ềh 1Vh 1Ph 1Ih 21P78 u1CB1CB10DCB1C1118101A111C1181 11 1 51 1 1111111 21 111111111111111111111111111111111111111111111111 DACB13415167189101211Ềh 21Vh 21Ph 21Ih 21P78 Ău101CB10DC1 2  110011A1C1CB1CBC1CB1Ềh 1Vh 1Ph 11Ih 1P78 Ố-u118110D1A18101112   11Ềh 1Vh 1Ph 21Ih 1P78 ỐỐu11C111C01    112   1 101A111    11Ềh 1Vh 21Ph 41Ih 1P78 Ố1u1 101A11CB10D1191A1CB1Đ11C10181CB111Ềh 24 Đ1Vh Đ1Ph 2 Đ1Ih Đ1P78 Ốu111A111B  11Ềh 2 C1Vh C1Ph 2 C1Ih C1P78 Ốu1 118C11A111 2  11Ềh 2 2 :  :    X 9 ) 9 )8 ( 8 (1Vh 1Ph 2 2 :  9 )8 (1Ih 2 2 4  3 2 1P78 Ốu11A1111C1C1C101CB1g11g1 11BA10101A111C110101A111C1CB1Ềh 21Vh 1Ph 21Ih 1P78 Ốu11C11 11 18101CB1B11CA1 C1010111 101 g g  1Ềh g1Vh g1Ph 4g1Ih g1P78 Ốu1111 11CB1C10C1C1A11 211211 111111111111111111111111111111111111111111111111 DACB134151671891012111CB101A1CB10DC1 11Ềh 1Vh 1Ph 21Ih 1P78 Ốu1111B 1181B10D1A1101 gg: 9 )8 (11g1CB1Ềh 21Vh 1Ph 1Ih Bgg1P78 ỐĂu101112 1101CB11101101CB1CB1410A10181001C11C10DC1111CB1A1C1Ềh 1Vh 4 1Ph 1Ih 24 1P78 1-u101CB10DC1 B B 2   11A1CB12 1 C12 1Ềh 2 1Vh 2 2 1Ph 2 2 1Ih 2  1P78 1Ốu1 10CB100111B10D1A10A110131818101112 311A18CB101C1011A10D10A18101C11C011C101CB11Ềh 1Vh 1Ph 1Ih 1P78 11u1111 211 31C1011B10D1C1C011B10D1C1C01A1110DC18C11 C1 31Ềh 3 1Vh 2 3  1Ph 3 1Ih 2 3 1P78 1u110111 2   11A18110D1010C1010A1B11C101CB1Ềh 21Vh 1Ph 1Ih 21P78 1u1111 11811C1010DC1111CB1011811C1A111 CB1B10D1001118110D1A1112   11Ềh 1Vh 41Ph 1Ih 21P78 1u11C10D1191A1CB1C1C1811CB11601CB11CB1CB110D101C10D10101001C1111CB141CB110118C10D1A1C10D1CB1Ềh 1Vh 11Ph 1Ih 1P78 1u11C11 118111C11C01 g1C1C11CB1B1181 g1CB11BA1A18CB10CB1111CB1Ềh g1Vh g1Ph g1Ih g1P78 1u11(1101100111B10D1CBC1CB1A10A11318111 2 3  18CB1C10DC1CB11 CB1B10D11C101A1(1CB1Ềh 1Vh 1Ph 1Ih 41P78 1u11111C0110101191A1C101CB1g11g1011181C1C01CB1A1C1 111111111111111111111111111111111111111111111111 DACB1341516718910121Ềh g1Vh 2g1Ph 4g1Ih g1P78 1Ău1111  1C1CB18C18CB1Ềh 118CB1C10DC1CB1C  1Vh 118CB1C10DC1CB1 C 1Ph 118CB1C10DC1CB1 C  1Ih 118CB1C10DC1CB1C 1P78 -u111012g11g101g1CB1Ềh 1Vh 41Ph 241Ih 21P78 Ốu11 g t 111101110A17C1 4g t  1181 g t1CB1Ềh 21Vh 241Ph 41Ih 1P78 1u11A11010112 2g t 112 11A1CB1A1CB10DC122 g t1 DCB10DCB111012 2 2 4 4 :   9 )8 (1801B10D1C1C01CB18C1C11818CB1Ềh g t1Vh 4g t1Ph 2g t1Ih 24g t1P78 u110A1B1CB1C111 g 11A10A1B11AC18CB10CB181A111CB1CB1110A101810110DC1A110101CB1Ềh g 1Vh g1Ph g1Ih g 1P78 u11A11CC11CB10D1 11CC101C0118C11811C10DC1011C1C111CC101A118C11811C10DC1011C1C11 101C1CB1A1A11CC18711CB1Ềh 1Vh 1Ph 1Ih 1P78 u110011A1C1B10D1CBC1018C11A10A1131818CB10CB1 3 101810111 2101A181C101Ềh 41Vh 1Ph 1Ih 1P78 u11B B g t  11g t111101C1C12110D1A1Bgt1CB1Ềh 1Vh 1Ph 41Ih 41P78 u11C11 1 111i  p1111C1011C11A110DC111C1C11A1011CB101C1111Ềh 1Vh 21Ph 21Ih 1P78 u11111C1 118110DC1110111 1C1C1111111  1CB1C10DC1CB1C1181 5111 111111111111111111111111111111111111111111111111 DACB1341516718910121Ềh 1Vh   1Ph 221Ih 21P78 Ău11111A1 118101C1C1 1110011A1C1B10D1CBC1A10A113181CB10DC1 C 4 2 3 3  11CB1Ềh 1Vh 1Ph 1Ih 21P78 -u1101C1891191A1CB1Đ1 101A1101C187111Ềh Đ1Vh Đ1Ph Đ1Ih Đ1P78 Ốu11C11    1 110101A1101C1  110101A111    1CB1Ềh 1Vh 241Ph 21Ih 21P78 1u1 CB1111A1001111CB10DCB1A10DC1C101 11Ềh 21Vh 1Ph 21Ih 21P78 u1111    3 3110011A1C1B10D1A10A11013181810111010D1C101181C1011C18110C10111CB1Ềh 21Vh 1Ph 1Ih 1P78 u11C11 11 g10A1B118910A1B11CB1C10111C10DCB101CB1CB1B1101CB181CB110118C101CB11CB1Ềh g1Vh 2g1Ph 2g1Ih 4g1P78 u110011A1C18CB10CB101810111 2101A181C1011A1BA181811C18110CB181111CBC1Ềh 21Vh 1Ph 41Ih 21P78 u11011C111C01A181C1B10A11A11CC1BCB1CA11001C1A10D11010A1B1CB1C111A11A18CB10DC181A1A11CC181010111CA1011CC118CB10DC1811011101C1A18111A11CC1C1118CB10DC1810111A1C1A1811A181CB10A1801C18C1A1A11CC101111C1C1CB11C1C1C181A11CC10111A181CB10DCB1C110CB1CB1C10D1DA11 101 101C1AC18110DCB1101CB1C11818C110C101Ềh 2221Vh 2221Ph 221Ih 221 111111111111111111111111111111111111111111111111 DACB14341516718910121P78 u11311 B1 0D1 C1 C01 A1 0A1 1 0131A1 1 CB1 0DC12 2 2 B B 2 3 3 3       11CB101CB11CB18C1C11818CB11Ềh 23: R 9 )8 (1Vh 23: R9 )8 (1Ph 24: R9 )8 (31Ih 3: R 9 )8 (1P78 u11CB10D1    118111C11C01141 1101CB1  1CB1B11018101A101CB1  1  1011CA1B1-110AC-1 101A11CB10D1    11Ềh 21Vh 41Ph 1Ih 21P78 Ău1 A111C11 2 3    1 CB100111B10D1CBC1A10A11318 8101A11101CA10101811C01CB1Ềh 41Vh 2221Ph 1Ih 221P78 -u11C11 1181110A1B1CB1C1011 g10A1B1110A1B11C101CB101111CB110118C101CB11CB1g1C1A1B1BA1A101CB1111CB1Ềh 1Vh 21Ph 1Ih 11555555555551 15555555555111 1 SỞ GIÁO DỤC VÀ ĐÀO TẠO NINH BÌNH ĐÁP ÁN ĐỀ THI THỬ THPT QUỐC GIA LẦN THỨ 1 - NĂM HỌC 2019 – 2020 Môn: Toán (Đáp án gồm 03 trang) Câu hỏi Mã đề thi 001 002 003 004 005 006 007 008 1 D A C A B A D A 2 D A A C A B C A 3 A B A D D D B B 4 C C D B C D D D 5 A C B A C B D C 6 B D C B D A B B 7 C D A A D C A D 8 A D B A B C C D 9 B B D C C C A A 10 C B A C D A B D 11 A D D C B A C C 12 D D D B C B B C 13 C A C A B C D B 14 D B B C A C C B 15 C D B A D B A D 16 C C A A C D B C 17 B B C B B A A B 18 B B C A B A A C 19 D A C C D C A C 20 A D C C A D A D 21 A C C D A A A B 22 D A D D B C C D 23 D B C D B C A A 24 A D D B A D D A 25 D C A D D A B B 26 A B B A B B B D 27 B B D D A D B D 28 B C A D B A C B 29 A C B A A C C B 30 C C A C C A B D 31 D A D B D D A B 32 C A A B A A A A 33 C C B D B B C D 34 C B D B D B D A 35 A A A B A B D B 36 A B A C A D C A 37 C C B A D B D B 38 B D B B B B B C 39 B A D D A D C A 40 B A A D C D D C 41 D C B B C D B B 42 D A B C D C D C 43 B C C C C D B A 44 A D A A D D C A 45 C C C D C C C C 46 B A D A C B D D 47 D C A D A C B B 48 C B B A D B D A 49 B D D C C D A C 50 C D C B D A B C 2 Câu hỏi Mã đề thi 009 010 011 012 013 014 015 016 1 D D A C B B D D 2 A C C A D B A A 3 D A D C B C B D 4 A D A B C D C D 5 A D B A C D D A 6 A D C A B B C A 7 C A C B A A A D 8 A D C C D A C C 9 B B D A C D B C 10 A C A D A B A A 11 D B B D C D D B 12 D D D C C A D B 13 C B C D B B B B 14 B B C A C C D B 15 C A A C A D D D 16 C B B B D D D D 17 C D B C B B A A 18 B B D A B D A C 19 D A B D D D B C 20 C D A D D B B C 21 C C A A A C D D 22 D C A C D A D A 23 C A B B A A C C 24 B B D C A D B B 25 B D B D A A B C 26 D C A C C B A D 27 C D C D A C A D 28 B D B B D A C B 29 A A C B C C D D 30 C A D D D D D B 31 C C A B B C A B 32 B C C B B D C C 33 D B D D C C B C 34 A C B B C C A A 35 C A D A D B B A 36 B B B A A B C B 37 A B C B D D B A 38 D B C B A C D C 39 B C D C D A C A 40 B D A A A A A B 41 C D C B B D A D 42 B A B A D A B B 43 A A D C C A D B 44 A B B D B D D D 45 D A D B B C C C 46 D C B D D C C C 47 D C A D B B C A 48 A C D A A A C C 49 B A A B C C A C 50 D B B C C B B A 3 Câu hỏi Mã đề thi 017 018 019 020 021 022 023 024 1 B A D D D C D C 2 A A A C B B B B 3 B A D C D A D C 4 B A D C C A A A 5 D D C B B B C B 6 B A C A A C A B 7 C C B C A A D C 8 C B A D C D B A 9 B A C D B D A D 10 D D B C D A B B 11 C A C B A D B B 12 C C D D C B C A 13 D D A B B B D D 14 B D D B B C D D 15 D C A B D B C A 16 A B B A A A C D 17 A C D D D D B C 18 C B D B A D A A 19 A B B C C C C B 20 C D B B C D B B 21 B D D A C D C D 22 D D C A B C C C 23 A D B C C D A A 24 C C A A A A B C 25 D A D B C B C D 26 B A C B D D D B 27 D B A B B B A C 28 D D B D D A B B 29 D B D A A B D C 30 D B B A A C D B 31 C A B A D A A A 32 A C C A C B B C 33 A C A A D C D D 34 B B C C A D A C 35 C B A A C C A A 36 A A C C B C D A 37 C C D B A A B D 38 B A A D B B A B 39 B D C B D A B C 40 B C D D C B D C 41 A B A C C C C A 42 D B C C B C D A 43 C D A D B D A C 44 A A C C B A B C 45 B C B A D D B D 46 B B B A B D A B 47 C C B A D B C D 48 A C A D A A D A 49 D D D D A C C D 50 A A A D B D C D --------Hết-------- .Có họnđápDứ cứnran ưhượ 2iượ cể pợmượrợâ pbệưt hrV ACbBlaSxqẳirợh rg rưóợhạlrợ lràAgpóh,ưB sẽoẽo 2véơkéV rnọ ư yDNLxLa SRdIPdLm 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G(ÁO(D Ở(ÁỤ C(ỤO(Về có 0 óĐÀ ĐT ( GC( VDpkậảH x ả,ụ píădụ 0861 3;()4,56 [4] /7n1o; g]* s61i;\1{; }c !"s{#s1 c;11)]$%6s& '1l6$%6s & }c ()*6. s+s6 , =S O  6t ((Á  $%6s&2pkậảH x ảpụ píăqụ pkậảH x ả, Ạ(  Đ (  r a ng ) T Đ  píăìụ 01]-.6)# 6/)61.n Ở(I À (V  V S (VpkậảH x ả2ụ píăốụ 0]*76) ;;\);\ =S Ạ( )   =S Đ( pkậảH x ảpụ píăựụ 013;4l nl;2 píăịụ 0. 1% 2 & h34ờ5i2 & ờ34i&ờ6Đ&l4[26 àm (pkậảH x ả2ụ píătoụ 0. =S bẠ(hĐ  47] 8+.'1]9.) =S Đ((:$) *! ; 4  <  = pkậảH x ảpụ píăttụ 0. I NbÁbỤbObN ( À bÁbỤbObN NbN( bÁbỤbÁbNNbN(  ( pkậảH x ả,ụ píătẽụ 0.s1 /s>61 }?;\s(@ =S I(ỤỞpkậảH x ảđụ píătổụ 0. bẠ( hĐ 6 t b hĐ 6 t   (h hĐ 6 t   ( Đh 6 t   pkậảH x ảpụ píătdụ Ac7!; Bl64=;1 i Đ h:iĐ: 9 pkậảH x ảđụ píătqụ 0. I H I B (  Ì H   Ì B (Ì H Ì B r a ng ) T  (V V r a ng ) T  ( pkậảH x ảpụ s7Hò ơké oot/Ӡ, *,Ҧ, &H, T,ẾT ___________________________________________________________________________________________________________________ píătìụ hr1 8ì3l; Sí ï8 3l; ì83Sí ï í = l3ìl;C V ÀĐ T Ạì83Sí öì9ì9ì ï9Sï 9 ï ö ìpkậảH x ảpụ píătốụ hr 1 f x DỤ SfL DỤ S NLlVD&Ckfq :CVCE*xr VyRCkVRM I pkậảH x ả2 píătựụ hr fLC a C V ÀĐ T Ạ1f a D Ụ Se#k 2 LC aC V ÀĐ T Ạ1e#k 2 aSe#k 2 LC a aC V ÀĐ T ẠS2 pkậảH x ả2ụ píătịụ tBC'>CV V:CVfqDC fVREf8R MC 2 ] ø;÷ S] = N/ = ø7-øo÷÷SíÞ = Nö = SçìhV. $RV: CVfqDC 2 = ] ø;÷ Síö ìpkậảH x ảđụ píăẽoụ hr e#k LL xNL 41e#k L x D Ụ N4S5Be#k L L xNLe#k L xN2S5Be#k L xS N4 e#k L xS2K ỐỆỆBxS4 L xS3K ỐỆỆỆ (:$) * 2 6 í 6 = S(ìpkậảH x ả,ụ píăẽtụ 0RMIx) CbSCk 2 6Sgî+α}ö) îîë÷îë 2 )SíìA÷Î(-F îë+α} ö)îî>*ö Gîë îí}I+í:îíöí Jîí)+ö ì8αî +>öí 2 íPgS£BgS Ầ£ìpkậảH x ả,ụ píăẽẽụ hr Ứ)S£ ø61=÷ = QÞ-U6ÌÛÞîïÎỞ*1gS)øï÷1)øÞ÷SíNí =Sí =ìpkậảH x ảđụ píăẽổụ hr Ứ)S(6 ï N(6SÞB6SÞî6S ẦíỞ3Þîí D Ụ -lNíîÞ D Ụ -;íîÞ D Ụ Ở8 3l; Sí =l;ì/ø3-l;÷Sí =ì=ìíSíì pkậảH x ảđụ píăẽdụ AEIg; 2 sø6÷öìÛ (Î{ }ö<ö+ø= 2 6SÞî6S=ìú:<) *í>}å ;DỤB(aKB(B (y fx S N1 xrVyR bRd I x61s1 ,H6 2 =. = N6 2 P0. ()*xV.$R V:CV xVJ CV)feE = =. = N6 = 10 DỤ S=öB6S. = NíïN0 = C V ÀĐ T Ạ = S£ = NíïN£ = C V ÀĐ T Ạ = SïìpkậảH x ả đụ píăẽìụ hr 2 3l{{,;Ở3l{{ø8;,÷Ở/ø3l-8,÷S/ø3l-ø8;,÷÷S/ø3-ø8;,÷÷ì PQ 2 3iA8,Ở3iAø8;,÷ìÞì 2 í 3i 2 Sí 3, 2 1í 38 2 Sí (9 2 1í (9 2 Ở3iS=9ìpkậảH x ả,ụ píăẽốụ hr 1 ỨyS6x 2 N6mxĂ5OUxQ2Bx L NmĂ5OUxQ2BmHx L OUxQ2BmHK 07Ckx Bxg;Ck.* RC8D&Ck xSCf:ImTCk \ \Síè M S(£ìpkậảH x ả2ụ píăẽựụ aRMCf >xVbB*x/y VIleE 8S1 0 Sè9 ï 9 Sè9 = ì P>xVfVDH xVIleE6P)P{Ở6)Sè9 = {S9GI O Áì hV. $RbB *VIl 2 =61) DỤ Ă=ì=6)S( è9 = Sí=9ìpkậảH x ả 2ụ píăẽịụ hr Ứ)S=} =6 NíQÞB} =6 Qí =B=6Q)æí =Ở6Qí =)æí =SNí =)æ=SN)æ= ìpkậảH x ả,ụ___________________________________________________________________________________________________________________ píăổoụ hr 2= Sè Vö (Sè Áö 6t Á (SGS Á (IĐGpkậảH x ảpụ píăổtụ hr 6 (K  (U  (ể5i (e#k 6  (e#k K  (e#k U ề có 0 óĐ p (e#k U  e#k 6  pe#k U  e#k K  (e#k U 6pe#k U K(e#k U 2U(L pkậảH x ảđụ píăổẽụ hr &  & L h4 (4i&eC pV& L h4WeC(5i& L peC eC&h4(5Đ& 4 p& L ( heC eC & 4 & L ( h4ề có 0 ó (:$) *ç(hI h ỤCè ỤCD r a n nng ) T TT S hĐhỤCè ỤC( r a ng ) T(ỤC( ỤCèr a ng ) T S pĐỤCè ỤC((ỤC( ỤCè r a ng ) T S pÁỤCè ỤC(pÁỤCè ỤC(5ÁỤC( ỤCè r a ng ) T S GÁỤCè ỤC(GÁỤCè ỤC( Á (Á V Á G a+.mTCk bzffz R 2 ỤC( ỤCè r a ng ) T S (ÁỤCè ỤC(iỤCè ỤC((I Á Á iỤCè(ỤC( I Á Á iè(( I Á Á GpkậảH x ảpụ píăổổụ hr í( ThöẠ S hI{ö hI Ở u p ö S hI{ö Ở I u (D Á GpkậảH x ảpụ píăổdụ 0VREf8RM C9.yfq@x'VRx Nf VyR'V;RCrCCVD CVDV: CV$KmRCX YVSCxV.Ckx /yVyR CrCeEVyR CrCx rxV.Ck mBC'>CV 2= ì £ (}>(}"OxVR-.xy# fD&CkS CkeE 2= }çO}bçG bIỀ1 âĐIĐSO1 âOILTOỐí( }> S G}ç Á p }> S G}bç Á ( }> S Gçbç Á (ì }> S G 0VO# 0VyeOgxr 2= >ç >ï (îç ï bç(I SĐç> çï(I ÁĐ}> b ç ï(ç> çï(I ÁĐ}>(Sì ÁG ()*í(ì S Á ìr a ng ) T S (D V ì Á GpkậảH x ảpụ___________________________________________________________________________________________________________________ píăổqụ hr L&-2 &-4 =&+7ÛL&-2=V&-4WV&+7WÛ& L +V7-2W&-7+2=5V4W ZR-.'RMC xNffzR VyRbRdIlVPC mRMfeE S( D>Û :- - -:+>Û:> :<-é ëê (:$) * S( :Î -  -    {} xrf +fx, S  !; .*R pkậảH x ả,ụ píăổìụ hr e#k 9 L o &=4 e#k & 9 L o æ è çö ø ÷ =4 Le#k & 9-e#k & o =4 L e#k 9 & -4 e#k o & =4 L L -4[ =[ 6 pkậảH x ả,ụ píăổốụ hr pkậảH x ảpụ píăổựụ hr S( ¢p g=, g+  g-  g-,> () ì (:$) *¢T= g¢p g + = ,gg + + ( ) g + - ( ) g + - ( ) = ,g  g + ( ) g - ( ) <Û5$E fqRCC\y'V#, Ck S +¥! ;]E V ]=R W (:$) * 1% 'L gRC& () =' 7 L +U7+45 () ÛL gRC&=7 L +U7+45 YVD&Ckfq: CVxrCkVRM I S( Û£: +ö:+íÞ£=Û-(£:£-=ìhr2 g; Ck.*RC fV#,I^C pkậảH x ả2ụ píădoụ 0Vdf >xVfS8RM Cb-.x zCV ( aôểc= a    hVR-.xy# fS8RM Cb-.e EĐ=c n= a   æ è çö ø ÷  a  = aÞa= Đ (:$) *c=   Đæ è çö ø ÷  = Đ  pkậảH x ả2ụ píădtụ hr c ó0¢t¢h = c ót0h¢ó¢t¢0¢h pkậảH x ảđụ píădẽụ hr &-L( ( ) L5L5 =) L5L5{ & L5L5-{ V-L(W { {=5L5L5 å =V-LW { ) L5L5{ & L5L5-{ ( { {=5L5L5 å 07Ckf +fx,xB xVMg; x/y'VyRfqRdC mTCk V-LW { ) L5L5{ {=5L5L5 å bDNxf >CVmTCkxV# S( g=T=$# $E /    R $T S -  () =Þ=Þ =pkậảH x ảđụ _hV`a xB xOImb9.yxP .VbR CE* O': fVRC?ICy*'VcCkfVR CV=fV SxCOdf&C ___________________________________________________________________________________________________________________ píădổụ hr 6 í 16 = 16 = Sïg 6 í 16 ï S=6 = G IÁ O ÁỞ=6 = 16 = SïgB6 = SgỞg ï Nïg ï 1=gSÞBgSÞîgS Ầíì 0V\e zRxVeCV)C (2 gSẦí({ æí-ùîë+ø: îí (2 6 ï NÁg6 2 1SgSỞxr2 CkVR MIlVPCmRMf pkậảH x ả2ụ píăddụ fAR 2 i}cs(.;\ 4]/ 76{;1 (2 3lỞ8iAø3l;÷Ở/øl-ø83;÷÷S=/øi-ø83;÷÷ì 0ç8Î ±î (2 iì83;$.c;\ s{]iỞí / = øi-ø83;÷÷ Sí i3 = 1í i; = 1í i8 = Sí = 9 = C V ÀĐ T Ạ = 1í ï = =9C V ÀĐ T Ạ = 1í = 9 = C V ÀĐ T Ạ Sí( ï9 = ì (:$) */øl-ø83;÷÷S=ï í(9S(= 9 } ìpkậảH x ả,ụ píădqụ hr (SL&12 &N4SLV&N4W1[ &N4SL1[ &N4 (HR (2 6-)ÌỞ6NíÌ Ầí-Ầ£ ỀÍ Ở(bRdIfqR Cb]fV=xr f#zbI eE xBxg ;Ck.*RC ( :$)*x rf+fx, (2 ; (= SöbDFCkfV HCkbR9.yLfq#Ck 6bR dI fqR Cx Nf b]fV=fzR VyR bRdIlVPC mRMfx rf#zbIeE xBxg ;Ck.*R C pkậảH x ảpụ píădìụ 0Vdf >xVCDHxfqE CqymTCk fVdf>xVx S. gfVdf >xV2CrC $EmTCk (2 ( ï ] ï 1ïí ï. = 0C V ÀĐ T ẠS( ï ( ï .C V ÀĐ T Ạ ï 1ïí ï. ï C V ÀĐ T ẠSïï}  =( B.Sï =ì (HRkR ,fVREfb^xV# fV:bB*x/y VIl$E 2bDF CkfqDCbB *x/yCrC xr$=fq >CVD V:CV $KmR CX P>xVfVDH xbB*VIleE (.P=.1ï. DỤ ì0VREf8RM C9.yfq@xx/y CrCe EIIffyI kRBx $.cCkx PCCRC bI8ER bDFCkgRCV x/y CrC nS=. = S=.ìhVR-.xy# x/yCrC (2 0S.ì___________________________________________________________________________________________________________________ Khốit âmcótâ mI,R=4 3r.Cácnónc óđỉ nh S S 1 ,S 2 ,S 3 .Theo giảthiết chiềucaoc ủahộpbằng h+d(I,(S 1 S 2 S 3 ))+R=h+R+R 2 -R (S 1 S 2 S 3 )2 =h+R+R 2 -R (O 1 O 2 O 3 )2 =r+4 3r+4 3ræ è çö ø ÷ 2 -2r 3 æ è çö ø ÷ 2 =3r. Vậythể tíchc ủahộpbằng4r´2r+3r ( ) ´3r=81(2+3 ) 2 .rfC,A1 B1,i@ r>?8;@ Đặt t=log 1 2 (x-2)=-log 2 (x-2)Î(-1;+¥),"xÎ(2;4).Phươngtrì nhtrởthành: (m-1)t 2 -(m-5)t+m-1=0Ûm(t 2 -t+1)=t 2 -5t+1Ûm=g(t)=t 2 -5t+1 t -t+í ø±÷ì Cầntì m mbd V_Wx rCkVRM I S tÎ(-1;+¥),'1,#! ls$c})n g,;\ g]E;s1]R; S g(t)!.* ()-3£m<7 3. Vậy S m 0 =-3.rfC,A1 B1,6@ r>?8<@ Kẻ S ¢AH^ACÞ¢AH^(ABCD);¢AH=h>0.0) 6. S V ABCD.¢A¢B¢C¢D =S ABCD .¢A H=AB.AD.¢A H=18h. Xétchóp S ¢A.ABC.PQ S HK^ABÞAB^(¢AHK),AB//HK. 0. S V ABCD.¢A¢B¢C¢D =6V ¢A.ABC .Ta cóS ¢A AC =1 2AC.¢A H=3 2hvà HC= ¢A C 2 - ¢A H 2 =9-h 2 ÞAH=3-9-h 2 ;A¢A=h 2 +3-9-h 2 () = ì Vìvậ y HK BC =AH AC=3-9-h 2 3 ÞHK=3-9-h 2 3 Þ¢A K=¢A H 2 +HK 2 =h 2 +3-9-h 2 3 æ è ççö ø ÷÷ 2 . Suy raS ¢A AB =1 2AB.¢A K=1 26h 2 +3-9-h 2 3 æ è ççö ø ÷÷ 2 . Vậy V ¢A.ABC =2S ¢A AB .S ¢A AC .sin(¢A AB),(¢A AC) () 3A¢A =2 1 2 6h 2 +3-9-h 2 3 æ è çö ø ÷ 2æ è ç ççö ø ÷ ÷÷3 2 hæ è çö ø ÷3 5 æ è çö ø ÷ 3h 2 +3-9-h 2 ( ) 2 =18 6 hÞh=4 2 3 .___________________________________________________________________________________________________________________ VậyV ABCD.¢A¢B¢C¢D =18 4 2 3 =8.Chọnđá pánC. Câu49. Phươngtrì nhhoànhđộgia ođiểm: lnx-2 x=3 x-2-1 x+4m-2020Ûm=g(x)=1 4lnx-2 x-3 x-2+1 x+2020é ëêù ûú. Điềukiện: Î\{0,2}.x Rhr¢g(x)=1 4 2 x(x-2) +3 (x-2) 2 -1 x 2 æ è çö ø ÷=0Ûx= ±1. Bảngbiế nthiên: Cắtnhaut ại 1điểmkhivà chỉ khi(*)cóđúng1 nghi ệm. Từbảng biế nthi ênsuyra m=505;m=506;m=505+ln3 4.Tổngc ácsốnguy êncầ ntìmbằng 505+506=1011.Chọnđá pánB. Câu50. Gọi H=h/c(S,(ABC))ÞSH=a. 0. C= BA^SA BA^SHìíîÞBA^(SAH)ÞBA^AH;BC^CS BC^SHìíîÞBC^(SCH)ÞBC^CH. Kếthợp tamgiá c ABC$.cCk xPCfzR BÞBAHCeEV: CV$.cCkxz CV2a. 0) 6.V S.ABC =1 3S ABC .SH=1 3 1 2 2a2aæ è çö ø ÷a=a 3 3 và V S.ABC =2S SAB .S SBC sin(SAB),(SBC) () 3SB =21 2 BA.ASæ è çö ø ÷1 2BC.CSæ è çö ø ÷sin(SAB),(SBC) () 3SB =1 6 2a3a2a3a 5 a sin(SAB),(SBC) () =a 3 5 sin(SAB),(SBC) () . Vậy a 3 5 sin(SAB),(SBC) () =a 3 3 Ûsin(SAB),(SBC) () =5 3 Þcos(SAB),(SBC) () =2 3.Chọnđá pánC.________________________________________________________________________ ___________________________________________-------------------- HẾT --------------------
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