Chifdtig: II TO H d P VA XAC SUAT Phan ^mifsG vA9f D £ CUA cm/dii^G I NOI DUNG Ndi dung chfnh eua chuong II: Quy tae ddm: Gidi thieu quy tac cgng va quy tac nhan va nhiing iing dung cua cac quy tdc Hoan vi - ehinh hgp - td hgp : Day la ba quy tac ddm cu thd nhdm dd dem cac phdn tii cua tap hgp hffu han theo cac quy luat thii tu ggi la hoan vi, chinh hgp va td hgp Nhi thiic Niu-ton : Nhdm tim he sd ciia mdt khai tridn (a + b)" Phep thit va bidn cd: Day la nhflng khai niem quan trgng cua xac sudt, Trong bai cdn dua nhiing quy tdc tfnh xac su& Xae sudt cua cac bidn cd II MUG TIEU Kie'n thurc Nam dugc toan bd kidn thiic co ban chuong da neu tren, cu thd : Hinh nhflng khai niem mdi cd hen quan ddn cac quy tac ddm Tfnh dugc sd cac td hgp, sd cac chinh hgp va sd cac hoan vi cua mdt taj hgp gdm n phdn tit Phan biet dugc su khac cua chinh hgp va td hop • xay dung dugc khdng gian mdu, each xac dinh bidn cd va xac suat 118 KT nang Suf dung thao cdng thiic td hgp, chinh hgp va cac cdng thfle vd xae suat • Ap dung tinh dugc cac bai toan cu thd Thai Tu giac, tfch cue, ddc lap va chu ddng phat hien cung nhu luih hdi kien thfle qua trinh hoat ddng Cdn than, chfnh xac lap luan va tfnh toan Cam nhan duge thuc td eua toan hge, lihdt la ddi vdi xac sudt III CAU TAO COA CHUONG Ndi dung eua chuong gdm hai phdn du kidn duge thuc hien 21 tiet, phan phdi cu thd nhu sau : PM/iA.Tdhgp(8tidt) § Hai quy tac ddm co ban tidt §2 Hoan vi, chinh hgp va td hgp tidt Luyen tap tidt §3 Nhi thfle Niu-ton tidt Luyen tap tidt Phdn B Xac sua't (11 tidt) §4 Bidn ed va xac sudt cua bien cd tiet Luyen.tap tidt §5 Cac quy tac tfnh xac sudt tiet Luyen tap tidt §6 Bidn ngdu nhien rdi rac tidt Luyen tap tidt 6n tap va kiem tra chuong 2 tidt 119 Phan 2, CAC B A I i§iOAIV A T HOP §1 H a i q u y t a c deiii cd b a n (tiet 1) I MUC TIEU Kien thurc HS nam dugc: • Hai quy tac ddm co ban : quy tac cdng va quy tac nhan • Bidt ap dung vao tiing bai toan : nao dung quy tdc cdng, nao diing quy tdc nhan KT nang Sau hge xong bai HS sfl dung quy tac dem thao Tfnh chfnh xae sd phdn tfl ciia mdi tap hgp ma sdp xdp theo quy luat nao dd (cgng hay nhan) Thai Tu giac, tfch cue hge tap Bidt phan biet rd cac khai niem quy tac cdng, quy tdc nhan va van dung tflng trudng hgp cu the Tu eac vdn dd cua toan hge mdt each Idgic va he thdng II CHUAN Bj cClA GV VA HS Chuan bj cua GV - Chudn bi cac cau hdi ggi md - Chudn bi hinh 2.1 Chudn bi phdn mau, va mdt sd dd dung khac Chuan bi cua HS Cdn dn lai mdt sd kidn thfle da hge vd td hop d ldp dudi 120 III PHAN PHOI THOI Ll/ONG Bai chia lam tidt IV TIEN TRINH DAY HOC A DAT VAN DE Cdu hoi Cd thd lap dugc bao nhieu sd cd chft sd khac tfl cac chft sd 1, 2,3,4 GV Cho HS liet ke Cdu hoi Cho 10 chft sd, 0,1, ,9 Cd thd liet ke dugc tdt ca cac sd lap tfl 10 chft sd tren dugre khdng? GV: Ta thdy : Rdt khd liet ke Do dd phdi cd mgt quy tdc didim sd cdc phdn tit cua mdt tap hgp B BAIMCII HOAT DONG M6DAU • GV neu bai toan SGK GV dat mdt vai cau hdi nhu sau: ?l| Hay vidt mdt sd mat khdu GV chia ldp td, mdi td viet mdt so mat khdu, sau dd cho mdt ban trinh bay xem cac td cd trung khdng? • Thuc hien [HIJ 3' Hoqt dpng cda GV Cdu hoi Hay vidt mgt sd mat khdu Hoqt dpng cua HS Ggi y tra Idi ckn hdi Ir64j5, abcdeh, 123456, 121 Hoqt dpng cda GV Hoqt dpng cda HS Cdu hoi Ggi y tra ldi cau hdi Cd the liet ke duge cap ki tu Khdng thd Uet ke mdt thdi gian khdng? nhdt dinh Cdu hoi Ggi y tra ldi c^u hdi Du doan sd m^t khdu? Khdng du doan dugc HOAT DONG Quy tac cdng • GV neu va thuc hien vf du Hoqt dpng cda GV Hoqt dpng cda HS Ggi y tra ldi cau hdi Cdu hoi Cd bao nhieu each chgn tai Cd31 each chgn ldp 11 A? Cdu hoi Ggi y tra ldi ckn hdi Cd bao nhieu each chgn tai Cd 22 each chon ldp 12B? Cdu hoi GcA y tra Idi c^u hdi Tdt ca cd bao nhieu cadh chgn 31+ 22 = 53 each chon • GV neu khai niem quy tac cdng Gid sit mdt cdng viec cd the dugc thuc hien theo phuang dn A hodc phuang dn B Cd n cdch thuc hien phuang dnAvdm cdch thuc hien phuang dn B Khi dd cdng viec cd the dugc thitc hien bdi n + m cdch Quy tac cdng bdi nhieu phucmg an Gid sic mdt cdng viec cd the dugc thuc hien theo mdt k phuang a« AJ, A2, , A|j Co nj each thuc hien phuang dnA^, n2 cdch thue hien phuang dn A2, vd n^ cdch thuc hien phuang dn A.^ Khi cdng viec cd the dugc thue hien bdi nj + n2 + • • • + n,j cdch 122 GV thuc hien vi du Vf du chi mang tfnh minh hoa • Thuc hidn [H2j 5' Muc dich Kidm tra xem hge sinh da biet van dung quy tac cdng hay chua Hoqt dpng cda GV Hoqt dpng cda HS Ggi y tra ldi cau hdi Cdu hoi Cd bao nhieu de tai Cdu hoi + + + = 31 (each chgn) Ggi y tra Idi cau hdi GV ddi sd va hdi xem cd HStutraldi bao nhieu each chgn • GV neu each phat bieu khac cua quy tac cdng neu chu y Sdphdn tit cua tap hgp hitu hgn Xdugc ki hieu la |x| (hodc n(X)) Quy tdc cdng cd the dugc phdt bieu dudi dgng sau : Neu Ava Bid hai tap hgp hitu hgn khdng giao thi sdphdn tilt cua A uB bdng sdphdn tit cua A cdng vdi so phdn tif cua B, tUc Id |AuB| = |A| + |B| Quy tdc cdng cd thd md rdng cho nhidu hanh ddng - Neu AJ, A2, ; A^ Id k tap hOu hgn vd A; n Aj = vdi i ^ j (vdii,j = 1, , k) thi\AiKj A2^ yJ A^.\=\AI\ + |A2| + " - + |A,^| - Hai tap hgp A, B bdt ki thi\A u B| = | A|+|B| - | A n B| HOAT DONG 2 Quy t^c nhan • GV hudng ddn HS thuc hien vf du 3, sfl dung hinh 2.1 Hoqt dpng cda GV Cdu hoi Hoqt dpng cua HS Ggi y tra Idi cau hdi Gia sfl tfl nha An ddn nha Cd 6.1 = dudng Binh cd dudng thi tfl nha An ddn nha Cudng cd bao nhieu each chgn? 123 Hoqt dpng cda GV Cdu hoi Hoqt dpng cda HS Ggi y tra ldi c^u hdi Hdi An cd bao nhieu each Cd = 24 each di tfl nha An qua nha chgn dudng di den nha Binh ddn nha Cudng Cudng? • GV neu quy tdc nhan Gid sit mdt cdng viec ndo bao gdm hai cdng dogn A vd B Cdng dogn A cd the ldm theo n cdch Vdi mdi cdch thuc hien cdng dogn A thi cdng dogn B cd the ldm theo m cdch Khi cdng viic cd the thuc Men theo nm cdch f Thue hien |H3| 5' Muc dich Kidm tra xem hge sinh da biet van dung quy tac nhan hay ehua Hoqt dpng cda GV Cdu hoi Hoqt dpng cda HS Ggi y tra ldi cau hdi Mdi each dan nhan cd bao Viec lap mdt nhan ghd bao gdm cdng nhieu cdng doan, hay kd ten doan Cdng doan thfl nhdt la chgn chft cae cdng doan dd cai 24 chft cai Cdng doan thfl hai la chgn sd 25 sd nguyen duong nhd hon 26 Cdu hoi Ggi y tra Idi cau hdi Cd nhidu nhdt bao nhieu Cd nhidu nhdt la 24.25 = 600 chide ghd chide ghd dugc ghi nhan dugc ghi nhan khac khac nhau? • GV cho HS md rdng quy tdc nhan cd nhidu hanh ddng Gid sit mdt cdng viec ndo dd bao gSm k cdng dognA^, A2, , Aj^ Cdng dogn Aj cd the thue hien theo Uj cdch, cdng dogn A2 cd thi thuc hien theo n2 each, , cdng dogn A^ cd the thuc hien theo ny cdch Khi dd cdng viec cd the thuc hien theo nin2 •••ny cdch 124 • Thuc hien vf du Hoqt dpng cda GV Cdu hoi Hoqt dpng cda HS Ggi y tra ldi cau hdi Mdi each lam mdt bidn sd xe Cd cdng doan: Chgn chft cai may cd bao nhieu cdng doan, 26 chft cai; cdng doan chgn hay kd ten cac cdng doan dd: chft sd, ed each chgn, va cdng doan cdn lai mdi cdng doan chgn chft sd va cd 10 each chgn > Ggi y tra Idi c^u hdi Cdu hoi Thed quy tac nhan, ta cd tdt ca Cd bao nhieu each lam mdt bidn 26 10 10 10 10 = 2340000 sd xe may? (bien sd xe) • Thuc hien Vl du Hoqt dpng cda GV Cdu hoi Hoqt ddng cda HS Ggi y tra ldi cau hdi Cd bao nhieu day gdm kf ttt Vl ed 26 + 10 = 36 each chon nen mdi kf tu hoac la mdt chft cai theo quy tae nhan, ta cd the lap duge (trong bang 26 chft cai) hoac la 36 day gdm kf tu nhu vay mdt chft sd (trong 10 chft sd tfl ddn 9) Ggi y^tra ldi cau hdi Cdu hoi Cd bao nhieu day gdm kf tu Vi mdi kf tu cd 26 each chgn nen ndi d cau>a) khdng phai la mat theo quy tac nhan, sd day gdm ki tur khdng phai la mgt mat kh'du la 26 khdu? Cdu hoi Ggi y tra ldi eSu hdi Cd thd lap duge nhidu nhdt bao cd36^-26^ nhieu mat khdu? 125 HOATD6NG4 TOMTATBAIHQC - Gia sfl mdt cdng viec co thd dugc thuc hien theo phuong an A hoac phuong an fi Cd n each thuc hien phuong an A va m each thue hien phuong an B Khi dd cdng viec cd thd dugc thuc hien bdi n + m each - Gia sft mdt cdng viec cd the dugc thuc hien theo mdt ^phuong anAj, A2, , Ajj Cd Wj each thuc hien phuong anAj, n2 cdch thuc hien phuong an A2, va n^ each thue hien phuong an Aj^ Khi dd cdng viec ed the dugc thuc hien bdi Uj + n2 + • • • + nj^ each - Gia sfl mgt cdng viec nao dd bao gdm hai cdng doan A va B Cdng doan A cd thd lam theo n each Vdi mdi each thuc hien cdng doan A thi cdng doan B cd the lam theo m each Khi dd cdng viec cd thd thuc hi6n theo nm each - Gia sfl mdt cdng viec nao dd bao gdm k cdng doan Aj, A2, , Aj^ Cdng doan AJ cd thd thue hien theo Uj each, cdng doan A2 cd the thuc hien theo n2 each, , cdng doan Aj^ cd thd thuc hien theo n^ each Khi dd cdng viec cd the thuc hien theo njn2 Uj^ each BOAT DONG Cdu Cdu 126 MOT S6 CAU HOI TRAC NGHI£M Mdt bai tap gdm cau, hai cau cd cac each giai khdng lien quan den Cau cd each giai, cau cd each giai So each giai dd thuc hien eac cau bai toan tren la tren la (a) 3; (b) 4; (c)5; (d)6 Trd ldi Chgn (c) DQ giai mdt bai tap ta cdn phai giai hai bai tap nhd Bai tap cd each giai, bai tap cd each giai Sd cac each giai dd hoan bai tap tren la (a) 3; (b)4; (c)5; (d)6 Trd ldi Chgn (d) Cdu Mdt Id hang duge chia phdn, mdi phdn duge chia vao 20 hop khac Ngudi ta chgn hop dd kidm tra chdt lugng Sd each chgn la (a) 20.19.18.17; (b) 20 + 19 +18 + 17; (c) 80.79.78.77 ; (d) 80 + 79 + 78 + 77 Trd ldi Chgn (e) Cdu Cho cac chft sd: 1, 3, 5, 6,*8 Sd cac sd chan cd chft sd khac cd duoc tfl cac sd tren la: (a) 12; (b) 24; (e) 20; (d) 40 Trd ldi Chgn (b) Cdu Cho cae chft sd: 1, 3, 5, 6, So cae sd chan cd chft sd khac cd dugc tfl cac sd tren la: (a) 4.3.2; (b) + 3+ 2; (c) 2.4.3.2; (d) 5.4.3.2 Trd ldi Chgn (c) Cdu Cho cac chft sd: 1, 3, 5, 6, Sd cac sd le cd chfl sd khac cd dugc tfl cac sd tren la: (a) 4.3.2; (b)4 + 3+2; (c) 3.4.3.2 ; (d) 5.4.3.2 Trd ldi Chgn (e) Cdu Mdt ldp hge cd td, td cd ban, ba td cdn lai cd ban a) Sd each chgn mgt ban lam ^ trudng la (a) 17; (b) 35; (c) 27; (d) Trd ldi Chgn (b) b) Sd each chgn mdt ban lam ldp trudng sau dd chgn ban ldp phd la (a) 35 34.32; (b) 35+ 34+ 33; (c) 35.34; (d) 35.33 Trd ldi Chgn (a) 127 Phdn Tu ludn (6 diem) Cdu Gieo hai sue sac can ddi a) Tfnh xae sudt dd tdng hai mat xudt hien bang b) Tinh xae sudt dd tich hai mat xuat hien la sd le c) Tfnh xac sudt dd tfch hai mat xudt hien la sd chan 0^2 Phdri Trde nghiem khdch quan (4 diem) Cdu Hay didn dung, sai vao d trdng sau day Cho tap hgp ed n phdn tfl (a) Sd cac hoan vi eua n phdn tfl ldm hon sd cac td hep chdp k eua n [j (b) Sd cae hoan vi eua n phdn tfl ldn hon sd cae ehinh hgp chap k cua n [j (c) Sd eac chinh hgp chap k cua n phdn tfl ldm hon so'cac td hgp chap k eua n |_1 (d) Sd cac td hgp chap k eua n phdn tfl ldm hon sd cac ehinh hgp chap k eua n Cdu [j Hay didn dung, sai vao d trdng sau day Cho tap hgp gdm n phdn tfl Cdu (a) Sd cac ehinh hgp chap k eua n phdn tfl la A„ jj (b) Sd eac ehinh hgp chap k cfla n phdn tfl la C„ [j (e) Sd cae chinh hgp chap k eua n phdn tft la AJ^ n Q (d) Sd eac ehinh hgp chap k cfla n phdn tfl la A^ [] Hay chgn cau tra ldi dung cac eau sau: Trong mgt ldp hge Xet bidn ed A : Chgn mgt ban hge sinh gidi van; bidn cd B: chgn mdt ban hge sinh gidi toan Bidt n(A) + n(B) = n(AuB) Khi dd A va B la hai bidn ed: 211 Cdu (a) Ddc lap; (b) Xung khdc; (c) Ddi; (d) Cd giao bang rdng Hay chgn cau tra ldi dung cac cau sau: Gieo mdt eon ddng xu hai ldn Sd eac phdn tfl cua khdng gian mdu la : (a) 4; (b)2^ (c) 1+2 ; (d) Phdn Tu ludn (6 diem) Mdt ldp hge ed 25 hge sinh, dd cd 15 em hge kha mdn toan, 16 em hge kha mdn ngoai ngfl a) Tinh xae sudt dd chgn duge hai em hgc kha ca hai mdn; b) Tfnh xac sudt de chgn duge em hge kha mdn toan nhung khdng kha mdn van Phdn Trde nghiem khdch quan (mdi cdu diim) Cdu (a) (b) (c) (d) S D (a) (b) (c) (d) S S S D S Cdu Cd«3 (b) D Cdu (b) Phdn Tu ludn (6 diem) a) Ta cd n(Q) = 36 Cac bidn cd thudn lgi cho A la {(2, 6), (6, 2), (3, 5), (5, 3), (4,4)} Ta ethdy n(A) = Vay P(A) =^: 36 212 b) Xae sudt dd mdi eon sflc sdc xudt hien mat le la— Vay dd hai mat • ddu le thi xac sudt la —.— = — (do hai bidn cd mdi mat xudt hien 2 mat le la ddc lap) c) Xae suat dd tfch hai mat la mdt sd chan va tfch hai mat la mdt sd la la hai bidn cd ddi Vay kdt qua la — = — 4 Dfi2 Phdn l.Trdc nghiem khdch quan (mdi cdu diim) Cdu (a) (b) D D (c) D (d) S (a) (b) (c) S (d) Cdu D Cdu (a) S S Cdu (b) Phdn Tu ludn (6 diem) Ggi A la bidn ed: Ban dd hgc kha mdn todn Ggi B la bidn ed : Ban dd hge kha mdn van a) Ta cd n(A nB) = n(A) + n(B) - n(AuB) = 15 + 16 - 25 = VayP(AnB)=^b) Ta cd sd hgc sinh kha todn nhung khdng kha van la : n(A) - n(AnB) = 15 -7 = vay xae sudt cdn tim la : —^ •^ 25 213 x* MOT SO CAU HOI TRAC JSTCAIIl^M ON TAP HOC KY I CAU HOI DUNG SAI Hdy khoanh trdn y md em cho Id hgp ly Cdu Tap xae dinh eua ham sd y = sinx la R (a) Dung; Cdu Tap gia tri cua ham sd y = cosx la doan [1; 1] (a) Dung; Cdu (b)Sai Ham so y = cotx vfla la ham sd chan vfla la ham sd le (a) Dung; Cdu (b) Sai Ham sd y = tanx vfla la ham sd chan vfla la ham sd le (a) Dung; Cdu (b) Sai Ham so y = cosx vfla la ham sd chan vfla la ham sd le (a) Dung; Cdu (b) Sai Ham sd y = sinx vfla la ham sd chan vfla la ham sd le (a) Dung; Cdu (b) Sai Chu ki cia ham sd y = tanx.eotx la bdt ki (a) Dflng; Cdu (b) Sai Chu ki cia ham sd y = tanx cotx la TI (a) Dung; Cdu (b) Sai (b) Sai Trong doan [0; TI] phuong trinh sinx = sina cd nghiem (a) Dung; (b) Sai CdH 10 Trong doan [0; Tt] phuong trinh cosx = cosa ed nghidm (a) Dflng; (b) Sai Cdu 11 Trong doan [0; Tt] phuong trinh tanx = tana cd nghiem (a) Dung; 214 (b) Sai Cdu 12 Trong doan [0; Tt] phuong trinh cotx = cota cd nghiem (a) Dflng; (b) Sai Cdu 13 Hai bidn ed ddi la hai bidn ed xung khde (a) Dflng; (b) Sai Cdu 14 Hai bidn ed xung khac la hai bien cd ddi (a) Dflng; (b) Sai Cdu 15 Ndu A va B la hai bidn cd ddc lap thi P(A nB) = P(A).P(B) (a) Dung; (b) Sai Cdu 16 Ndu A va B la hai bidn cd ddc lap thi P(A) + P(B) = (a) Dung; (b) Sai Cdu 17 Ndu A va B la hai bien ed xung khdc thi P(A uB) - P(A) +P(B) (a) Dung; (b) Sai Cdu 18 Cho P(A) =0,3; P(B) = 0,5; P(AB) = 0,2 dd hai bidn ed A va B ddc lap (a) Dung; (b) Sai Cdu 19 Cho P(A) =0,4; P(B) = 0,5; P(AB) = 0,2 dd hai bidn ed A va B ddc lap (a) Dflng; (b) Sai Cdu 20 Cho P(A) =0,3; P(B) = 0,7, P(AuB) = Khi dd hai bidn cd A va B xung khac (a) Dung; (b) Sai Cdu 21 Cho P(A) =0,3; P(B) = 0,6, P(AuB) = Khi dd hai bidn cd A va B xung khdc (a) Dung; (b)Sai Cdu 22 Chg P(A) =0,3; P(B) = 0,7 Khi dd hai bidn ed A va B ddi (a) Dung; (b) Sai Cdu 23 Cho P(A) = 0,4; P(B) = 0,7 Khi dd hai bidn ed A va B dd'i (a) Dung; Cdu 24 (b)Sai Cho P(A) =0,3; P(B) = 0,5 Khi dd hai bidn cd A vd B ddi (a) Dung; (b) Sai 215 n DI£N DUNG, SAI VAO THICH HOP Hdy dien dung, sai vdo cdc trdng sau ddy md em cho Id hgp li nhdt Cdu 25 Ham sd y = sinx: • n (a) Ddng bidn tren khoang (0; Tt).(b) Nghich bidn tren khoang (0; TI) Tt , (c) Ddng bidn tren khoang (0; — ) Tt D (d) Nghich bidn tren khoang (0; — ) a S b c d 00 Trd ldi D S Cdu 26 Ham sd y = cosx: (a) Ddng bidn tren khoang (0; Tt) • (b) Nghich bidn tren khoang (0; Tt) Tt (c) Ddng bidn tren khoang (0; —) • Tt (d) Nghich bidn tren khoang (0; —) Trd ldi a b c d S D S D Cdu 27 Ham sdy = tanx: (a) Ddng bidn tren khoang (0; Tt) (b) Nghich bidn tren khoang (0; TI) Tt , (c) Ddng bien tren khoang (0;—) 216 D • • n TI, (d) Nghich bidn tren khoang (0; —) Trd ldi Cdu 28 a- b c d S a D S Chgn em hge sinh nam de di da bdng Sd cae each chgn la (a) Sd cae hoan vi eua |_J (b)Ai D D D (c)C^ (d) Ca ba cau tren deu sal Trd ldi a b c d S S D S Cdu 29 - Chon em hge sinh nam dd di da bdng vao vi trf khac Sd cae each chon la U (a) Sd cae hoan vi eua (c)Ct n • (d) Ca ba cau tren ddu sal D (b)A^ Trd ldi a b e d S D S S Cdu 30 Chgn em hgc sinh nam de di da bdng vao vi trf khac So cac each chgn la (a) Sd cac hoan vi eua [_| 217 D D D (b)A^ (c)C^ (d) Ca ba eau tren ddu sal Trd ldi a b e d D S S S m CAU HOI DA LUA CHON Chgn cdu trd ldi dung cdc bdi tap sau: Cdu 31 (a) cos 1> cos 2; (b) cos 1< cos 2; (c)cosl - [J (e) Phuong trinh sinx = m ed nghiem - < m < [J (d) Phuomg trinh sinx = m ed nghiem vdi mgi m [j Hay didn dung, sai vao d trdng sau day (a) Ham sd y = sin2x cd gia tri ldm nhdt la £J (b) Ham sd y = sin3x ed gia tri nhd nhdt la - U (e) Ham sd y = tan2x ludn ddng bidn [J (d) Ham sd y = cot3x ludn ddng bidn [J Hay chgn cau tra ldi dung cac cau sau: Cho diem mat phang Sd eac doan thang cd dugc tfl didm dd la: Cdu (a) 10; (b) 5; (e) 15; (d) 20 Cho hinh binh hanh ABCD va mdt didm E ^ (ABCD) dd giao diem cua hai mat phang (ABCD) va (EAC) la (a) A; (b) C; (e) AC; (d) CE Phdn Tu ludn (6 diem) Cdu Giai cac phucmg trinh sau day a) sin2x + tan2x = 0; 220 b) eos2x -i^cos3x = Cdu Gieo hai sue sac can ddi Tfnh xac sudt dd tdng hai mat eua hai eon sue sac la mdt sd chan Cdu Cho hinh chdp S.ABCD, day ABCD la hinh binh hanh a) Hay xae dinh giao tuydn d cua hai mat phang (SAB) va (SCD) b) Ggi M la mdt diem tren SA Mat phang (BCM) mcdt SD tai N Chumg minh BM, CN va d ddng quy D£2 Phdn Trde nghiem khdch quan (4 diem) Cdu Hay didn dung, sai vao d trdng sau day (a) Phuomg trinh cosx = m ed nghiem m < LJ (b) Phuong trinh cosx = m ed nghiem m > - |_1 (e) Phucmg trinh cosx = m ed n^ghiem - < m < \_j (d) Phuong trinh cosx = m cd nghiem vdi mgi m Cdu Cdu [j Hay didn dflngj sai vao d trdng sau day (a) Ham sd y = sin2x + cd gia tri ldm nhdt la [J (b) Ham sd y = sin3x cd gia tri nhd iihat la - [j (e) Ham sd y = tan2x + ludn ddng bidn [j (d) Ham sd y = eot3x - ludn ddng bidn LJ Hay chgn cau tra ldi dflng cae cdu sau: Cho v(l;l) va A(0; 2) Anh cua A qua phep tinh tidn theo vecto v cd Cdu toa la: (a)(l;l); (b)(l;2); (c)(l;3); (d)(0;2) Mdt ldp hge ed 20 ban nam va 15 ban nft Sd each ldy ban nam va ban nft di thi ddu thd thao la: (a)c^o; (b)cf5; (e)Cf5+C^o; (d)C35221 Phdn Tu ludn (6 diem) Cdu Giai cae phucmg trinh sau day a) eos2x + cot2x = 0; b) sin2x +cos3x = Cdu Gieo hai eon sue sac can dd'i Tfnh xac sudt dd tdng hai mat cua hai sue sac la mg sd le Cdu Cho tfl dien ABCD Chumg minh rang dudng nd'i trung didm Cac canh ddi dien ddng quy D£I Phdn Trde nghiem khdch quan (mdi cdu diem) Cdul (d) (a) (b) S S (c) D (a) (b) (c) (d) D D S Cdu Cdu (a) D D Cdu (d) Phdn Tu lugn (6 diem) Cdul a) Phuong trinh trd cos2x = - l kTt sin2x' sm2x + = 0"»