Đề thi chọn đội tuyển dự thi học sinh giỏi cấp tỉnh môn Toán Lớp 12 - Năm học 2019-2020 - Trường THPT Chuyên Lê Quý Đôn (Có đáp án)

so cD&Dr euANc rnl rci rnr cHeN EQr ruYEN DU rHr HSG LdP TRUcTNG TIrPT cuuv0n t2 cLP rixn NAnn HQC: 2019 - 2o2o Ln eu.f BON Mdn thi: To6n (DA thi gim cd 01 trang) Thdi gian ldm bdi: 150 philt Cflu 1(6 tli6m) a) Gi6i phucrng trinh (x -2)' + J.. 6 = 67 +Ji 1 -n b) Tir c6c chfi s5 0, 3, 4, 5, 6,7, 8, glpp dugc bao nhi6u s5 chin, c6 ba cht s5 kh6c nhau Chn 2(a ili6m). Trong mflt phhng tqa d0 Oxy, cho hinh vu6ng ABCD v]r cic di6m M, N thoa mdn: fu+zMe --d ; zfii+ wD =6 a) Chimg minh tam grttc BMN vu6ng chn. b) Tim tqa dQ di€m A,UietNp;21, dvdngthingBM cd phuong trinh x-2y-3:0 vit di6m A c6 hodnh d0 nh6 hon2. Cflu 3 (a tli6m). a) cho c6c s5 thqc a,b,c th6a man ili6u kign a+b+c>1. chimg minh ring A4 +D14 rC o-at+b'+c'1 Ja b) Tim GTLN, GTNN ctra him s(i: /(x) = x(l0 *Jtz-71 Cflu 4 (a tli6m). Cho hinh ch6p S.ABC, c6 SA : SB : SC vd d6y ld tam gi6cvu6ng cdn v6i c4nh huyAn AB : oJi.tvtqtb6n (SBC) hqp v6i mflt d6y mQt g6c (p saocho cosp =+. Tinh theo a th6 ' J13 tich kh6i ch6p S.ABC vi khoing cdch gifia hai duong thFngAB vd SC. cffu 5 (2rli6m). Tim t6t cil cilchim si5 /:(0,+m) + (0;+oo) th6a mftn ding thirc .f(x+ y)+ f(xy) : x * ! * xY, Yx,Y e (0;+m) Hq vi t6n SBD urldNc nAN cuAilr uoN roAx DE THI CH D ruvnx gsc THr Hsc cAp rixu Di0m CAU D an I a) Gini phuong trinh (x-2)' +J*+e =67 +Jir-, 1t), x e [-o;t t] (6d) (r) o (J, + 6 -4) + (r - Ji r-,)+ *' - +* -60 = 0 1 11 e (x-10) _: + x+6 -0 Jx+6 +4 t+.,/t 1-x 11 1 Qx:10 ( Vi _: +x+6>0,xe[-6;11] ) Jx+6+4 l+Jl1-x b) Gqi ot" As6 cAn t\m: a,b,c d6i mQt kh6c nhau, a+0 ' c lir sti chin. o, 3, 4, 5, 6,7 . {a ,b , c\ c { ,8,9\ . c=0,m6ic6chchon cs6c6 7 cdchchon a (kh6c c:0),m6ic6chchqn 1 c,a sdc6 6 c6ch chon b (kh6c c,a), n€nc6 7.6:42 s6loai ndy' . c * 0, c6 3 c6ch chon c chian, mdi crich chon c sE c6 6 c6ch chgn a 1 (a * 0,a+ c), m6i c6ch chon c,a c66 c6ch chgn b (kh6cc,a ) ndn c6: 3.6.6=108 sd loai ndy. V{y, tong cong c6: 42+lO8:150 s6 thoa m6n dO bei' 1 7 A N F (4d) B E C a) E[t canh hinh vu6ng bing 3m. Qua M k6 dunng vu6ng g6c v6i BC cEt BC, AD frnp:Tr,Ttr-- mn tuqt t4i E, F. Khi 416, tt de bdi ta c6: ] :: ,^: ^' =ABEM:AMFN' 1 [BE:MF:2m 1 +6ME+ vay ABMN wdng cdn t4i M' = fr: fr,us:vnt fr =9oo " D @M) :x -2y -3 = O+(MN) :2x + y - 6 = 0 . 1 rqa d0 di6m M rd nghi€m cua h6 ,(g; o) . {;;1; _', =or= IvF2 +FN2 :MN2 e4m2 + m' = 5 e m- 1 > AN : l,AM :2J, (o-2)'*(b-2)' =r c6 h€ Gqi A (a;b) , v6i a 12 , ta 1 (o-3)'+b2 =8 Giai hQ, voi a <2 , ta duqc A (1;2 ) 3 a) Vx,-y € R, ta c6 xo + yo ) x3 y+rlr' ( * (4d) ) 0,5 * o (, y)' (r' + xy + y')> o, dirng Vx, y e tR. * Ap dgng ( ), ta duoc o,*( !)'=1r,*( !\'o \3, 3 (3.r I o' *( !l' la,*(1)'u \3, = 3 (31 .' *[1)' ,1.,'*(1)'. \3, 3 \3/ /r\3 r r,r (r) (o +b+.)>l ; I . cQng ve theo vC ta co: = \J/ 13 3 J 14 4_ A-.3 +D +c'+C a' rb-+c- ) ( dpcm ). Ddubangxay^r ra o:6- 0,5 33 e s=! b) TXD: D = [-.E,Jir], r{-,):-/(x),vx e D /(r)>otr€n [0,.D], /(,) < o tron [-Jo,o] 0,5 max J (x) = (,) (,) : (*) ,{,1fr,f = X.T / - ,{,?fr,f Theo BDT C-B, ta c6 +4- x2 ,r (r) < rJ0.r f0.(r2-r') = *.Ji.Jzz- *' . ^1n.x: :r rJil : 1 =,r(r)<ttJiT, dau " " xay ra <+, = JtT YQy mal f (,)=1lJ[, khi -r="'fi. r ;.f (*)=-1lJ1, khi x=-fi 0,5 4 (4d) ^s I A B K C Isa t(ABCI I H ld tam clucrng tritn (ABC). }dd, LABC vu6ng can tai c H ld trung lsr: sB:.rc = = di€m AB. Goi K ld trung iti6m cira BC. suy ra HK ld dudng trung binh cua a,4BC. lruxtttc n ri 1"" "'- uo oo :#:;tr" : -fuxtac - BC )' (sHK)=> SKH =Q' I \rr ea ali)=t* L BC __1 Cosrp :i + tanp : 2Ji + SH = HK.tane = aJI. ./l 3 1 I .6 so..,a. = !,qc.ac a')Vr.uu. s^,{BC'sF1 a 1 2 J 6 CH LAB Gt= > AH r (SCH). SH LAB 1 Trong LSCH kdtludng cao HI-d(SC,AB)=nt 11 ,t I I I .tn'' l 1 I a CO = -----'---:- T ---------= = T d( AB:SCl= Hl: a. HI, SH' CH" -;3A" o3a 7 a - r 5 f(x+y)+"f(xl,)- x+y+ry,Yx,y >0 (1) (2d) . x: !=)a f (4)=4. LAn luot thay (x;y) . {{t;t);{z;1);(3;1)} vdo (1), ta c6: I tzl+ rD:: =: tlf / [/1:) i/t:lo f Q)=5-\l \2)=2- 0.5 [l r+i * /(3):; [7'trl : r : > O) vio ( l) ta thu dugc: Th6 x !.! =l1rt' ' I oo,*1=z''l. 0.5 1(r+-1+J(l):t+-t'',=Jtt+!)=t+1 " t. t =J'(x)=x.vx>z(1 / ) Ti6p tuc th€, y = 2. Tt (1) ta suy ra lf lr*2)+ f (2x): x+2+2x 0.5 I -f12x):2x.Yx >0 haY./(x)=x.Vx>0 [,ftr* 2)= **2.Vx> 0 = Thu lai,f(*) :x,Vx > 0 th6a mdn di6u kiQn bdi to6n.V{y :x,Vx e (0;+"o). "f(*) 0.5