BỘ đề THI HSG VÒNG TỈNH MÔN TOÁN lớp 12 năm 2013 2014 SGĐT tây NINH

SO GIAO DUC VA DAO TAO TAY NII\H Ky rrrr cHoN Hec srNH Gror Lop 12 THpr voNG riNn NAvl Hec 2ot3 - zot4 Ngay thi: 25 thing nlnr Z0l3 thi: ToAx - Btroi ttri thri,ntrdt Tho'i gian: 180 phut (khong kA thdi gian giao Mdn DE CHfI\H THTIC thi gont co t trang, thi ,sinh khong phai chep @A dA) di vao giay thi) Bii (4 dient) Cho ba s6 ducvnga,b, c tho6 mdn cli6u kiqn Chirngminhr6ng: a+b+c * b * t ,3 -L l+bc l+ca l+ab = Ri i (4 client) Cho harn s6 f(x) xac clinh vcyi rnQi gia tri x > rra thoi m6n + y) = f(x).f(y), Vx > vy> o Irfx Lrf28) - Q Chu'ng mi nh r6ng: f(x) = 0, Vx > Ilii 3, (4 diem) cho lrinh ch0'nh$t ABCD Tr0n c6c cluong thang BC va cD, l6y lAn lugt c6c di6m di = e00 Gqi H rd hinh chi6u vu6ng g6c cria A tr6n MN rirn qu! fl:il*"x,,il;i?r:n" ffir Bni g aiam) cho cludng thing d cri clua trtrc tam H cria ta,r gi6o ABC Gei ;, dz, d3 lan lucyt la oac dudng ttring dtii xung vdi d qua BC, CA, AB Chring rninh rang ba dtrong th[ng dr, d2, d-, d6ng quy Bei s G aiiim) tno't minh rf,ng I I s5 thuc khirc thu6c croan 10001 o6 the chon cluoc hai [l; , so x vdy cho: < x _y 28 thi f(*o)=f(xo -28+28):f(xo-28).f(28)=0 t(28) = f(l +14): f(l4).f(14) = =+ f(l4) = Lim tuong tU nhu trOn thi f(x) LAp lupn tucrng thi co: Vay f(x) f(x) - 0, Vx > 14 0, vx >7; f(x) = 0, vx a! - 0, Vx > trang I rlei g diem) Cho hinh chfr nhQt ABCD TrAn cdc itudng thhng BC vd CD, ldy tdn lw,qt cdc ifiAm di itQng M, N c/ro = 900 Ggi H ld hinh chi6u vuAng gdc ctia A tr€n MN Tiim qu! tich ctic iti6m H ffi M(a; ma) B(a; 0) DUng hQ trpc toa d0 Oxy nhu hinh ve Gi6 sri B(a;0), D(0; b), y = mx thi M(a; ma) l)Ni5u & ) v6i A = O(0;0) 0, b > vd phuong trinh duong th[ng AM ld m*0: Phuong trinh cludrng thdng AN Phucmg trinh dudrng thing h y= -l* m Suy N(-mb; b) MN lA (b - ma)x + (a + mb)y - ab(l + m2) : g (1) Phuong trinh dulne th6ng OH le (a + mb)x GillhA;il""s;'i"h TU d6 thu dugc * * aoI (1, = ;irfiil d;;;; l - (b - ma)y = (2) Phusng trinh ndy phuong trinh cfia dudng thing BD 2) NOu rD = : Khi d6 M = B, N Vfly quy tich cdc - D Suy H eBD di6m H la dudrng thang BD Gtti chfi: Ndu thl sinh s* dung ki6n ththc itudng thiing Simson gidc CMN vd di€m A dd gidi thi: - Phdn thudn: 2il - PhAn ddo: 2d diSi vdi tam trang Bni G dieln) Cho itudng *dng d iti,qua trgc tdm H cfia tam gidc ABC Ggi d1 , dz, dj lhn lwgt ld gdc itudng thdng ddi x*ng: vt6,i d qua BC, CA; AB Ch*ng minh rdng ba dwd'ng thdng & , dz, dj cl6ng quy Ggi A', B', I' lAn luqt ld c6c thi phuong trinh 4xn + (x+1)2 =yz adml Cho tam gi6c ABC c6 ba g6c A, B, C deu nhgn vd AB > AC Ggi D, E, F hn luqt ld chdn c6c ducrng cao cta tam gifucABC vE ttr A, B, C xuiSng c4nh di5i diQn.Ducrng thlng EF cfu BC t4i P, dudng thang qua D vd song song vdi ef c6t c6c dudng tfrang AC va Ae tuong img tpi Q R Gqi M ld trung ei6m cpnh BC Chr?ng minh ring b6n Oi6* P, Q, R, M ctng thuQc mQt dudng trdn I H6t a {/ri, SO GIAO DUC VA DAO TAO TAY NINH xi Tm CHQN HQC SINH cror LOP 12 THPT voNG riNn NAvt Hec 2or4 -zors TTUONG oAN cnAu rm vrON roAN - su6r rnr rrrtl NnAr trang I Bili = Chrfrng minh rlng f(x + y) = f(x) + f(y), Vx f (0) Cho v Vdi f(; e R, Vy e IR ' f (Zxy), Yx e v x*0,dflt x-u; y= 2u e lR'' VY € R = f(2x) Vai, f(x +2xY)'Vx thi f(u + v) R,Vy e R - f(u) + f(v), Vu * 0, Vv e R (1) V6i x =0 thi f(x+y) =f(y)=f(0)+f(y) =f(x)+f(v), Vv e]R' (2) (1) vd (2) suy ra: f(x + y) - f(x) + f(y), Vx e R, Vy e IR i; Bili Q diam) Cho hai tlulng trdn (O1), (Oz) cit t4i hai Oi6m A, B vi PlP2li mQt fi6p tuy6n chung cira hai dudng trdn d6 (P1 thuQc (Or), Pz thuQc (Or).Ggi Qr vi Qz fa" foqt n trintr chi6u vu6ng g6c cria Pr, Pz l6n tlulng tneng OrOz.Dudng tning AQr cit (Or) t+i aiOr" tfr1i hai M1, Dulngtneng AQz cit (O2) tli aiam thrri hai M2 Chrfrng minh ring ba di6m Mr, Mz, B thing hing (hl) (h2) trang r 1., ( I trang trang so cIAo DUC YADAo rAo rAvNIi\H xV rHr cHeN Hgc sINH cr6r Lop NAna 12 rHpr voNG riNn Hec zor4 -zots i\giy thi: 25 thfng nlm 20L4 M6; thi: ToAll - Buoi thi thfr hai Thli gian: 180 ph fit (kh6rg kA thdi gian giao ai) DE CHINHTHTTC @A gom cd 0I trang, thi sinh kh1ng phai chdp di vdo gidy tht) Bei L (5 diem) Giei h0 phuong trinh [*'(3v L*V Bili2 + $'+ s 5):64 3y + 3) : l2+5 lx (5 diAm) (1 Cho day sO (*,) th6a mdn: J"'=5 Lr'+1) ' *n = r2Xn*r + ( 4n+2)XnXn+1 Vn > 1, n e N Tim limxn Bei (5 diem) Chrlng minh rang vdi n nguy6n duong vd n > thi phucrng trinh 4xn + kh6ng c6 nghiQm nguy6n ducrng (x; y) Bni (x+l)2 =y2 aidml Cho tam girlc ABC c6 ba g6c A, B, C ddu nhgn vd AB > AC Ggi D, B, F lan luqt ld chdn c6c dudmg cao cta tam gifucABC ve tt A, B, C xuiSng c4nh e16i diQn.Duong thing EF cat BC tai P, dudng tfring qua D vi song song v6i EF c6t c6c dudng thing AC vd AB tuong img t4i Q vi R Gqi M ld trung eiem cpnh BC Chrlng minh ring b6n AiCm P, Q, R, M ctng thu$c mQt dudng trdn , oo HCt SO GIAo DUC VA DAo TAo TAv NINH KY THI CHQN HQC jsrNu cror Lop 12 THpr voNc NAM HQC 2014 -201s HrroNG nAN csAvr riNu rur vroN roAN - BUor rur rntruar CACH GIAI Bni Q diem) Giei h phucrng trinh I.'(ry L*y Vd'i x = 0, h.6 + ss) = 64 (y'+ 3y + 3) = lZ + 5lx ph"""g Dod hQ tuong duong uoir f_ 64 llv+5s-:' *3 I [r'+3 DAt t- A ),thi x co hQ: y2 +3y - 9+51 x lZv+ 55 = t3 0,5 tr'+3y2+3y-3t+51 C0ngth.o;cffiipt,"ffic.iir,ethi.;6' +6y+ 55 - y3 +3y2 13 1,0 +3t+51 €> (y+1)3 +3(y +1)+ 51 - t3 +3t + 51 (*) ii;=ffi;'e, ffi;tio Hdm'6 (Ho[c tii5n a6i (*) e (y+l-rl[fr+1)2 +(y+l)t y+1=t) Vdi y=t-1 Vfly hQ d5 *r, 0r5 "ry." t y+l =T*' ] = o di5 suy 1r0 thi t3 -3t-52=0t-4 2,ta"6 *(m-l)' =[1+(m-1;1('-rt2 = (mod(m-t)r) 0r5 0r5 .- - MAt khSc T6m lpi m/l (mod(m -t)') Bdi to6n kh6ng tho6 man ffr:1 ld c6c gi|tri cAn tim 0r5 0r5 trang Bni ft diem) Cho tsm gidc it1u ABC cgnh o M vd N ld hai iti6m di ctQng ldn lwgt trhn cgnh AB vd AC hui N*4N "6 MB NC = a) Ch*ng minh rdng MBCN ld tir gtdc ngogi ti6p itwgc mQt itwdng trbn b) Ttm gid tr! td,n nhdt crta diQn tich tam gidc AMN theo a ,A XA, B A\ C a) Dat AM = x, AN = y Ggi E, F l6fl luqt ld trung rti6m cria AB, AC ri, suy r" {Y:Y **g MB NC= LANI < NC vay M, N ran ruqt thuQc c6c do4n AE vd AF Ttr gi6thi6t 0,5 ta.o, u YB * = u, IC = I =+ =L*j= NC MB NC = MB :i1YP:XE3P_I9:_ I > f MN2 = AM2 + A}-.{2 _ AI\4.ANI = (a - MB)2 + (a - NC)' - (a - MB)(a - NC) 0,5 = a2 +MB2 + NC2 - 2a(MB + NC) + a(MB + NC) - MB.NC = a2 +MB' +NC2 -2a(MB + NC) + 3MB.NC - MB.NC - a' +MB2 +NC2 -Za(MB + NC) + 2MB.NC - (MB+NC -a)' 0,5 + MN = MB +NC - a )MN +BC = MB +NC + MBCN ld trl gidc ngoqi.ti6p dugc Dudng trdn nQi ti€p tti gi6c MBCN ctng ld [...]... TAv NINH KY THI CHQN HQC jsrNu cror Lop 12 THpr voNc NAM HQC 2014 -201s HrroNG nAN csAvr riNu rur vroN roAN - BUor rur rntruar CACH GIAI Bni 1 Q diem) Giei h 0 phucrng trinh I.'(ry L*y Vd'i x = 0, h.6 + ss) = 64 (y'+ 3y + 3) = lZ + 5lx ph"""g Dod 6 hQ tuong duong uoir f_ 64 llv+5s-:' *3 I [r'+3 DAt t- A ) ,thi x co hQ: y2 +3y - 9+51 x lZv+ 55 = t3 0,5 tr'+3y2+3y-3t+51 C0ngth.o;cffiipt,"ffic.iir,ethi.;6'... thrri hai M2 Chrfrng minh ring ba di6m Mr, Mz, B thing hing (hl) (h2) trang 2 r 1., ( I trang 3 trang 4 so cIAo DUC YADAo rAo rAvNIi\H xV rHr cHeN Hgc sINH cr6r Lop NAna 12 rHpr voNG riNn Hec zor4 -zots i\giy thi: 25 thfng 9 nlm 20L4 M6; thi: ToAll - Buoi thi thfr hai Thli gian: 180 ph fit (kh6rg kA thdi gian giao ai) DE CHINHTHTTC @A gom cd 0I trang, thi sinh kh1ng phai chdp di vdo gidy tht) Bei L... TAv NINH KY THI CHQN HQC jsrNu cror Lop 12 THpr voNc NAM HQC 2014 -201s HrroNG nAN csAvr riNu rur vroN roAN - BUor rur rntruar CACH GIAI Bni 1 Q diem) Giei h 0 phucrng trinh I.'(ry L*y Vd'i x = 0, h.6 + ss) = 64 (y'+ 3y + 3) = lZ + 5lx ph"""g Dod 6 hQ tuong duong uoir f_ 64 llv+5s-:' *3 I [r'+3 DAt t- A ) ,thi x co hQ: y2 +3y - 9+51 x lZv+ 55 = t3 0,5 tr'+3y2+3y-3t+51 C0ngth.o;cffiipt,"ffic.iir,ethi.;6'... xi, rnr cHeN Hec srNH cror Lop tz rHpr voNG riNn NAvr Hgc zot4 -zots i\gny thi: 25 thfng 9 nIm 2AL4 Mdn thi: TOAX -Budi thi thfr hai Thbi gian: 180 phrit (khdng kA thdi gian giao dA) DE CHINHTHTIC @A SOm cd 0l trang, thi sinh kh\ng phai chdp di vdo gidy tht) Bni 1 (5 diAm) Giai h0 phuo-ng trinh 5)=64 I"'(3y+s L"V (y'+ 3y + 3) = 12 +5 lx Bei 2 (5 dihm) (l Cho day sO (",) th6a mdn: J*'=5 Lt" +r) ' *n=...trang 4 so cIAo DUC YADAo rAo rAvNIi\H xV rHr cHeN Hgc sINH cr6r Lop NAna 12 rHpr voNG riNn Hec zor4 -zots i\giy thi: 25 thfng 9 nlm 20L4 M6; thi: ToAll - Buoi thi thfr hai Thli gian: 180 ph fit (kh6rg kA thdi gian giao ai) DE CHINHTHTTC @A gom cd 0I trang, thi sinh kh1ng phai chdp di vdo gidy tht) Bei L (5 diem) Giei h0 phuong trinh [*'(3v L*V Bili2 + $'+ s 5):64... DB=a+d; pC-p-a; Thay vdo (1) thi dugc: (p + a)(p - a) = (p = (a+dXa-d):6;t; - CD =a-d; Dp=p_d d)p Suy ra a2 = dp +DB.DC _DP.DM (3) Tt (2) vA (3) suy ra: DQ.DR VOy P, Q, R, - DP"DM M cirng thuQc mOt dudng trdn , HGt eooooo trang 3 so GrAo DUC vA nAo rAo rAv NII.IH xY rnr cHeN rrec sINH ctor Lop 12 THpr voNG NAna Hoc zot4 - 2ors rixn NSi,y thit 24 th6ng 9 nf,m 2014 MOn: TOAN ru6i thi th* nh6t Thli gian:... pC-p-a; Thay vdo (1) thi dugc: (p + a)(p - a) = (p = (a+dXa-d):6;t; - CD =a-d; Dp=p_d d)p Suy ra a2 = dp +DB.DC _DP.DM (3) Tt (2) vA (3) suy ra: DQ.DR VOy P, Q, R, - DP"DM M cirng thuQc mOt dudng trdn , HGt eooooo trang 3 1, sO crAo DUC vA DAo TAo rAv Nrr\H Ky rHr cHeN Hec srNH Gror Lo'p 12 THpr voNG NAtu Hoc zot3 -zot4 rixu I\giy thi: 26 thftng 9 nf,m 2AI3 MOn thi: TOAN BtrOi ttri thir hai TIrd'i gian:... cci thi m=1(modn) g aie4 Cho tam gi6c dOu ABC canh a M AC sao ho AM * AN = l MB vi N ld hai cli6m di dQng lAn lucrt tren hai canh AB vd NC ring MBCN la tf gi6c ngoai ti6p dugc m6t dudng trdn b) Tim gi6tri lcyn nh6t cria diQn tich tam gi6c AMN theo a a) chring minh IIeI He ve ten thi s inir: So bao danh: s0crAo DUC vAEAo rAo rAy NrNH rY rnr cHeN Hec jsrNH cr6r lop 12 Trrpr voNG riNn NAM HQC 2013 -2014. .. phut (khong ke tho,i gian giao dA) or csixn rsrlc @a thi gom co 0l trang, thi sinh khong phai clt6p di vdo giay thi) Bai 1 (4 diem) Giai phrrong trinh: (Z*- 3) +2x+2 = x2 -2x+ 4 Bi i 2 (4 diLrn) Giai Bni 3 hQ phucmg trinh: I*' - 6x + y + 1 1 lv'-v-x-3 g arcm1 * t I vol rnoi n nguyOn clucrng =1[r, z\ un) Chimg minh ring v6i moi sO nguyOn ducrng n Z 2 thi ,n , ,6 Tim limu,, Cho ddy s6 (r,.,) xac dinh boi:... DUC VA DAO TAO TAY NINH xi Tm CHQN HQC SINH cror LOP 12 THPT voNG riNn NAvt Hec 2or4 -zors TTUONG oAN cnAu rm vrON roAN - su6r rnr rrrtl NnAr trang I Bili = 2 Chrfrng minh rlng f(x + y) = f(x) + f(y), Vx f (0) Cho v Vdi f(; e R, Vy e IR ' f (Zxy), Yx e v x*0,dflt x-u; y= 2u e lR'' VY € R 0 = f(2x) Vai, f(x +2xY)'Vx thi f(u + v) R,Vy e R - f(u) + f(v), Vu * 0, Vv e R (1) V6i x =0 thi f(x+y) =f(y)=f(0)+f(y) ... )(=xi,Y=xj thi 0