PGS TS U~ Van Hi~n (Chu bien) Le Hai Trung Nguyen Khac Nghia Tong Van Phuc Tran NgQc Hoan Lo van Thai Nguyen Do Chien 5r qu ~et chinh phlJ c NHA XUAT BAN DJ;1 HQC QUOC GIA HA NQI PGS TS U~ Van Hi~n (.
PGS.TS U~ Van Hi~n (Chu bien) Le Hai Trung - Nguyen Khac Nghia - Tong Van Phuc - Tran NgQc Hoan - Lo van Thai - Nguyen Do Chien 5r qu ~et chinh phlJ c NHA XUAT BAN DJ;\1 HQC QUOC GIA HA NQI T938 ~ ~ L IN I ~ •···························0···························• Toan hQC Ia m()t nhO'ng man khoa hQC tl)' nhien nam ChU'O'ng trlnh giao dl)C thong (y Vi~t Nam tCP b~c Tieu hQc den THPT Ve m~t kien thCPc, Toan hQc cung cap cho cac em cong Cl) de tinh toan va nghien cCPu cac man hoc khac nhU' V~t li, H6a hoc, Ve m~t kT nang, hoc toan giup cac em rem luy~n kha nang tU' duy, suy lu~n logic, suy nghT m~ch l~c dong thai g6p phan hlnh va phat trien kT nang giai quyet van de cu()c s6ng V&i d~c thu man hQc doi hoi c6 nen tang kien thCPc va kT nang th~t vO'ng chac, chung toi luon tran tra tim nhO'ng phU'ang phap t6i U'u nhat giup cac em tiep c~n va hoc t~p v&i taan hoc 8U'Q'C due rut tCP kinh nghi~m giang d~y va nghien CLPU cac an pham ca va ngaai nU'&c, cu6n sach Bi quyet chinh ph~:~c di~m cao Toan - T~p dU'Q'c chung toi thiet ke nham ho trq t6i da cac thay co va cac em hoc sinh trang qua trlnh hoc t~p, on luy~n, kiem tra va danh gia kien thCPc H~ th6ng kien thCPc bam sat chU'ang trlnh hoc tren l&p ·········~ Mvc tieu ro rang thea tCPng bai ·~ Thea sat ho trq cac em hoc t~p tren l&p, on t~p, kiem tra danh gia kip thai -~ Giup cac em c6 cai nhln tong quat, d~t mvc tieu de d~t dU'Q'c trU'&c hoc Lf thuyet dU'Q'C trlnh bay hai c()t chi tiet, ket hQ'p SO' h~ thong h6a l~i kien thCPc ·~ -~ Trinh bay kien thCPc chi tiet, li giai day du giup cac em hieu sau sau d6 h~ thong h6a l~i de cac em thay ro dU'Q'c sl)' logic cua kien thCPc, nhO'ng kien thCPc tam can nh& Bai t~p dU'Q'C phan d~ng chi tiet cac bai taan, kieu cau hoi thU'ang g~p, c6 phU'ang phap giai chi tiet thea tCPng bU'&c va cac vf dl) minh hQa ·~ ··~ Giup cac em nh~n biet dU'Q'c tung d~ng bai, kieu hoi va each tU' de giai d~ng bai, kieu hoi H~ thong cac m~o giai, kinh nghi~m lam bai, lei sai can tranh, kien thCPc bo sung va cac chu y di kern v&i cac vf dv ··~ ·~ Truyen thv cho cac em nhO'ng kinh nghi~m cua cac tac gia nhieu nam d~y hoc, giup cac em nam vO'ng kien thCPc m&i, on t~p l~i kien thCPc cO can H~ th6ng de kiem tra 15 phut, 45 phut, hQC kl bam sat theo qua trlnh hQC ······~ ·~ Ho trq cac em on t~p dung thai diem de tl)' tin lam cac bai kiem tra tren l&p Cau true Cl) the cua cuon sach nhU' sau: Phan 1: N9i dung bai hQc Trang d6 phan dU'Q'C chia cac n¢i dung l&n: D~i ChU'ang Can b~c hai, can b~c ba ChU'ang Ham so b~c nhat Hinh hQC ChU'ang H~ thCPc IU'Q'ng trang tam giac vu6ng ChU'ang DU'ang tron so Phan 2: ae kiem tra Ph an 3: Dap an V&i nhCPng U'U diem cua cuon sach chung t6i mang rang Bi quyet chinh phl:JC diem cao Toan - T~p se dem den cha cac em hQc sinh nguon cam hCPng hQc t~p, dong hanh cung cac em trang hanh trinh chinh phvc tri thCPc va Ia nguon tai li~u tham khaa hCPu ich cha cac b;~m dong nghi~p trang qua trinh giang d9y Trang qua trinh bien sa()n, chung t6i kh6ng the tranh kh6i nhCPng thieu s6t kh6ng mang muon, rat mang se nh~n dU'Q'C nhCrng d6ng g6p quy bau b) b()n dQC gan xa, cac thay CO cOng nhU' cac em hQC sinh than men de chung t6i haan thi~n cuon sach han trang nhCPng lan tai ban tiep thea Xin chan cam O'n! I •···························0···························• Sau thai gian d~ii no ll)'C trien khai, trU'&C Sl)' cha d6n cua cac em hQC sinh va bc;ln dQC mQi mien To quae, cuon sach Bi quyet chinh ph~Jc diElm cao Toan - T~p da chfnh th(fc phat hanh Moi cuon sach dai Ia qua cua ca mot t~p the l&n mc;1nh; d6 khong the thieu cong lao cua cac thay co giao - nhll'ng ngU'ai trl)'C tiep chap but viet nen cuon sach V&i tat ca tam huyet, tinh hoa dU'Q'C tfch lOy, SU'U tam suot qua trlnh giang dc;1y, cac thay co da tra nhCI'ng "ngU'ai truyen ll:ra" tham l~ng cho cac em hQc sinh qua tCPng trang sach Cong ty Co phan CCGroup Toan Cau xin dU'Q'C gl:ri lai cam an chan va sau sac den cac thay co: Hi~n (Chu Giang vien Khoa Toan - Tin PGS.TS LeVan bien) TrU'&ng D9i hQc SU' ph9m Ha N(Ji Le Hili Trung i6 Tong Van Phuc i6 Lovan Thai Tran NgQc Hoan ·Nguyen Do Chh~n Chung toi hi vQng se luon nh~n dU'Q'c nhCI'ng y kien dong gop tam huyet, t~n tlnh cua cac thay co de chung toi c6 the hoan thi~n han nCI'a nOH =2:::::>0H=J2 4 V~y khoang each tll' goc tQa dQ t&i dU'ang thang y = -x + Ia J2 o,5 diem c) GQi diem co dinh ma dU'ang thang (d): y = (m- 2)x +2m di qua v&i mQi m Ia D(x0 ;y0 ) suy Yo =(m-2)x0 +2m, Vm Yo = mx0 - 2x0 + 2m, Vm m ( X0 + 2) - 2x0 - Yo = 0, Vm o,5 diem X0 + = { -2Xo - Yo = ¢::> X0 = { Yo -2 =4 V~y dU'ang thang (d): Yo= (m- 2)x0 +2m luon di qua D(-2;4) v&i mQi m o,5 diem EJ6p 6n X -4 -6 GQi (d) Ia dU'erng thang di qua g6c toa d9 t&i dU'erng thang (d) o va vuong g6c v&i 00, suy khoang each tCP Ia oo GQi ( d') Ia dU'erng thang bat ki qua o khac (d), dl)'ng OH vuong g6c ( d'), suy khoang each tCP g6c toa d() t&i dU'erng thang ( d') Ia OH Xet tam giac OHO vuong t~i H ta c6 OH ~ V~y (d) oo Ia dU'erng thang can tim Vi dU'erng thang oo di qua g6c toad() nen phU'ang trinh dU'erng thang oo c6 ax (a =t= 0) d~ng y = M~t khac, dU'erng thang oo di qua diem o (-2; 4) suy = a (-2) ~ a = -2 oo Ia y = -2x (d) Ia y =a x + b Do d6 phU'ang trinh dU'erng thang Goi phU'ang trinh dU'erng thang Vi dU'erng thang th~ng (d) (d) c6 d;;tng y vuong g6c =ix +b Ma phU'ang trinh dU'erng thang Suy 2 oo (d) nen a -2) = -1 ~ a I ( 0,25 diem I = _:!_ , suy dU'erng Ia y = ( m- 2) x +2m m- =- ~ m =- V{Jy m = ~ I %thi khoang each tlr g6c lQa dQ t&i dU'cYng th~ng (d) 18 l&n nhM J DE SO PHAN TRAC NGHII;M (3,0 f>IEM) 1-C 2-B 3-C 4-D 5-C 6-D I o,25 diem PHAN TV' LU~N (7,0 £>1EM) cAu £>I EM NQIDUNG A 1,0 di~m ~~ ~ ~c Cau H X 1,0 di~m Xet tam giac ABC vuong t9i A, dU'ang cao AH BH.HC = AH => 2 25 x = HC =- = -· 4 AH + HC = A C => A C = 25 + 625 => y = A C = s.J41 0,5 di~m Ta c6: 1,5 phUt = Cau 4~ gi cao Ia 5,13km A C Cau ~ H M 0,5 di~m B a) Ta c6: C=goo- B=goo- 30° = 60° Xet MBC vuong t9i A, c6 AH BC Ta c6 AC = AB tanB = tan30° = 2.J3(cm) BC=~= cosB 1,0 di~m =4.J3(cm) cos30° b) Ve dU'ang cao AH va trung tuyen AM cua ~ABC Xet tam giac AHB, ta c6: 0,5 di~m sinB = AH => AH = AB.sinB = 6._:!_ = 3(cm) AB EJ6p 6n HB HB = AB.eosB = 6.-= J3 3J3 eosB =-~ (em) AB J3 o,5 diem BC MB=-=2 3(em) HM =HB-MB=3J3 -2J3 =J3 (em) Di~n tfeh MHM Ia: SAHM ~ = o,5 diem AH.HM = 3.J3, , 6( 2) , , , em 2 ~ DE SO cAu £>11:M NQIDUNG A B & a) Xet tam giac ABC vuong t~i H C A, dU'ang eao AH AB = BH.BC = 9.(9 + 16) = 225 => AB = 15cm Cau 2,o diem AH = HB.HC = 9.16 = 144 ~ AH = 12em b) Theo h~ th(fc IU'Q'ng MHB vuong t~i H , ta c6 1 1 1 HD = AH2 + BH2 => AH2 = HD - BH • ( 1) Theo h~ HE2 th(fc IU'Q'ng MHC vuong t~i H , ta c6 1 1 = AH2 + CH => AH2 = HE2 - CH • ( 2) T U'' (1) va' (2) suy HD a) Theo h~ - 1 BH2 = HE2 h~ • h) CH (d"A 1eu ph, a1 ch'U'ng mm th(fc IU'Q'ng tam giae AHB vuong t~i H, ta c6 AH2 = AM.AB Theo - 2,o diem (1) th(fc IU'Q'ng tam giac AHC vuong t~i H, ta e6 AH2 = AN.AC 3,0 diem (2) TCr (1) va (2) suy AM.AB = AN.AC Cau b) Theo cau a) ta c6 AM.AB = AN.AC => AM = AN AC AB Xet MNM va MBC e6 A Ia g6e ehung va Do d6 MNMC/?MBC (e.g.e) AM= AN (ch(fng minh tren) AC AB 1,5 diem c) Do MNMcnMBC => -/"._ SAMN SACB =(AN) AB - (4) • Ta c6 AHN = C (cung phl) v&i CHN) Trang MNH vuong N, ta c6 AN= AH.sinAHN = AH.sinC t~i H, ta c6 (5) => AN = AH sin C Trang MHB vuong - t~i 2 1,5 diem 2 2 AH = AB.sinB => AH = AB sin B => AB = , ( ) , SAMN Thay ( ) , ( ) vao , ta co - - = SACB A~ sin B • (6) 2C AH sin C = Sin Sin AH (dieu phai ch(rng minh) ~ - DE SO PHAN TRAC NGHieM {2,5 aiEM) I 1-C I 2-D I 3-B 4-B 5-A PHAN TV' LU~N (7,5 DIEM) cAu NQI DUNG aiEM a) Ta c6 A= J28 + J63- 217 = 2/7 + 317-217 = 317 0,5 diem -r , B 1 - J2 + j2 b) laOO Cau == c)Tac6 = a) Th ay Cau = + = + + 12 - j2 (3 + 12) ( - -!2) ( + -!2) ( - 12) o,5 diem 3-12 3+-!2 3-12 +3+-!2 + = =-· 9-2 9-2 7 )14-6)5 -)29-12)5 =~(3-Fs)' -~(2J5-3)' o,5 diem 13 - v!sl-12vts - 31 = 3-vis - 2vfs + 3= - 3vfs 31 ' b"1eu th U'C , B , ta d U'Q'C B = 36 - ~ ~ + = 6" x = 36 vao it , A 2x + 3-fX b) Ia co = + x.fX +1 x-.JX +1 o,5 diem fX +1 o,5 diem EJ6p 6n , P-_A8 -_ c) a co T · Xet P -1 = x + 5JX x- JX + - JX + (x-fX+1)(1X+1)' JX fX+1 o,25 diem £+5 JX -1 = JX4 > 0, \fx > X+1 X+1 o,25 diem Suy P > d)Ta c6 P=1+ JX4 +1 X Vi x > o nen M~t JX > o JX + 1> 1==> JX4 X +1 < P < khac thea ch(rng minh cau c) thi P > Suy < P < Ma P c6 gia tri Ia s6 nguyen nen P {2;3;4} E 0,25 diem Ta c6 bang sau p 4 fX+1 X - Ket lu~n Th6a man Th6a man Vf)y v&i x E a) V&i y Cau H;1;9} Th6a man o,25 diem thi P c6 gia trj 18 s& nguyen = thi x = -2 D6 thj ham s6 cat trl)c hoanh t~i diem A( -2;0) V&i X= thi y = Db thi ham s6 y= D6 thj ham s6 cat trl)c tung t~i diem 8(0;2) o,5 diem x + nhU' hinh ve yl 8(0;2) /Y=x+2 v ~ A(-2;0)/ / b) Vi ( d ) II ( d1 ): y = x + nen -1 o X {ab*21 0,25 diem = Do d6 phU'ang trinh dU'ang thang ( d ) Ia ( d2 ): y = x +b Vi dU'ang thang ( d ) cat trl)C hoanh t~i diem c6 hoanh dQ bang nen thay X= va y = vao phU'O'ng trinh y =X+ b, ta dU'Q'C = + b => b = -3 (thoa man) V~y h~ s6 can tim Ia a= 1; b = -3 o,25 diem D 0,25 di~m K a) Ta c6 ~MAB noi tiep dU'ang tron => ~MAB vuong t~i - (0) dU'ang kinh AB - M => AMB =goo hay EMF= goo Xet nll'a dU'ang tron ( o) c6 E Ia trung di~m cua MA (gia thiet) => OE l_ MA (quan h~ - vuong g6c dU'ang kinh va day) => MEO =goo Xet nll'a dU'ang tron ( o) c6 F Ia trung di~m cua MB (gia thiet) =>OF _i MB (quan h~ Xet tu giac MEOF co - vuong g6c dU'ang kinh va day) => MFO =goo EiliiF = iViEO = iViF6 =goo => Tu giac MEOF Ia hlnh chCP nh~t b) Xet MMO c6 OA =OM => ~OMA Ia tam giac can cau M~t khac oc t~i 0; Ia dU'ang trung tuyen => OC Ia dU'ang trung trl)'c cua MA => CA = CM Xet MCO va ~MCO c6 OM= OA; CM = CA; OC Ia c~nh chung Do d6 MCO = 11MCO ( c.c.c ) Suy GAO= CMO =goo=> CAl_ AB => CA Ia tiep tuyen cua nll'a (O;R) hay CA tiep xuc v&i dU'ang tron (O;R) - - Xet MEO vuong t~i E co EAO = 30° => EOA = 60° CA r;; Ta c6 tanAOC=-:::>CA=AO.tanAOC=3.tan60°=3-v3 (em) AO c) Ta c6 F Ia trung di~m cua MB va OF l_ MB => OD Ia dU'ang trung trl)'c cua MB => BD =MD - Ta c6 EOF =goo (do MEOF Ia hlnh chCP nh~t) - hay COD= goo Xet ~COD vuong t~i 0, dU'ang cao OM c6 OM = CM.MD => AC.BD = R • Chung minh tU'ang tl! cau b) ta dU'Q'c BD Ia tiep tuyen cua (O;R) Ta c6 CA lAB, DB lAB=> AC II BD => tu giac ACDB Ia hlnh thang => SACDB =-(AC+BD).AB EJ6p an Ap dvng bat dang th(fc Co-si, ta c6 ~ SACDB ( AC + BD ) AB;:::: -.2R.2R == 2 Dau II= II xay Khi d6 M V~y va chi SACDB;:::: 2R AC = BD = R Ia diem chinh giCra cung AB 2R2 Xet so hang tong quat · (2n + 1)(-Jn + Jn+1) = 1n+1 - -Jn = ln+1- -Jn 2n + J4n + 4n + Jn+1 Jn Jn+1 Jn ( 1 J = 2-Jn + 1.-Jn = -Jn- Jn+1 · Cau ~ AC + BD;:::: 2-J AC.BD = 2JR2 = 2R o,5 diem < J4n + 4n ~ DE SO PHAN TRAC NGHieM (2,5 DIEM) I 1-C I I 2-B 3-B 4-A 5-A PHAN TV' LU~N (7,5 DIEM) cAu DIEM NQIDUNG o,5 diem b) Cau 8=~12-613 +~35-12J6 = ~32 2.3 /3 + # = ~(3-.J3f + ~( /3- 2-J2f - + ~( /3f - 2.3 /3.2-J2 + (2-J2)' o,5 diem = 13-131 +1313 -2~1 = 3-13 + 313 -2~ (do > 13, 313 > 2~) =3+213 -2~ a) Dieu ki~n x::::: -2 Cau 2 -J4x+8 +5-J9x+18 =-Jx+2 +8 ~ ~2 (x+2) +5~3 (x+2) Jx+2 =8 ~ 2-J X+ + 15-JX+ - -J X+ = ~ 16-JX+ = ~ -J X+ =! 0,25 diem x + = _.:!_ x = _!_ (thoa man x ~ -2) 4 Vf}.y phU'ang trinh aa cho c6 b) Ta c6 ~ = x- { nghi~m x =-: x-2~0 X ( -3= X-2 )2 0,25 diem {x~2 X -3=X2 -4X+4 >2 {X~ { ; ; x (vO nghi~m) = =: Vf)y phU'ang trinh da cho v6 a) V&i y = thi V&i X= x = thi y = -4 o,25 diem 0,25 diem nghi~m £)o thi ham s6 cat trvc hoanh t9i diem A(2;0) £)o thi ham s6 cat trvc tung t9i diem 8(0;-4) 86"thi·h~~·~e;·~h~··h·,·~·h··~·~······················································································ 0,25 diem YH j;~2x-4 2;0) x"' -1 ~ o,25 diem -2 r -3 -4 I cau 8(0;-4) b) Ta c6 OA = lxAI = va 08 = IYal = Tam giac A08 vuong t9i Di~n tfch tam giac OAB o nen A8 = JoA + 08 = J22 + = 2Fs 18 S0 A8 = ~OA.OB = ~.2.4 = 0,25 diem {dvdt) 0,25 diem Chu vi tam giac OA8 Ia PoAa = OA + 08 + A8 = + + 2J5 = + 2Fs (dvdd) 0,25 diem V~y m = th6a man yeu cau bai toan 0,25 diem EJ6p 6n o,25 diem Tam giac ABC c6 AB + AC => MBC vuong t~i = g + 16 = 25 = BC o,25 diem A (dinh ly Py-ta-go dao ) a) Ap dvng h~ thCrc v~ c~nh va dU'erng cao tam giac vuong ABC, ta c6 AH.BC = AB.AC => AH = AB.AC = BC Cau4 V~y ' o,5 diem AH = 2,4cm b) Vi BC Ia ti~p tuy~n cua ( 0) va => ~ Ma = ~ = 4cm = ~ (cung phv v&i B;_) va B;_ + ~ = goo nen (!) nen OB _L BC, /C _L BC G; = B;_ (cung phv v&i ~ ) 0,25 diem 6; + ~ = goo Xet MOB c6 OA =OB nen MOB can t~i 0=> A=~ Xet MIG c6 /A = IC nen MIG can t~i I => ~ = f; 0,25 diem Suy A+ ~ = ~ + G; = goo => 6AJ = A+BAG + ~ = 180° => o, A, thang hang Tll' d6 suy Of= OA +/A o,25 diem V~y hai dU'erng tron ( 0) va (!) ti~p xuc ngoai t~i A ································································································································· c) Goi P va Q lan IU'at Ia trung diem cua AB va AC ' b'mh tam g1ac ·, ABC nen ~ {MP II AQ co: MP I'a dU'O'ng trung MP=AQ => APMQ Ia hinh binh hanh o,25 diem M~t khac 0,25 diem -r.1a J5AQ =goo=> APMQ Ia hinh chCP nh~t => J5MQ =goo ( *) Ta c6 OP lAB (quan h~ vuong g6c giCPa dU'erng kfnh va day cung) Ma MP _LAB nen 0, P, M thang hang ( * *) 0,25 diem Chll'ng minh tU'ang tl,J' ta cOng c6 /, Q, M thang hang TCP ( *), (* *) va ( * * *) suy d) Ta c6 AM= iac (AM ( * * *) 6Mi =goo=:> 11/MO vuong t~i M 0,25 diem Ia dU'Q>ng trung tuytm Lrng v&i c;;mh huyi'ln BC tam giac vuong ABC) =:>A thu MA Ia ban kfnh cua dU'ang tr6n dU'ang kfnh BC Xet hai tam giac OBM va OAM c6 OA = OB (cung Ia ban kfnh cua(O) ); OM chung; MB = MA (chll'ng minh tren) o,5 diem Do d6 !108M= 110AM =:> OAM = OBM =goo=:> MA _l 0/ Suy OJ Ia ti~p tuy~n cua dU'ang tron dU'ang kfnh BC Tac6x= ~4+213-13 = 13+1-13 (Fs +2)~17Fs -38-2 ~(Fs +2)" (17Fs -38) -2 0,25 diem Cau s = 17 ~( 17J5 - 38) ( V~YP=(x2 +x+1) 2019 [ J5 + 38) =-1-=-1 1- = (-1) +(-1)+1 ]2019 =1 0,25 diem EJ6p 6n NHA XUAT BAN ll~l HQC QUOC GIA HA NQI 16 Hang Chuoi - Hai Ba Trll'ng - Ha N9i Di~n thoc;1i: Bien t~p: (024) 39714896; Quan ly xuat ban: (024) 39728806; TcSng bien t~p: (024) 39715011; Fax: (024)39729436 Chju tnich nhi~m xu§t ban: Giam d6c- TcSng bien t~p: TS PH;:\M THI TRAM Bien t~p chuyen nganh: HOANG LE THU HI~N Bien t~p xuat ban: DANG TH! PHU'ONG ANH Sua bai: PHI TH! KHANH VAN Ch~ ban: NGUYEN TH! NGQC HA Trinh bay b1a: NGUYEN VAN CU'ONG 11111111111111 II 1111111111111111111111111111111111111111111 Ill !111!111111111111111 1111111111111111111111 D6i tac lien k~t: CQNG TY CO PHAN CCGROUP TOAN CAU Dja chi: S6 10 DU'O'ng Quang Ham, PhU'b'ng Quan Hoa, Qu~n C§u Giay, TP Ha N9i !IIIII II l l i l l i l l i l l i l l i l i l l l ! l i l l , I llill!llill!llil!illll!llillillillillillillillill!l!!lillilllllillillillillilliliillilillillillillillillilliliil!!l SACH LIEN KET I HU II Ma s6: 1L- 238PT2019 In 6.000 cu6n, khcS 22x27,5cm tc;1i Gong ty c6 phan in Ngoc Tram Dia chi: phong 107, E8 t~p the Thanh Xuan Bac, Qu~n Thanh Xu an, TP Ha N9i S6 xac nh~n DKXB: 4729-2019/CXBIPH/05-337/DHQGHN, 20/11/2019 Quy~t djnh xuat ban s6: 698 LK-TN/Qf)- NXB DHQGHN, 26/11/2019 In xong va n(>p IU'U chieu nam 2019 ... (1) va (2) ta c6J2 +J26 > > J3 +11 5 V~y)2 + J26 > J3 + 11 5 b) Ta c6 11 5 -1 < J16 -1 = -1 = ; J10 > J9 ~ J10 > ~M>3 >11 5 -1 V~y J10 > 11 5 -1 c) Ta c6 -f2 > )1 ~ J2 > ~ 1- -f2 < 0; 15 1 >14 9 ~15 1>7 ~15 1-7>0;... hay so sanh a) -13 +11 5 va J2 + -12 6 b) 11 5 -1 va J1Q c) 1- J2 va 15 1- HU''&ng dan giai a) Ta c6 -13 < -14 ~ -13 J25... va -14 3 c) -.J 11 va -3 HLP&ng dan giai J4 ma > => J4 > J3 hay > J3 V~y > J3 -14 9 ma 49 > 43 => -14 9 > -14 3 hay > -14 3 V~y > -14 3 c)Ta c6 3 =19 ma 9 19 -19 >-.J 11 hay -3>-.J11