Mỗi bài giảng trong tài liệu Phương pháp giải toán trọng tâm trình bày các phương pháp cơ bản, thông dụng và hiệu quả nhất để giải các loại bài tập thường xuất hiện trong các đề thi Đại học và Cao đẳng. Mời các bạn tham khảo phần 1 tài liệu.
PHAN HUY KHAI Cty TNHH MTV DVVH Khang Vigt L6IN6IDAU , , Trong tay cac ban la cu6n '"Cac bai giang luy^n thi it{ii IIQC mon Toan vao cac truing D(ii hpc vd Cao dSng" Toan bg chuang trinh on luyen thi dugc trinh bay hoan toan d i y du, c6 dgng va sue tich 20 bai giang Cac bai giang dugc due rut tiJ kinh nghiem chinh qua trinh luyen thi nhieu nam cua tac gia cuon sach Mue tieu ciia cuon sach la giup cae ban hgc sinh nam vung va su dung (hanh t h p cac phuang phap co ban va hi?u qua nhk de giai cac bai toan cac de thi tuyen sinh vao cae truong Dai hoe va Cao dang Ngi dung cua cuon sach bam sat chuang trinh ehuin, chuang trinh nang cao ve mon Toan da dugc Bg Giao due va Dao tao quy dinh cac ki thi tuyen sinh vao Dai hgc va Cao dang Nhu da biSt, moi d^ thi tuyin sinh mon Toan hien cua cac khoi A , B, D va Cao dang khoi A , B bao g6m cau Cac bai giang s6 16, 17 dung dS on luy?n cac kT nang giai cau s6 cua d^ thi Tuang t u cac bai giang so 18, 19, 20 danh cho cau so 2, bai giang so 12 cho cau s6 3, cae bai giang so 1, 2, cho cau so 4, bai giang so cho eau s6 5, cac bai giang so 3, 4, 5, 6, 13, 14, 15 cho cau s6 va eac bai giang so 8, 9, 10, 11, 17 cho cau so Trong m6i bai giang chimg toi trinh bay cac phuang phap ca ban, thong dung va hieu qua nhat de giai cae loai hinh bai tap (da dugc phan dang can than) thuang xuyen CO mat tron^ eac dh thi tuyen sinh vao Dai hgc, Cao dang nh&ng nam 2002-2011, va rk CO the se c6 mat nhOng ki thi sap tai (vi no ehua xuat hien eac ki thi vira qua nhung cae dang toan nam chuang trinh han ehe dugc Bg Giao due va Dao tao chinh thuc ban hanh quy ehe cua cac ki thi tuyen sinh) can nhSn manh rang toan bg chuang trinh on luyen mon Toan phue vu cho cae ki thi tuyen sinh dugc trinh bay day dii chi mgt cuon sach Vai phuang cham ehung toi lira chgn trinh bay mgt each dan gian va thiet thuc cac phuang phap giai toan ea ban, thong dyng lien quan true tiep den viec giai bai thi mon Toan cae ki thi tuyen sinh N h u vay on tap theo eu6n sach nay, mgt mat cae ban se dam bao cho minh cac k i l n thuc viJng vang \k mon Toan, mat khac cac ban se tiet kiem dugc thai gian danh eho viec on luy^n mon Toan va tir se c6 them nhiSu thai gian hen danh cho viec on luyen cac mon thi khac lai cac ki thi tuySn sinh sap tai Chung toi tin rang eu6n sach se dap ung dugc mgt s6 lugng Ian ban doc, t u eac em hgc sinh ehuan bj cho cac ki thi tuyen sinh vao Dai hgc, Cao dang sap tai, cho den cac thay, c6 giao day Toan a nha truong thong c6 the tim thay a day nh&ng dieu b6 ieh cho vi^e day va on tap mon Toan eho hgc sinh cua minh Mac dku rdt c6 gang va nghiem tiic luc bien soan cuon saeh, nhung dung lugng eiia cu6n sach qua Ian nen chac chan eac khiem khuyet ton tai la khong the tranh khoi Tac gia rat vui long neu nhan dugc sir gop y ciia ban dge xa gan de cuon sach hoan hao han cac Ian tai ban tiep theo Thu tu gop y x i n g u i v e theo dja chi: , i ; r !|i PHAN HUY K H A I ' \N T O A N HOC 18 aifo-ng Hodng Quoc Vi?t, qugn Cau Giay, Ha N{>i Tac gia xin chan cam an r TAG GIA Bai giang so 1: T H ^ TICH K H d l H A D I ^ N Cac bai toan thugc chu de c6 cac de thi tuyen sinh D a i hgc, Cao d i n g d cau so'4 H a l ngi dung chinh du'dc hoi den la: - T i n h the tich cua mgt khoi da dien (hinh chop hoac hinh lang tru) cho tril'dc nao - Sur dung phu-dng phap the tich de t i m khoang each gii?a mgt d i e m den mgt mat phang hoac khoang each giffa hai diTdng t h i n g cheo Cac n g i dung sau day chiTa du'dc de cap den cac de thi T u y e n sinh D a i hgc, Cao dang tijr nam 2002 den nam 2011, nhiTng rat cd biin va deu nam han che k i e n thiJc ve m o n Toan ap dung cho cac k i thi tuyen sinh B g Giao due va D a o tao quy djnh - Cac bai toan ve the tich k h o i da dien c6 ket hdp v d i viec t i m gia tri Idn nhat va nho nhat - Cac bai toan ve so sanh the tich B a i giang se de cap den cdc ngi dung 66 § T I N H T H E T I C H C U A M O T K H O I O A D I E N C a c kien thtfc coT ban c a n biet ( X e m sach giao khoa H i n h hgc 12) C a c dang toan thuTofng gap \i tinh the tich ^ Ta thirdng gap hai loai todn chinh sau day: '• Logi 1: T i n h the tich bang cac su^ dung triTc tiep cac cong thiJc toan PhiTcfng phap g i a i cac bai toan thugc loai dufdc tien hanh nhiT sau: - ' ' X a c dinh chieu cao cua k h o i da dien can tinh the tich Trong nhieu triTdng hdp chieu cao'nay diTdc xac dinh tiT dau b a i , nhiTng cung CO trirdng hdp viec xac djnh phai diTa vao cac djnh l i ve quan he vuong goc da hoc d Idp 11 (hay dung nhat la cac dinh l i ve ba diTdng vuong goc, cdc dinh l i ve dieu k i e n de mgt du^dng thang vuong goc v d i m g t mat phlng ) V i e c tinh cac chieu cao thong thudng nhd vao viec sijT dung d i n h l i Pitago, hoac nhd den phep tinh liTdng giac - T i m dien tich day bang cac cong thuTc quen bie't ' < i N h i n Chung cic bai toan thugc loai rat cd ban, chi d o i h o i viec tinh ioin can than va chinh xac Thidu 1: (De thi tuyen sinh Dqi hgc khoi A - 2009) Cho hinh chop S.ABCD c6 day la hinh thang vuong tai A va D ; A B = A D = 2a; C D = a, goc giffa hai mat phang (SBC) va ( A B C D ) bang 60" G g i I la trung I H ( A B C ) Theo dinh l i Talet ta c6: ^ ^ S H ^ = ^ ( d v t t ) Thidu 2: (De tuyen sink Dai hgc khoi B - 2009) Cho hinh lang tru diJng A B C A ' B ' C c6 canh ben B B ' = a va B B ' tao v d i mat phang A B C goc 60" Gia s\i A B C la tam giac vuong tai C va B A C = 60" H i n h chieu vuong goc cua B ' len ( A B C ) trung v d i tam tam giac A B C T i n h the lich t t f d i c n A ' A B C IA' Taco: A'M IH AA' CI CA' • CI I_ H = _ C L ^ j j ^ ^ ^ ^ , ^ a l A ' + CI AA' CA' V.^gc ==^SABcIH = ^ l A B B C I H = i a a y = 3 (dvtt) Thi du 4: (De thi tuyen sink D^i hQC khoi A - 2007) Cho hinh chop S A B C D day la hinh vuong A B C D canh a, mat ben S A D la Giai tam giac deu va n^m mat ph^ng vuong goc v d i day A B C D G p i M , N , P G o i G la tam tam giac A B C ta c6 B ' G Ian IiTdt la trung d i e m cua SB, BC, C D T i n h the tich tiJ dien C M N P Giai = 60" la goc ma B B ' tao v d i m a t phang ( A B C ) Trong tam giac G p i H la trung d i e m cua A D , thi S H I A D Do ( S A D ) ( A B C D ) nen suy ra: vuong B B ' G ta c6 ngay: B G = | ; B ' G - SHl(ABCD)va Dat A B = 2x, tam giac vuong A B C ta c6: A C = x ; B C = XN/3 (do A B C 9a^ 16 SH = — ( v i A B C la tam giac deu canh a) =60") Ke M K / / S H ( K e H B ) Gia sur B G n A C thi B N = - B G = — A p dung dinh l i Pitago tam giac vuong B N C ta co: BN^ = N C ^ + B C l : (^vtt) A C = V9a^-4a^ =aV5 T a c o : I H = I N c o s H I N = I N c o s M C B (do H I N va M C B la cac goc co canh ( A B C ) TCr E V B G 2Uo (2) d i e m cua A M va A ' C T i n h the tich tiJ dien l A B C G p i M , N ti/dng iJng la trung d i e m cua A B , BC V i I N la dirdng trung binh 9a-^ 12 Gia suf A B = a, A A ' = 2a; A C = 3a G p i M la trung d i e m cua A ' C va I la giao vuong S I H , ta c6 SI = I H tan 60" = I H 73 V a y VS.3C0 Cho hinh lang tru duTng A ' B ' C ' A B C c6 day la tam giac vuong A B C tai B Ta c : S H I = 60" la goc gii?a hai m a t Trong Thidit 3: (De thi tuyen sink Dai hgc khoi D - 2009) => SH ± BC (djnh l i ba diTcJng vuong goc) ( S B C ) va ( A B C D ) = i i A B B C ^ = -"-^x.xV^ = ^ T h a y (1) v a o ( ) , ta co: V A - A B C Ke I H BC phang ^""^ ^ — + 3x 9a^ = > x =• 52 => M K l ( A B C D ) va M K = — (1) Vay: VM.cNP-jScNp.MK = i ^ — iV3 aV3 96 Cty TNHH MTV DWH'Khang Vigt Thidu 5: D§ thi tuyen sink Dai hoc khoi B - 2006) Cho hinh chop S A B C D c6 d a y A B C D Ta c6: A B ± B C = > S B -L B C la hinh chi? nhat vdi A B = a, A D = aV2 , SA = a va SA vuong goc v d i m a i p h i n g ( A B C D ) G o i M , N Ian liTOt la trung d i e m cua A D va SC Gia suf I la giao d i e m cua B M va A C T i m the tich tiJ d i e n A N I B Gpi = > S B A la gdc tao b d i he mat phang ( S B C ) va ( A B C ) , nen S B A = 60" Qua M ke M N // B C = > N la trung d i e m cua A C Ta CO : M N = ^ B C = a ; M B = ^ A B = a Giai O la t a m cua d a y A B C D T r o n g tam _ ( M N + B C ) M B _ (a + 2a).a _ 3a giac S A C , ta c6 N O la diTdng trung binh nen N O / / S A , tiJc N O = > SMNBC - ( A B C D ) va N O = - (1) Thi du (Be thi tuyen sinh Bgi hgc khoi B - 2011) > Cho hinh lang tru ABCD.A|B|C,D| c6 day A B C D la hinh chu" nhat v d i A B = a; A D = a V H i n h chieu vuong goc cua A , tren mat phang ( A B C D ) trung v d i X e t hinh chu' nhat A B C D D o M A = M D MA BD: ^ ( A D D i A , ) va ( A B C D ) tao v d i goc 60" T i m the tich cua lang tru da cho AC^ AI =-AC=>AI^ — Do 66 giao d i e m O cua hai diTdng cheo A C , B D cua day B i e t rang h a i mat p h l n g AI = - I C Laicd: Giai 22+a^ G o i E la trung d i e m cua A D , ta c6 O E A D => A|E J A D (djnh l i ba du'dng vuong goc) Vi (ADDiA.) n (ABCD) = A D + BV- = a ' = A B ' , nen A I B la tam giac vuong d i n h I ^ = ^ (2) A|EO la goc giiJa h a i m a t phang ^ A ^ = 60" Ta c d : A , = O E t a n A ^ = | % ^ Thay (2) vao ( ) ta c6: V ^ N I B - (dvtt) 36 Thidu "^c, Ta CO A , ( A B C D ) B I = - B M ^ BI^ = - B M ^ Vay S „ e = l l A I B = l ^ ^ ' ^ 2 3a^ = - a > / — = a-^N/3 (dvdt) Tac6 V A N , B = V N A , B = ^ S A , B N O Ta tinh dien tich tam giac A I B : ~ V a y VsMNBC = ^ S A S M N B C - ^ A B t a n " AIB 2 (Be thi tuyen sinh Dai hgc khoi A-2011) Cho hinh chop tam giac S.ABC, day la l a m giac vuong can A B C tai B , A B = B C = 2a Gia suf hai mat phang ( S A B ) va ( S A C ) cilng vuong goc v d i mat ph^ng ( A B C ) G o i M la trung d i e m cua A B M a t phang qua S M va song song v d i B C ciit A C tai N B i e t rang hai m a t phang (SBC) va ( A B C ) tao v d i goc 60" T i m the tich hinh chop S B C N M Giai Do ( S A B ) va ( ( S A C ) cung vuong goc v d i ( A B C ) V a ( S A B ) n (SAC) = SA, n e n SA ( A B C ) , (SBC) n ( A B C ) = B C = ^ V A B C D A | B | C I D | =S/^QCQ.A^0 Thidu = a.a\l3 ^ ^ (Be thi tuyen sinh Bgi hgc khoi D - 2011) Cho hinh chop tam giac S.ABC day la tam giac vuong A B C t a i B va A B = 3a, B C = 4a Biet rang m a t phang (SBC) vuong goc v d i ( A B C ) Gia suf SB = IdiS va SBC = 30" T i m the tich hinh chop S.ABC Giai D o (SBC) ( A B C ) va (SBC) n ( A B C ) = B C , nen neu k c S H B C ( H e B C ) => SH ( A B C ) A Phuong ph^p giai JcAn trpng tam - Phan Huy Khii Cty TNHH MTV DVVH Khang Vi$t Ta c6 SH = SB s i n S B H = 2aN/3.sin30" = aVs hoSc tinh du'dc d i c n tich day nhu^ng cung khong de dang K h i nhieu triTcfng hdp ta c6 the lam nhu* sau: V a y V s A B C = j S A B c S H = ^ ^ a V = 2.alV3 (dvdt) Biiih ludii: Cac de thi noi t r e n q u a la rat c d b a n va dcfn g i a n ! Tin du (De thi tuyeii sink Dai h^c khoi A - - Phan chia k h o i can tinh the tich tdng hoiic hieu cac k h o i cd ban (hinh chop hoSc hinh lang tru) ma cac k h o i de tinh hdn >" " ^ v - 2010) Cho hinh chop S A B C D c6 day A B C D la hinh vuong canh a G o i M va N Ian Hoac la so sanh the tich k h o i can tinh v d i mot k h o i da dien khac da biet tri/dc the tich liTdt la trung d i e m cua A B va A D ; H la giao d i e m cua C N va D M B i e t SH - vuong goc v d i mat phang ( A B C D ) va SH = a^J3 Bai todn Vs.A'B'C _ SA' -^-^"^-ScDMN Vs ABC (2) 22 Thay (2) v a o ( l ) , ta CO : V^CDMN = r - a 73 (dvdt) 24 Ta c6: 2010) mat phang ( A ' B C ) va ( A B C ) bkng 60" T i m the tich k h o i lang tru ay Giai G p i M la trung d i e m BC, ta c6 A M BC = — -SB.SC.sina.AH SC SA' SB SC SA dpcm day a = B'SC' = BSC cheo A C = 4cm D o a n thang SO = lyjl tan60"= — 3a a^Vs tinh tdn}^ hodc hieu ciia cac 3a-^^ khdi c(f ban cm va vuong goc v d i day, d day O la giao d i e m cua hai du'dng cheo A C va B D G o i M la trung d i e m ciia canh SC Gia sijr mat phdng ( A B M ) cat SD tai N T i m the tich hinh chop (dvdt) Giai Ta CO AB//DC => A B // (SDC) vdo phan tich khdi can hodc bani^ each so sdnh the tich khdc Trong nhieu bai toan, viec tinh triTc tiep the tich kho'i da dien nhu" loai CO the gSp kho khan v i hai l i do: HoSc 1^ kho xac dinh va tinh du^dc chieu cao, 2^SBC-^H SB' Cho hinh chop S A B C D c6 day A B C D la hinh thoi canh bang Vs cm, diTdng todn tinh the tich khdi da di$n dua vdi mot khdi da dien ca ban -SB'.SC'sina.A'H' Thi du 1: (De thi tuyen sinh Dai hQC khoi A - 2004) Binh luqn: Cac de thi noi tren la qua cd ban va dcJn g i a n ! bai *^S.ABC ''A.SBC -Ssg.c^.A'H' d i e m A = A ' ; B = B ' ; C = C' = 60" TCrdo A ' A = A M t a n A m *^ Loai 2: Cac V A'.SB'C C h i i y : K e t qua tren van dung neu nhiT cac d i e m A ' , B ' , C co the c6 V a y i V M A la goc tao bdi hai mat phang ( A ' B C ) S ABC S.A'B'C V ^^^^ => A ' M BC (dinh l i ba diTcfng vuong goc) V ABC.A'B'C 'H'r*' — A A mat H thang hang 5a^ Cho hinh lang try tam giac deu A B C A ' B ' C c6 A B = a, goc giiJa he ^ ~ SA • SB • SC ph^ng (SBC) K h i A ' H ' / / A H va S, H ' , TItidu 10 (De thi tuyen sinh Dai hoc khoi B - va ( A B C ) va A SC K e A ' H ' va A H ciing vuong g6c a 5a' —a = — 2'2'2 SB' Chtfnt; m i n h Lai CO : S^Q^^N - ^ABCD ~ ^ AMN "S^^Bc = a C tiTdng uTng tren canh SA, SB, SC K h i Giai a a cifbdn: Cho hinh chop S A B C L a y A ' , B ' , T i n h the tich hinh chop S C D N M Ta CO Vs.CDMN -~^^-^CDMN V d i l o a i toan ngifdi ta rat hay sijf dung k e l qua sau day =^ (SAB) n (SDC) = M N // A B ( N e V i M la trung d i e m cua SC nen N la trung d i e m ciia SD Ta c6: V j 'S.ABMN - ^ S A B N + ^S.BMN Theo bai toan cd ban ta c6: (^) SD) giai To^n trpng tarn - PhiBng ph^p VS.ABN _ S N _ 'S.ABD SD _ Nh(^n xet: S.ABN - ~ ^ S A B D - ~ ^ S A B C D Ta uin n i l * ' CO A E I B C , SA B C BC i (SEA) = K e A H SE ( H e SE) => A H (2) ^ B M N = ^ ^ A^CD D g thay: V , B C D = ^ S , , , S = i.l.AC.BD.SO = 1.4.2.272 = 8V2 ^ (SBC) Trong tarn giac vuong SAE, ta c6: (3) AH' Nhan xet: Ta - H i n h lhang A B M N CO the tinh du'dc dien tich (luy khong de dang) - V i e c xac dinh chicu cao tiif S xuong A B M N va tinh no phiJc tap hdn ( B a n CO the tinh todn x e m no de k i e m Ira tinh phiJc tap) c6 SA' SB ^SBC i ^ : A H = 2a'^ 4a AE' SsMN _ SM _ - B M N C - - S B C S.K,., SN _ SB SC 19 3a^ SA'* 16 SB' 25 9a (xem d tren) Thidu 2: (De tin tuyeii siiih Dai hoc khoi D - 2006) Cho hinh chop tarn giac S.ABC c6 day la tarn giac deu canh bang a SA = 2a va SA vuong goc v d i mat phang ( A B C ) G o i M , N tiTdng iJng la hinh chieu 3a' 25-2 50J a ' + — = -r^S,„p = ^ - B C S E = — B^^NC = ^S.ABC ~ ^ S A M N Theo bai toan cd ban thi: Vi AB = AC ^ •' 'IS 9a' 100 — yfi9 Ta thu l a i ke't qua tren Theo ban each f^idi ndo la h(fp li? ThidiJ 3: (De thi tuyeii sink Dai hgc khoi A - 2003) Cho hinh hop chOT nhat A B C D A ' B ' C ' D ' day la hinh vuong canh b^ng a, chieu Giai Ta C6: u, t„ • • Cdch giai nhiT tren la hdp l i ! vuong goc cua A tren SB, SC T i m the tich khoi chop A B M N C y ( S B C ) ± (SEA) V a y A H la chieu cao cua hinh chop A B M N C TO (2) va (3) suy ra: Vg ^BCD = V2 (dvtt) - b a n h a y so sanh e a c h giai tren v d i e a c h giai b a n g "tinh toan triTc t i e ' p " ( t h e o p h i T d n g p h a p e u a l o a i 1) v d i t h i d u n a y = -V 'S.BCD TO (1) suy ra: Cac G o i E la trung d i e m cua BC SM V s B M N _ SN Cty TNHH MTV DVVH Khang Vigt Phan Huy Khii cao A A ' = b Goi M la trung diem cua canh C C T i m the tich tuT dien B D A ' M Giai (1) ^ S M ^ SN SB SC V,S.ABC SB = SC Ta c6: SA^ = SM.SB = SN.SC=> S M = SN y g y VS.AMN _ S M Vs.ABC (2) SB^ Ta c6 S A ' = SM.SB V a y tir (2) CO V,S.AMN SM = \2 4a^ + a^ TO(l)va 10 Taco: SB^ '^S.ABN (3)c6: ^A^CNM bb iH>n n diem cua C C , nen la c6 CE = A C = aV~ SB SV _ 4a' Trong ( A C C ' A ' ) : A ' M n A C = E G o i O \k tarn ciia day A B C D v i M la trung SA' SB^ ^A'BDM = ^A'BDE - ' ^ M B D E = 16 25 =^'^S.ABC 'S.AMN 25 a^Vs =:^-:^ J ^^ 25 f = -.-BD.EO b (3) S.ABC = 3a^73 50 B D E - A A ' - ^ S g D E - M C = ^SBDE ( ^ A ' - M C ) b^ I r-b — = -.aV2.-aV2.—= 2 a'b,^., (dvtt) (dvtt) 11 Phitang ph^p giai To^n trpng tflm - Cty TNHH MTV DWH Khang Vigt Phan Huy KhSi Thidu 4: (Be thi tuyen siiih Cao dang khoi A - 2009) Cho hinh chop ttf giac S.ABCD c6 canh ddy AB = a, canh ben SA = aVI Goi M, N, P Ian IiTdt la trung diem ciia SA, SB, CD Tim the tich tiJ dien AMNP Giai DoMS = MA::^ d(A,(MNP)) = d(S,(MNP)) (*)=>VA.MNP = V.S.MNP Theo b^i toan cd ban, ta c6: = i i s ABP SO = - - - A B H P S O V s M N P _ S M S N _ _ ^ ^'S.MNP , 32 SA SB 'S.ABP (O \k H tifdng ijTng la tarn cua day ABCD va trung diem cua AB) a-^V6 (2) = —.aa 48 24 V TCr(l)va (2) suy ra:VA.MNP = aV6 (dvtt) 48 §2 sCr DUNG PHUONG PHAP TH^ TfCH TJM KHOANG CACH Cac bai todn tim khodng each: Khoang each tir mot diem den mot mat phing, khoang each gii^a hai diTdng thang, nhieu triCdng hdp c6 the quy ve bai ioin the tich khoi da dien Viec 3V tinh cac khoang each difa vao cong thtJc hien nhien sau: h = — day V, S, h Ian li/dt la the tich, dien tich day va chieu cao cua mpt hinh Y ch6p nao (hoSc h = — doi vdi hinh lang tru) PhiTdng phap ap dung diTdc triTdng hdp sau: Gia sii' ta c6 the quy bai toan tim khoang each ve bai toan tim chieu cao cua mot hinh chop (hoac mot lang try) nao DI nhien cac chieu cao thiTdng la khong tinh triTc tiep diTdc bang each suf dung cac phiTdng phap thong thifdng nhiT dung dinh li Pitago, diing cac cong thuTc liTdng giac Tuy nhien cac khoi da dien lai de dang tinh diTdc the tich va dien tich day NhiT vay chieu cao ciia n6 se dUdc xdc djnh bSi cac c6ng thilc ddn gian tren LiTdc suT dung phiTdng phap the tich de tim khoang each nhU" sau: 1/ SiJf dung cac dinh li cua hinh hoc khong gian sau day: + Neu AB // (P) mat phang (P) chiJa CD thi d(AB,CD) = d(AB,(P)) + Neu (P) // (Q), cac mat phing (P), (Q) Ian liTdt chiJa AB, CD thi (AB,CD) = d((P),(Q)) Ta quy bai toan tim khoang each theo yeu cau dau bai ve viec tim chieu cao cua mot khoi chop (hoac mot khoi lang try nao do) 21 Gia suT bai toan quy ve tim chieu cao ke tiT S cua mot hinh chop (hoac mot lang try) nao d6 Ta tim the tich cua hinh chop (hoac lang try) theo mot difcJng khac ma khong diTa vao dinh S (thi du tinh the tich hinh chop ay vdi quan niem no la hinh chop vdi dinh S V S) Tinh dien tich day doi dien vdi dinh S TuT ta c6 chieu cao ke tuT S can tim Xet mot vai vi dy minh hoa sau day ^r ) q: Thi dv 1: Be thi tuyen sinh Bai hgc khoi A - 2004 Cho hinh chop S.ABCD day la hinh thoi ABCD c6 SO vuong goc vdi day va O la giao diem cua AC va BD Gia suT SO = 2%/2 , di/dng cheo AC = 4, canh day AB = Vs Gpi M la trung diem cua SC Tinh khoang each gii^a hai du'dng thang SA va BM Giai Gpi M la trung diem cua SC Khi ta c6 OM // SA => SA// (0MB) => d(SA,MB) = d (SA, (MOB)) = d(S, (MOB)) = d(C,(MOB)) (1) (do MS = MC) Ta c : B ' = A B ^ - A ^ = l OB = Ke MH (ABCD) MH = - S = V2 = ^.ioB.OC.MH = ^1.2.72 = Vay V,M.OBC=^SoBcMH Taco: OM = - S A =-VsTI = >/3 (2) Gpi h la khoang each tif C tdti (MOB), ta c6: M O B h = ^.^OB.OM.h = hV3 *^C.MOB Tir (2) va (3) suy V3h 72 =^h = 2V6 Tir (1) va (4) di den: d(SA, MB) = (3) (4) Cty TNHH MTV DVVH Khang Vi$t Phuong phAp giii To^n trpng tam - Phan Huy KhSi Thidu 2: (De thi tuyen sinh Dai hgc khoi D - 2009) Cho hinh lang tru du-ng A B C A ' B ' C ddy la tam gidc ABC vuong tai B Gia sit AB = a, A A ' = 2a; AC = 3a Goi M la trung diem cua A ' C va I la giao diem cua A M va A'C ^ Tim the tich ti? dien lABC Tim khoang each tiT A tdi m3t phang IBC Trong tam giac vuong VJ^BC = 4a^ Ta Mat Ke IH AC => I H (ABC) Ke HE BC CO: Vc B C D ' khae: Ta c6: Do vay AB CA IE = N/IH^ „ ^ ^ „ 2a = -=>HE =-AB = — CA' 3 = 16a^ 4a^ 2aV5 Ke A K ( I B C ) , ta c6 V , A B C = V ^ I B C - - S , B C A K Tiif cau suy ra: S A ' = S H S B => S H = "3 BP aV3 SBCD BCD = T ^ S C D B F I 2aV5 ,/ 2aV5 = - - a — — A K => AK = d ( A , ( I B C ) j = 3 4a2 Thi du 3: (De thi tuyen sinh dai hgc khoi D - 2007) Cho hinh chop tuT giac S.ABCD eo day ABCD la hinh thang vuong Tinh khoang each tir A tdi mat phang (SCD) Giai Do AB BD => SB ± BC (dinh l i ba diTdng vuong goe) Di/a vao dinh l i Pitago cac tam giac vuong SAB, ABC, SAD de dang tinh diTde: SB^ = 3a'; SC' = 4a'; S D ' = 6a' Tiif d6 eo: DC' = 2a' Vay suy ra: S D ' = S C ' + S D ' SCD la tam giac vuong tai C => dpcm Ke HK va BP ciing vuong goe vdi mat phang (SCD) Ta c6: 14 BP SB (1) t Ti^ (2) va (3) suy ra: H K - d ( H , ( S C D ) ) = | Thi du 4: (De thi tuyen sinh Dai hQC khoi A - 2006) Cho hinh lap phiTdng ABCD.A'B'C'D' cd canh bang I Goi M , N Ian liTdt la trung diem eua AB va CD Tinh khoang each giffa hai du'dng th^ng A'C va M N Giai Ta CO MN//BC MN//(A'BC) => d(MN.A'C) = d(MN.(A'BC)) = d (M,(A'BC)) (1) DS thay V^-MBC =^S^BC-A'A -^•{•^•^•^ = ^2 (1) Vi CB (BAA'B) =^ CB BA' Tirdd: ChiJng minh SCD la tam giac vuong 6 (3) ABCD va SA=aV2 Goi H la hinh chieu cua A tren SB SH = -.2a.aV2.BP =^ B P = ^ => A'BC la tam giac vuong tai B HK = -a.aV2.—aV2 = ABC = BAD = 90"; BA = BC = a; AD = 2a Gia suT SA vuong goc vdti day ^ HK // BP va S, K, P thang hang (2) - 7SC.CD.BP = -BP -SA = - - BC.CDsin 135".SA Tir dd ta cd: HK ! CB' +HE^ HK =- => IE BC (dinh ly ba diTcJng vuong goe) HE CH c6: 2aV3 ' Vay tijf (1) C O Giai I Theo thi du 3, loai 1, §1, ta eo SAB V A M B C = VM A ' B C - (2) ^SA'BC-^ Tilf(l), (2) ta cd - i =UA'B.BC.h iV2h=^h = ^ 12 Tir ( I ) , (2), (3) suy ra: d ( M N , A ' C ) = Thid^5(De (3) 72 thi tuyen sinh Dai hoc khoi D - 2011) Cho hinh chdp tam giac S.ABC day la tam giac vuong ABC tai B va AB = 3a; BC = 4a Biet rSng mat phang (SBC) vuong gde vdi (ABC) Gia sur SB = 2aV3 va SBC = 30" Tim khoang each ttf B den (SAC) Giai ; itoi, TheothiduS, loai l , § l , t a c d V g s A C ^ ' ^ s A B C =2a^V3 PhiBno phap giai Toin trpng tarn - Phan Huy KhSi Cty TNHH MTV DWH Khang Vi$t Mat khac goi h la khoang each tit B tdi mat ph^ng (SAC), ta c6: T i r ( l ) ( ) ( ) s u y r a : Vg A , B D = ^B.SAC - - h S s A c - Theo(l) Goi h la khoang each tif Bi xuo'ng matphing A , B D , ta c6 : (2) S ^SAC (4) ^SAC VB,.A,BD=^h.SA,BD=^h.iBD.A,0 (5) Ke SH BC => SH (ABC) v i c S H = SB.sin30"= aV3 Thay (4) vao (5) ta c6: — = - h a - ^ ^ o BH = VSB^ -SH^ = N/l2a' - V = 3a ^ HC = a Ke BE AC va H K AC HK // BE Thco he ihijTc liTdng tam giac vuong ABC, ta c6: 1 1 BE' AB- BC' 9a' 16a' Theo dinh l i Ta-let, ta c6: BD = V A B ^ + A D ^ A 12a • BE = HK CH BE CB 4a Tilf suy ra: h = Thi du (De thi tuyen sinh D^i hpc khoi Vay S s A c = - A C S K = - a ^ ^ = a'N/2T 2 9a' 25 (3) Giai s\ De thay SA (ABC) va SBA la goc tao bSi Thay (3) vao (2) vii c6: h = hai mat phlng (SBC) va (ABC) => SBA = 60" Ke NK // AB, d6 K la trung diem cua BC Thidu 6: (De thi tuyen sink Dgi hijc khoi li - 2011) Cho hinh lang tru A B C D A , B , C D , c6 day ABCD la hinh chff nhat vdi AB = a ; AD = diS A-2011) Cho hinh chop tam giac S.ABC, day la tam giac vu6ng can ABC tai B, AB = BC = 2a Gia suf hai mat phing (SAB) va (SAC) cung vuong goc vdi mat phang (ABC) Gpi M la trung diem cua AB Mat phdng qua SM va song song vdi BC c^t AC tai N Biet r^ng hai mat phIng (SBC) va (ABC) tao vdi goc 60" Tim khoang each giffa hai diTdng thing AB va SN theo a Theo djnh l i Pytago, thi: SK = VsH^ + HK^ = ^ 3a'^ +ã (xem thidu 7, Đ1 loai 1) = Va^+Sa^ = 2a; A , = Hinh chieu vuong goc cua A, trcn mat phang (ABCD) Ta CO AB // NK AB // (SNK) V i the' d(AB,SN) = d(AB;(SNK))=d(M,(SNK)) (1) trung vdi giao diem O ciia hai diCdng chco AC, BD ciia day Bict rang hai mat phang (ADD|A,) va (ABCD) tao vdi goc 6()" Tim khoang each tiJT B, tdi DO mat phang (A|BD) / (1) A|BD =^D.AiB|B -^'^D.AiBiBA Mat khac ta CO : VUA,B|BA (2) =r'VDDiCiC.AA,B|B 3a Ta lai c6 : VDD|C|C.AAIBIB - ^ABCD,A|B|C|D| - (theo thidu 7, loai §1) - S'NKC SA|B|B - 2^A|BIBA ^ -ãJSBMNC (2) Theo thi du loai Đ1, ta c6 V j g ^ M N = Nen tir (2) ta c6: Ma y^SNK = VS.^NK = ^ (2) Ta co: VM.SNK = \ M N K = ^ (theo (2)) j h S s N K , day h = d ( M , ( S N K ) ) ^-^^ Nen suy h = 3V,M S N K SsNK 16 '^MNK ^S.MNK - g^S.BCMN Giai TaCO S M B K -^MNK (3)_.^4tfa~va:&^^SNK ii\iJ ^lEii TiNH iilNH THUAX Cty TNHH MTV DVVH Khang V i j t K e A H ± N K (tuTc la A H // BC) => A H = B K = i f i C = a Theo djnh l i ba diTdng vuong goc ta c6 : SH ± N K ^ Tacd M N = Ta c6: SH = VsA^ + A H ^ = Vl2a^ +a^ = aN/lJ (do SA = 2atan60" = ) V i t h c SsNK = - N K S H = i a ^ (4) TiJf (3) (4) ta c6: h = ^ ^2aV39 2 a^ /— 13 \ • r\,,., • ! • /I-!,-.'.: yVl3 §3 CAC BAI TOAN v i T h i TICH KHOI OA DIEN CO KET HOP D a y C O the x e m nhiT mot bai toan rat cd b a n mac du no chu'a mot Ian x u a l 2002 2a sma sma T t f d d SO = O N t a n a = D o Vs A B C D = - sma sma cos a cos a 2a sin a cosa sin a cos a Tuf (1) suy VS.ABCD he nha't va chi sin' a c o s a Idn nhat X e t bieu ihu'c P = sin^ a c o s a = cos( - cos^ a ) = c o s a - cos^ a vol VIEC TiM GIA TRj LON NHAT VA NHO NHAT hien cac bai thi l u y e n sinh vao D a i hoc, Cao dang tif nam MH dc'n TxS (2) dan dc'n xet ham so y = x - x ' v d i < x < Ta cd y ' = - 3x^ va ta cd bang bie'n thien sau: 73 x (cho du bai toan vc gia trj Idn nhat, nho nha't v d i ham so hau n h U n a m nao cling Cac b a i toan n a y cd noi dung rat cd b a n nhiT sau: The tich kho'i da dien y' cac dang toan n a y phu thuoc mot tham so nao (lham so c d the la gdc, hoac y V d i cac bai toan du'dc g i a i theo cac biTdc sau: iS 73 O X= V a y : VS.ABCD nhan gia tri be nha't bhng - Bu'dc 3: D e n day n h i c m vu cua bai toan hinh hoc x e m nh\i da "ke't thuc" Ta cd mot ham so f(x) v d i x € D ma can t i m gia t r i Idn nhat, nho nha't ciia no Thidu Thidu 1: Cho hinh chop tuT giac deu S A B C D ma khoang each tif d i e m A den mat phang (SBC) bSng 2a V d i gia trj nao cua a , v d i a la g d c giffa mat phang Giai ' G o i M , N Ian lUdt la trung d i e m cua A D , BC Ta cd S N M = a D o D A // B C => A D // (SBC) ^ d ( A , (SBC)) = d ( M , ( S B C ) ) = M H = 2a d diiy: M H SN ( H e SN) (Chu y v i ( M N S ) nenMHl(SBC)) 18 ,^ ^ v • M^v\i::,V,.~- -\-/fV;^-""^' (SBC) chi k h i cosa = -7^ gdc v d i day ( A B C ) Gia silf SC = a Hay t m gdc giffa hai mat phang (SBC) va ( A B D ) cho the tich k h o i chdp la Idn nhat Giai ^ Ta tha'y SCA = a, SA = asina va A C = acosa S u y r a : V ,S.ABC ben va day cua hinh chop, thi the tich cua k h o i chop la nho nha't? T i m gia trj nho nha't 273a-^ k h i va 2: H i n h chdp S.ABC cd day la tam giac vuong can dinh C va SA vuong Dung bat dang thtfc hoac suf dung tinh dong bie'n va nghjch bien cua ham so thong qua vice khao sat ham cua ham so ^ = 273 - Bu'dc 2: V d i an so du'dc chon d biTdc 1, ta coi nhU la cac yeu to' da cho de tinh the tich V cua k h o i da dien theo cac phUdng phap da bie'l ^ ^1 - Birdc 1: Chon tham so, thiTc chat la chon an A n c d the la gdc a thich hdp k h o i da d i e n , hocic la mot yeu to dai nao + la dai canh), bai toan d o i h o i xac dinh gia tri c u a tham so de the tich nhan TiJf dd ym„ = y = V3 C O mat cac de thi tuyen sinh) gia tri Idn nhat hoac nho nha't (2) la^cos^a a' •) asina = — c o s a s i n a (1) Tir (1) suy ra: VS.ABC nhan gia t r i Idn nha't va chi k h i bieu Ihu'c P = c o s a s i n a nhan gi^ t r i Idn nhat V i sin a > 0, nen P^,, o P,?„, < ^ ( \ sin^ a)^ sin^ a dat gia trj Idn nhat T, /, \ , (l-sin^a)(l-sin^a)f2sin2a) d i ' - j M Tacd l - s i n ^ a s i n a = -^ '- TCr d6 c6 bang bien thien sau: T£r ta c6 bang bien thien sau: -1 f(t) f ~ V a y m a x y = y ( l ) = \/2; m i n y = m i n | y ( - l ) ; y ( ) } = m i n - 0;-j=r • = [-l;2] Thidfi [-l;2] I 10: (Be thi tuyen sink B K h i < x < 2, ta c6: (4 - x^) - x ^ = - 2x' y ' > < V4 l (ab + be + ca)^ + 3(ab + be + ca) + j l - ( a b + bc + ca) - (do = (a + b + c)^) y V a y m a x y = y(V2) = 2V2 ; M i n y = m i n {y(-2);y(2)} = m i n { - ; } = - Dat t = ab + be + ca, ta Nhan xet: Ta c6 the giai nhif sau: X e t h a m s o f ( t ) = t ' + t + 2sfrTt CO : 0 -2 va y(-2) = -2 V a y miny =2 De t l m gid t r i Idn nha't ta se suf dung bat d^ng thtfc Bunhiacopski T a c o : f ( t ) = 2t + - - ^ • V t e Cho x, y, > v^ X + y = T i m gia tri Idn nhat va nho nha't cua bieu thtfc: ' Do dd f(t) la h a m dong bien tren Giai T a c y = l - x , t i r d P = 3^'' + ' ' " = ^ " + — v d i O < x < I 150 0;i' t f'(t) = 2t-4 t' = 2t'-3 t' o -1 N h i r t h e ' f ( t ) > f ( ) = Vt> ;M=2o ab = bc = ca < ab + be + ca = [a + b + c = l pp.- D a t t = 3\i d6 < t < X6t h a m so: f(t) = Vithe'M>f(t)>2Vte °1 (a ; b ; c) la mot cac h$ so ( ; 0; 0), (0; 1; 0) (0; 0; 1) V a y gid t r i be nhat cua M = IY- 151 2t^ + 12l §3 C A C PHLTONG PHAP KHAC CHUNG MINH BAT DANG THUG VA TiM GIA TR! LON NHAT, NHO NHAT CUA HAM S O Trong muc nay, chiing t o i de cap den m o t so' phiTcfng phap khac de chiJng minh bat d i n g thiJc va t i m gia t r i Idn nhat, nho nhat cua ham so' M a c du cdc bai Goi m la gia t r i y cua ham so f(t) = ;iChi dd phiTdng trinh sau (an t) : tap sijr dung nhiJng phi/dng phap hoac la chi/a c6 mat, hoac la c6 mSt chi mot, hai Ian cac k i thi tuyen sinh vao D a i hoc 2t^ +12t (2) o t ^ + t = m t ^ + m t + 3m i : i ( m - ) t ^ + ( m - ) t NSu m = t h i ro rang (3) cd nghiem nho nhat cua h a m so' V i e c suf dung no de giai cac b a i toan cac k i thi N^'u m tuyen sinh s i p tdi la kha nang hoan loan hien thifc va c6 tinh kha thi idn tri chiing minh bait dang ^.^(m_6)^-3(m-2)m>0 om^+3m- " ^' " < o - < m < (m^2) x2 +6xy) T o m l a i ta cd: - < m < D i e u cd nghla l a : - < PhU'dng phap dac biet hSu hieu de t i m gia trj Idn nha't va nho nhat cua cac ham so c6 dang sau day (hoSc cac dang khdc ma c6 the dU'a ve chiing): a, sinx + b, cosx + C| f ( x ) = —! • a2sinx + b2C0sx + C2 f(^)^ a , x ^ + b , x + C| ' ' a X ^ + b X + C2 + 3m = (3) 2, k h i d d (3) cd nghiem va chi k h i thiic va tim gid iihd't vd nho nhat cua ham so (2) = m t^ + t + day la rat c6 ich viec chiJng minh bat d i n g thiifc va l i m gia t r i Icln nhat va mien gid trf ham so'de vdi t cd nghiem D o t^ + 2t + > 0, V t e IR, nen Cao d i n g nhiJng nam gan day Nhu'ng chdng toi nghl r^ng phiTdng phap ma s i p diTcfc gicti thieu d a PhUcfitg phap t^ + t + x^+2xy+3y^ dpcm Dau b^ng ben trai xay k h i : m = o t ^ - l + = o t = X = o i^ D e g i ^ i cic b a i toan nay, ta t i e n hanh theo lufdc sau day: Gia sur yo la mot gid t r i tily ^ cua ham so" K h i dd phiTdng trinh sau (an x) =3 x= — r = \ -1 f(x) = yo (1) cd nghiem TCr dang cua (1) m ^ ta cd cac dieu k i e n thich hdp N d i chung cic dieu kien CO dang a < yo < p (2) Phan l a i (dau b i n g ben phai xay ra) x i n danh cho ban doc _ , s i n x + cosx + l Thidfi 2: CMng m i n h rSng v d i m p i x e R , ta co: - - < g - „ ^ _ c o s x + Tir (2) suy max f ( x ) = P va m i n f(x) = a Thidtf 1: (De thi tuyen sinh Dgi hpc khoi B - 2(\^+6\y] Cho x^ + y^ = ChijTng minh r^ng: - < - i Giai ;f lb 2008) , s i n x + cosx + l ^ ,„ ' Goi m la gid t r i tiiy y cua h a m so: f(x) = - : ; — v d i x eIR ^ • •' •' sinx-2cosx +3 2sinx + cosx + l (1) K h i dd phiTdng trinh sau (an x): — =m sinx-2cosx + ' - ( x y + y z + z x ) ^ B a n doc tiT xet k h i nao dau b^ng ben phai xay ra! b PhU(fitg phdp sut difiig bat dang thutc Bunhiacopski = : > ( x ^ + y ^ + z ^ ) ^ > ( x y + y z + z x f = 16 (2) (do x y + y z + z x = 4) Ba't d^ng thiJc Bunhiacopski la m o t bat dang thiJc cung rat thong dung chi sau ba't d^ng thiJc Cosi M a c dil no c h i xuat hien m o t Ian cac de thl, nhung DS'u b i n g (2) x a y r a o —=^ =— y z X , khong VI the' ma ta x e m nhe cac dang bai tap thupc loai Tir (1) (2) suy r a : x"*+y'*+z'* > — => d p c m Thi du (De thi tuyen sink Bai hgc khoi B - 2003) Tim gia t r i Idn nhat va nho nhat cua ham so: i(x) = \ ^|4-x^ x y + yz + zx = 16 tren m i e n - < X < DS'u b^ng x a y Giai o D o x > - n e n hien nhien ta c6: f ( x ) > - v d i V x e I R Mat khac f(-2) = - x =y =Z=^ x2=y2.z2 y \ y -Z = — z - X)l /; V 2^/3 X f(x) = - Ta se siJf dung ba't d^ng thitc Bunhiacopski de t i m gia t r i Idn nhat cua ham so Thi du 3: Cho x, y , z > Chifng minh: A p dung Bunhiacopski v d i hai day: x;V4 - x ^ va 1; ta c6: =>8>(x + ^ ) ^ => !- i lai i x2+(4-x^) (l2+l^)>(x + ^ f ^ + ^ + — - — > y + 2z z + 2x x + 2y f(x) ^ nhiTng khong ton tai Xo X ma f(X(,) = - N ^ ! T a i sao?) Da'u bang (1) x a y x ( + 2z) Thidu 2: Cho x, y , z, la cac so' thoa man dieu k i e n xy + yz + zx = Glal A p dung ba't d^ng thuTc Bunhiacopski cho hai day s^ x ^ y ^ z ^ v i 1, 1, Dau b^ng (2) xay o z=>3(x'*+y^+z'*)>(x2 + y + z ) ' Y2 x = y = z c PhMitg phdp svC dung lUffng gidc (1) ^2 Trong m o t so triTdng hdp ngiTdi ta c6 the ddi bien de diTa bieu thtfc bat thtfc, hoac h a m so can la'y gia t r i Idn nha't, nho nha't ve dang li/dng S'^c Sur dung cac ph6p bien ddi lifdng giac c6 the l a m cho viec chtfng m i n h ba't ''^'^g thiJc hoac t i m gia t r i Idn nha't, nh6 nha't de dang hdn D a u b^ng (1) xSy o x = y = z Tfif (1) (2) suy dpcm Da'u b^ng xay Ta c6: ( x ^ + y-* + z - * + ^ + ^ ) ^ (x^ + y^ + z ^ ) ' X^ y z vBH 2x _ x + 2y z+ o X = y = z y(2 + 2x) z(2 + 2y) ( x + y + z)^ , , , T a c o V P ( ) =-7^ ^ > (2) d o x ^ + y ^ + z ^ ^ x y + y z + zx ( x y + yz + z x ) ChiJng minh: x'* + y'* + z'* > ^ > o y + 2z ^, — = _ = — o x^ = y^ = z^ 155 154 ?'tii/(i!uj |)!iai) giii Toaii tioiuj !am - Phan Huy Khii Hai phep d d i bien thong dung, hay dilng la: DSt x = sin t (hoac x = cos t) X = tan t (hoac cot t) fhidif J ; Gia sur x va y khong dong nhat b^ng Churng m i n h ,^ , -2V2-2«'>'' (1) +1 (xV; ^ /z- ^ -2 o o -2V2-2 0, y > va X + y = ChiJng minh: + fo ^v2 • ,, D e n day suf dung phiTdng phdp m i e n gia tri ham so' (giong nhiT thi du 2, mu^ a) se suy ra: - < P fidi 5; Cho a, b c > va a' + b S c" = 48 Chtfug m i n h ab' + be' + ca' < 24 i 157 Cly TNHH MTV DWH Khang Vigt PhuongphSp giji T o i n t3tn - Phan Huy KhSi Bai 6: Cho x, y, z \k ba so' diTdng vax + y + z = l 19: Cho ba so dtfdng x, y, z v^ xyz =1 ChiJng minh V l - x + ^ I - y + V l - z < V6 v2 Bai 7: Cho x, y, z >0 va xyz = ChiJng minh : 1+y + ^2 +-i— > - 1+z 1+x ChuTng minh : , fifli 5: Cho X, y, z la ba so' dufdng thoa man x + y + z = Si dung IWng gidc de gidi cdc bdi loan sau: pai21:Choa,b,c Dilng phiTdng phdp chieu bie'n thien ham so gidi cdc bai loan sau: Bhi 11: Cho X > y > ChiJng minh — X-1 In>4 1-x e\R a-c Chtfng minh Bai 9: Cho x, y, z la ba so diTdng va x + y + z > ChiJng minh: ^ + X + _ L > w vz Vx In^:^ 1-y Vl + a^ Vl + c^ ~ Vl + a^Vl + b^ y[\ 3S Bai giang so 8: Dap so: 16 Bai 74; Cho < X < va < y < Tun gia tri Idn nhat va nho nhat cija bieu thtfc: 2^ X y +- + y^ x'* y X y^ V _, 4249 „ 1083 Dap so: max P = ; mm P = 16 54 Bai IS: Tim gid tri Idn nha't v^ nho nhat cua h^m s6' f ( x ) = V3 + x + V - x - V l + x - x ^ v d i - < x < — Vl + b^Vl + c^ ' si dung phUffng phdp mien gid tri ham so gidi cdc bdi todn sau: , ^ • x + 2y + l ^ Bdi 22: Cho x, y e IR ChiTng mmh - — < Z = x - y i => Z Z = (x + y i ) ( x - y i ) = x ^ + y ^ • N e u b = 0, thay vao (1) ta c6: a^ = => a = • N e u a = - - , thay vao (1) ta co : b" = - => b = ± ^ V a y CO ba so phOrc can t i m : Z i = ; Z j = - Thidi^ I: (De thi tuyen sink D^i hgc khoi A - (a + bi)" = a"^ + b ' + a - bi < Tinh cac bieu thiJc c6 chtfa modun ciia so phiJc - A-2011) D a l z = a+bi => z = a - bi va |z|^ = a^ + b^ - ve modun Giai 2'' Taco I Z - ( + i ) l = ,j(x +2 f + ( y - l ) ^ i Tilf Iheo b a i ta co he phU'dng Irinh: (x-2)'+(y-l)'=10 ^.j • x = 3;y = j x + 2y = 10 IZI = -JyF+y^, o ( x + yi)^ + ^/x^"+y^ = o fx = 2xy (2)' l Z - i l + IZ + 4il = 10 o M A + M B = ( * ) x^ y^ PhiTdng trinh ciia elip c6 dang: — + — = 25 (x^ + y ^ j + 7x^ + y ^ + x y i = 0) cdc so'phiJc - i va i K h i d : tieu diem la A; B va true Idn b^ng 10 Z^+ IZI = x^-y2+7x^+y^ = ^' Tir (*) suy M n ^ m tren clip c6 hai Giai { MA = MB jj^ng trifc ciia doan AB so phuTc Z = X + y i n i m tren du'dng tron t a m Ta c6 the giai bai loin ' K h i d o (1) Ttf (2) suy tap hdp nhi?ng diem M bieu dien so phiJc Z n^m tren duTcfng ( x - f + ( y + 4)^ = Tif (1) suy cdc d i e m M ( x ; y) hiiu Thidv (1) s (1-2; 0), B la d i e m bieu d i e n so I (tiJc la B = (0; 1) Tim tap hdp cac d i e m bieu dien so phtfc Z biet rang IZ - (3 - 4i)l = Gia sur Z = IZ - (-2)1 = IZ - il Qoi M la d i e m bieu dien so' phuTc Z, A la d i e m bieu d i e n so' - (tiJc la (3) -y^+lyl=0 (4) y =0 (5) x^+lxNO (6) x = O;y = G i a i he (3) (4) va (5) (6) (de dang) ta d i den h? (1) (2) c6 nghiem x - ( > , y = ' ' Thidvi 6: Tim tap hdp cac d i e m mSt p h i n g phtfc bieu dien cac so phiJc Z * a man dieu k i e n sau: 2IZ - il = IZ - Z + 2il Giai Dat Z = X + y i , thi 2IZ - il = IZ - Z + 2il 2lx + ( y - l ) i l = l ( y + l ) i l ^>/x'+(y-i)'=>/(y^ tox^+y^- 2y+ 1= y^+2y+ o y = _x=a,y = l ' V§y ba so phtfc can tim Z = 0; Z = i ; Z = -i Thi dif 5: Tim tap hdp cic d i e m bieu dien so phiJc Z neu nhiT thoa man mO' cic dieu k i e n sau: 1/ IZ + 21 = li - Z l 2/ IZ - 4il + IZ + 4il = 10 V $ y tap hdp nhiyng d i e m M Ik parabol y = ' ^hqn xet: V d i c^c thi du 3, 4, nen suf dung bieu dien hinh hoc cua so phiJc C6n t h i du 6, t h i sijf dung b^ng phep tinh ve modun -f-5 rinil qiirfq J;;"" • , " 165 § DANG LUQNG GIAC CUA S O PHUC z -1 =1 Z- i Thidu 7: Tim so phiJc thoa man he: Z-3i Z +i J X6m t ^ t l i thuyet jsjc'u la mot acgumcn cija so phtfc z thi moi acgumen cua no c6 dang F z = cp+k2 7t, k e Z Giai ">.*."jt'^'''a' I>Jeu r > la modun cua so' phufc z, (p la mot acgumen cua no, thi Vdi hai so Z, Z ' (Z' ^ 0) thi z = — - , do he da cho liTOng u^ng vdi: Z' I ^ I |Z->l = | Z - i | 0) |Z-3i| = |Z-(-i)| (2) :v v "'^Jj z = r(cos(p + isin cp) la dang lUdng giac cua sophuTc z Neu Z| = ri(cos cp I + i sincpi), r, > Z2= r2(cos (p2 + i sin(p2), r2>0, MC = MD (4) £ 'ti'' 2, Cac dang bai t § p cd ban i?ng bieu dien cac so phuTc 3i va - i thi: (3) Nhin chung cac loai bai tsip c6 each giai chung nhU sau: Giii su" phai tim mot acgumen cua so' phiJc z Ta can bien doi cho z cd dang: z = r(cos (p + isin (p) vdi r > De tha'y y = x va y = tU"cJng ufng la hai dU'dng trung triTc cua AB va CD Vay Z = + i la so phiJc can tim Thidv 8: Tim so phiJc thoa man h0 ^ Z - 12 z - 8i Z-4 Z+8 Khi la m^t acgumen cua z Thidu 1: Cho so phuTc Z = - sin cp + i cos cp (0 < (p < - ) Tim mot acgumen cua so phiJc Z Giai = Tacd: Z = -sincp + isin(p = 1-cos Giai D a t Z = x + yi do: Z +8 o y - y + 136 = (1) = sin = sin 'n (p^ 2) K _ (p^ o + 2isin y = 17 y = Nhun xet: Hai thi du va cung loai nhiTng ta dung hai phi/dng phap giSi thich hdp khac (phifdng phap bieu dien hinh hoc cua so phiJc vdi thi du 7, phu'dng phap tinh toan modun vdi thi du 8) Do < (p < sm n + isin 71 (p ^7t_(p' cos U"2j (p 4"2 n 4- icos - 2j n cos - + — + i s i n — + — v4 u 4~2, Vay CO hai so phiJc can tim: Z = + 8i va Z = + 17i 166 (p 4~2 o 1-6 + yil = -16 + (y - 8)il (do x = 6) L n = sin = lx-4 + yil = lx - + yil c:>(x-4)' + y = ( x - ) ' + y ' c ^ x = z - 12 = ^ » l x - 12 + y i l = | l + ( y - ) i l z - 8i « + y2 = ^^ 36 + ( y - f " -.1 i ; VVV / Loai 1: Cac bai toan xac dinh acgumen cua so phiJc Tif (3), (4) suy M la giao diem cua du'ilng trung triTc cua AB va dU'dng trung triTc cua CD Z-4 V2 Cong thiJc Moivre: Neu z = r(cos(p + isincp), r > thi z" =r"(cosn(p + isinn(p) cac so va i ; C (0; 3) va D = (0; - ) tiTdng M A = MB mtf-,/ Z2 A = (1; 0), B = (0; 1) tiTdng i^ng bieu dien o X ' A " - thi Z i Z = rir2[cos((p, +(p2) + isin((P| - ) ] ; — =—[cos((P| - ( p ) + i.sin((P| -(p^)' Gpi M la diem bieu dien so' phufc Z, (1),(2) ^ n cp ^ • 2sin ~ > Vay tir(l) suy ™ O vM -'^i; + ^ la mot acgumen ciia Z ^Itidii 2: Cho so phuTc Z c6 modun b^ng v^ (p l i mpt acgumen ciia nd Z 1/Tim mpt acgumen cua so phiiTc — 2/ Tim mpt acgumen ciaa so phuTc Z + Z neu cos (f> ^0 167 Giai fhid^ 2: T i m so phufc cua Z cho Tflr gia thiet suy ra: Z = cos cp + isin (p ,, „ Z cos9-isin(p cos(-cp) + i s i n ( - ( p ) • •r ^ ^ 1/ T a c o — = - = — ^ = cos(-2(p) + i s i n ( - ( p ) Z cos(p + isin(p cos(p + isin(p ' + J N e u (p>0, k h i Z + Z = c o s - l = 2cos(p(cos0 + isinO) ' + N e u (p - Thuoc cac cong ihiJc nhan, chia va cong IhiJc Moivre doi vdi so' phiJc dudi dang lu'dng giac Thidit 1: Tim phan ihiTc va phan iio ciia moi so phiifc sau: *- 1/ Z, = 2/Z, = 71 COS — 71 1/ Taco: cos—+ ism— V3 + i = 4 l + i=>/2 = 2 Thco cong ihiJe Moivrc la c6: Zi = (l + i y (^/2)'^(cos37I + isin37l) cos 371 + isin ] 371 71 TT^ cos—+ isin— 6 cos— + isin — 2-^ 2 Tac6:- = cos + = COS 7t = COS — isin 7t 20107: = cos ( —n^ 7t f — cos — -isin — = cos + isin J 3 I 3y 37r r 371^ V3.1 V3i i'' = - i = cos = — H Tuf Ihco cong thufc Moivre la c6: 5n\ +1 ( 5nsin Z = cos — cos 371^ Z, 571 > f 5n^ 170 + isin 71 ( 3, + i sin \ 7t ' 3; + — \ — 3j (1) \ -isin—= cos n — isin 3J 20\()n\ (2) = 2cos(6707t) = ,2010 n — ' 71 71 r72()|{) 3J ,2010 + , -1 =2 7+i 4-3i; ( yK \ 25 • + —^1 42 ' = 71 sin71— cos—+ 3 In cos + isin > I 6; Vay phan thifc cua Z^ la -2^\yj3 = -64\/3 , phan ao cija Z2 la -64 = ' 201071 = cos I 2010 ' I 3, => —= c o s - + i s i n - => Z Tom lai ta luon cd : Z^'"" + -^TTT = 4-3i In + isin — 2' c o s — + isin 3 3J Thco quy t^c nhan so phiJc diTdi dang lu'dng giac la c6: ( / Giai 7t ; 1+ + isin + isin V + isin _1_V3 Tim m nguyen du'dng de Z 1^ so ihi/c; li so ao V§y phan thifc cua Z\a 0, phan ao cua no la — 7t • n 2 r/»' m = 4k + 2, vdi k = 0, 1, : ^hidi^ (De thi tuyen sink Dqi hgc kiwi li - 2011) Tim phan ihi/c va phan ao cua so phuTc: z = i + iV^^' 1+i V.)f! 171 Cty IMHH M I V DVVH Khang Viet S.'] Giai 3i V COS —3 + isin —3 J = Tac6: \ iS = —+— Theocong thdrc Moivze, thi (l + iVs)"^ =:8(cos7t + isin7i) (1) 2 ^ 7t 71^ Laic6: l + i = V2 cos—+ isin— 4 371 371 Theo cong thufc Moivre, Ihl: (l + i)^ = 2>y2 (2) cos 1- isin — 71 71 = 2V24 = + 2i Tir(l), (2) suy ra: z = 2V^ cos—+ Sin — , 4; Vay phan thiTc cua z la 2, phan cua z la § GIAI PHLTONG TRJNH TRgN T A P S O PHLTC Trong muc se x6t viec giai phifdng Irinh an so cua moi phUdng trinh la mot so phiJc Z De giai du'dc phifdng trinh tren tap so phiJc can n^m vifng cdc kie'n thiJc sau: 1/ Bie't each khai can bac hai cua mot so phtfc Z = a + bi 21 Biet each giai phifdng trinh bac hai ax^ + bx + c = 0, d6 cac h$ so a, b, c n6i ehung la cac so phifc 3/ Thanh thao each giai phifdng trinh va he phifdng tnnh hffu ti Can nhd rhng viec giai phu^dng trinh tren tap so phuTc ve cd ban giong nhir giai phUdng Irinh tren tap so thifc, chi c6 cai khac la phep tinh d day la cac phep tinh cpng, trijf, nhan, chia va khai can so phiJc Cac dang bai tap cd ban / Loaf 1: Giai phu'dng trinh bac hai tren tap cac so phiJc: Phu'dng phdp giai: SuT dung cong thifc nghi^m nhif cong thtfc nghiem doi vdi phu'dng trinh bac hai tren tap cac so thifc Thi du 1: (De thi tuyen sinh Cao dang khoi A,B- 2009) 4Z 7i Giai phifdng tnnh sau tren tap so phifc: Z - i = Z - 2i Giai Vdi dieu kien Z ^ i , phifdng trinh da cho tifdng difdng vdi phifdng trinh sau: 4Z - - 7i = (Z - i)(Z - 2i) o Z ^ - (3i + 4)Z + + 7i = (1) Ta CO A = (31 + 4)^-4(1 + 7i) = - i Gia sijf Z = X + yi (X, y e R ) cSn bac hai cua A d6 ta c6 he sau di xac 'x-2;y =-l dinh X, y: x = -2;y = l 2xy-4 «» Vay(l) +4+(2-l) = 3+i Vay Z|, Z2 la nghiem cua he phifdng trinh da cho '"'^ fhidu 2: Giai phifdng trinh tren tap so phifc: (l - i)^ - 2(l + 2i) Z - = Giai ' f ' ' ! - • r Do - i ^ O n e n t a c d : ( l - i ) Z ^ - ( l + i ) Z - - Z ^ - ^ ! ^ Z - ( l + i) = o Z ^ - ( l + 2i)(l + i ) Z - ( l + i) = o Z ^ + ( l - i ) Z - ( l + i) = (1) Ta cd: A = (1-31)^+8(1 + i) = 2i Goi X + yi (x, y e IR) la can bac hai cua 2i ta cd „ Z, = i - l + l + i 2i x = l;y = I Vay (1) o xy = X:3-l;y=._l Z2 = i - l - l - i = - l + i Thidu 3: Giai phifdng trinh sau tren tap so phufc: (2-3i)Z^ +(4i-3)Z + l - i = Giai Ta cd: (2 - i ) + (4i - ) +1 - i = -/ Vay phu'dng trinh da cho cd nghiem Z|=l va nghiem ! z,=^=i::i=-L(i-i)(2-3i) = -li^ -1^ ' a - i 13^ ^ 13 Nhgn xet: - Cac quy lac nham nghiem va dinh li Viet van dung trufdng hdp xet phifdng trinh bac hai tren tap so phuTc - Xet thi du sau: Xet phu'dng trinh bac hai d thi du Z ' + ( l - i ) Z - ( + i ) = (*) Tacd: 2i + (-1 + i) = 3i - o 2i(-l + i) =-2(1 + i) Vay theo dinn ly Vi-t suy ra: (*) cd hai nghigm Z, = 2i va Z2 = - + iTa cd each khac giai thi du Loai 2: Phifdng trinh quy ve bac hai va he phu'dng trinh: , Thidif 1: Giai phifdng trinh vdi Z la so phufc: (z^ + z)% 4(z^ + z) - - 0, Giai Datt = Z^+Z, tacd: t- + t - = o 17 Cty TIMHH MTV DVVH Khang Viet Phuong phap giii Toan trpnfl tarn - Phan Huy Kh^i Z| V a y phiTcing trinh da cho tiTdng di/cJng v d i : V a y d i f a he da cho v e dang imJng dUciiig: „ + Z+6 = -1+^23 i ^ ^1 - Z ' + Z - = () ; ^2 - Z3 =1;Z4 _-l-V23i z — ( t 'i V aV sS'phiJc): j ^ ^ ^ t n ^ i f l g o si t'-(4 + i)t + + i = () du : G i a i phi/(' 10 *-|()0 156200 Tirdo: P(B)= Thidu " =10 Co hai each chon thoa man yeu cau tren - Hoae la ca hai ta'm the mang so' chan ''^1 co so chan khoang tif den 9, nen so'each chon d kha nang la: C = ' 1/ Gieo dong thdi hai xuc s^c T i n h xac suat de: a/ T o n g so cham xuat hien tren hai la • Giai 1/ G o i Q la tap hdp tat ca cac kha nang xay O day co hai xuc s^c, m o i CO kha nang xuat hien, vay IQI = 6.6 = 36 a/ G o i A lii bie'n co' " t o n g cac cha'm xuat hien tren la " Ho2ic la chon mot ta'm the mang so c h a i , mpt ta'm the mang so le Theo quy Tir theo quy tac cong ta co: IQAI = +20 = 26 2/ Gieo dong thdi ba xuc sac T i n h xac suat de tdng so' cha'm xuat hien tren ba la 10 , • t3'c nhan so each chon la: c ! j C = b/ So cha'm xuat hien tren hai hdn k e m 178 •' Theo dinh nghla xac suat suy ra: P ( A ) = Q 26 13 Q 36 18 'Fhidu 5: Co 30 ta'm the danh so t i f den 30 Chon ngau nhien 10 ta'm th T i m xac suat de co tam the mang so l e , ta'm the mang so chan chi Co dung ta'm the mang so' chia he't cho 10 •J 179 ... nhU" sau: 11 11 11 11 = (15 ;11 ; -17 ) n = MN,u, V -1 2 3 -1 R6 rang (P) di qua M (1; - ; -1) nen (P) co phiTdng trinh: 15 (x- 1) + l l ( y + ) - 17 (z+ l) = o 15 x + l l y - 17 z -10 =0 Thi du 5: (De... (12 ; 0; 10 ) TCrdd: MN = (11 ;2 ;11 ) Mat phang (P) chuTa (d,) va (dj) nhan cap vectd MN = (l 1; 2 ;11 ) va u, =(3;-l;2) lam cap vectd chi phiTdng nen vectd phap n cua (P) xac dinh nhU" sau: 11 11 11 ... phap cua ( Q ) lam eSp veetd chi phiTdng -1 3 2 -1 = ( - ; ; ) / / ( - l ; l ; l ) ['' ; -1 -1 1 - Vay vcctd phap tuyen n cua (P) 1^ : r - — 1 -1 -1 = {4;3 ;1) n= -2 / V -2 2 Mat khac (P) chtfa (d,)