Học và ôn luyện theo cấu trúc đề thi môn toán

510.76 rS VO THE HirU - NGUYEN VINH CAN H419V • • HOC a ON LUYIN T H E O C A U T R U C D E THI MON I TS VU THE HlfU - NGUYEN VINH CAN nioc & ON LUYEN T H E O C A U T R U C D E THI THi; VIEN TINH BINH THU*N ON THI DAI HOC Ha NQI N H A X U A T B A N D A I HQC QUOC G I A H A N O I H O C vA NHA XUAT B A N DAI HQC QUOC GIA HA NQI 16 Hang Chuoi - Hai Ba TrUng - Ha Npi Dien thoai: Bien tap - Ciie ban: (04) 39714896; Hanh ctiinii: (04)3 9714899; Tong Bien tap: (04) 39714897 Fax: (04) 39714899 Chiu trdch nhiem xuat ban: Gidm doc - Tong bien tap: T S P H A M T H I T R A M Nha sach H O N G A N Che ban: THAI VAN Sica bai: H O N G SON Bien tap: Trinh bay bia: THAI HOC Thj^c hi?n lien kit: Nha s a c h H O N G A N SACH LIEN KET O NLUYEN THEO CAU TRUC D ET H I M O N TOAN THPT IVla so: 1L - 65DH2013 In 2.000 cuon, I h 17 x 24cm tai Cong ti Co phan Van hoa VSn Lang - TP H6 Chi IVIinh ) t r o n g eo d i i n g k d i e m n S m t r e n m o t d i i d n g t h S n g (3 < k < n ) H o i c6 bao n h i e u t a m g i a c n h a n cac d i e m d a cho l a d i n h CHIDAN Cuf d i e m k h o n g t h S n g h a n g t a o t h a n h m o t t a m g i a c So cac t a p h o p c o n d i e m t r o n g n d i e m l a : C^ So cac t a p c o n d i e m t r o n g k d i e m t r e n diTcfng t h i n g l a : C^ So t a m g i a c c6 d i n h l a cac d i e m d a cho l a : N = C^ - Cl t a m g i a c ; TS Vu The Hi/u - Nguyen VTnh Can a) Co b a o nhieu so t i i n h i e n l a so chan c6 chiif so doi m o t khac va chuf so dau t i e n la chOf so le b) Co bao nhieu so t i i n h i e n c6 chuf so doi mot khac nhau, c6 dung chuf so le, chuf so chSn (chuf so dau t i e n phai khac 0) CHI D A N a) So can t i m c6 d a n g : x = a^agaga^agag t r o n g a i , ae l a y cac chOf so 0, 1, 2, 8, vdfi a i ?i 0, aj v d i < i ?i j < - V i X la so chSn nen ae c6 each chon tiT cac chuT so 0, 2, 4, 6, - V i a i la chuT so le nen c6 each chon tiT cac chuf so 1, 3, 5, 7, Con l a i a2a3a4a5 l a m o t chinh hop chap eiia chuf so l a i s a u k h i da chon ae va a i Theo q u y t^c n h a n , so cac so can xac d i n h l a : N i = S.S-Ag = 5.5.8.7.6.5 = 42000 so b) M o t so theo yeu c a u de b a i gom chuf so tii tap X i = |0; 2; 4; 6; 81 va chuf so tCr t a p hop X2 = I I ; 3; 5; 7; 91 ghep l a i va loai d i cac day chuf so CO chuf so dufng dau So each lay chuf so thuoc t a p X i la : Ci? = 10 each So each lay p h a n tuf thuoc X2 l a : Cg = 10 each So' each ghep p h a n tuf l a y txi X i v o i p h a n tuf l a y tii X2 l a : C^C^ = 10.10 = 100 each So' day so' eo thuf t i f eiia p h a n tuf diioc ghep l a i l a : 100.6! = 72000 day Cac day so c6 chuf so a dau g o m chiJ so khac ciia X i va chuf so' ciia X2 : So cac day so nhif t r e n la : C C ! = 7200 day So cac so theo yeu cau de b a i la : N2 - C ^ C ^ ! - C ^ C ^ ! = 72000 - 7200 = 64800 so M o t hop diing v i e n b i do, v i e n h i t r a n g va v i e n b i vang NgLfofi ta chon r a v i e n b i t i f hop H o i c6 bao nhieu each l a y de t r o n g so b i j a y r a k h o n g dii ca mau CHI D A N Cdch : So each chon v i e n b i k h o n g d u mau b a n g so' each chon v i e n b a t k i trir d i so each chon v i e n c6 ca mau N = Cjg - (C^ C^ C^ + C^ C\ + Cl C\) = 645 each Cdch : So each chon v i e n b i k h o n g d u m a u bang so each chon v i e n m o t m a u (4 do, t r a n g va vang) cong v d i so each chon v i e n hai mau ( do, t r a n g hoae do, t r a n g hoac do, t r a n g hoae do, v a n g hoac do, v a n g hoae do, v a n g hoae t r S n g , v a n g hoae trSng, v a n g hoac t r a n g , vang) N = c : +Ct +CI+ ClCl + ClCl + C^C^ + + C^C^ = 645 caeh Hoc va on luyen theo C T D T m o n loan T H P T SI 9 Co 15 n a m va 15 nuT k h a c h du l i c h dijfng t h a n h vong t r o n quanh ngon lijfa t r a i H o i c6 bao n h i e u each xep de k h o n g eo triTcfng hop hai ngi/6i eCing gidfi canh CHI DAN ThiTe h i e n sap xep bang each d a n h so 30 cho t r e n di/orng t r o n tii den 30 va cho n a m dufng so le nuT dufng cho so chSn hoac ngi/gc l a i (2 each) Co 15! each sSp n a m dufng cae cho so' le (hoac chSn) va 15! each sdp nuf dufng t r o n g cae cho so ch^n (hoac le) V i diidng t r o n 30 cho nen m o i each sSp xep nao xoay tua 30 cho theo dung t r a t tiT ta cung chi eo mot each sap t r e n dirofng t r o n (xem b a i so 4) Do so each sSp xep theo difcfng t r o n 30 k h a c h du l i c h theo yeu cau 2.(15!)(15!) , , de la : N = = 14!.15! each 30 10 Chufng m i n h cae dang thufc : a) + + + = ——- (1) t r o n g A^ la c h i n h hop chap eua n Ag A3 A„ n b) CHI C;; = C;;:; + Cl;}^ + + Cl:\) t r o n g C; la to hop chap r ciia n DAN a) V(Ji k e N, k > ta c6 : A', = k ( k - 1) ^ = = - ' ^ A^ k(k-l) k-1 k Thay k = 2, 3, n vao (*) ta c6 ve t r a i ciia (1) la : (1 (I V f 1^ — — — — — — — + + [n-1 nj u l 3v b) Theo t i n h chat eua to hop ta c6 : = Cn_3 (*) + C^^g Cong ve vdi ve cae dang thufc t r e n ta diTOe : c:;-c::;+c-^3+c::u +c:-uc: Do C;; = C;::} = l n e n thay C;: d dang thufc cuoi bori C^:} ta dugfc dang thufc (2) can chufng m i n h 11 Chufng m i n h bat dang thile : t r o n g k e N, k < 2000, C^ooi + ^ C\Z + CfZ la to hop chap k eua n p h a n tuf t4l TS Vu The Hi/u - Nguyin Vinh CSn T i m d i e m d o i xufng c i i a m p t d i e m q u a m p t m a t p h a n g De t i m d i e m d o i xufng ciia d i e m M q u a m a t p h a n g ( P ) , t a l a m n h i i sau: - T i m t o a d o h i n h chie'u H ciia d i e m M t r e n m a t p h S n g (P) - H S u d u n g h e thuTc vectcf M M ' ^ M H hoac c o n g thufc t o a t r u n g d i e m de s u y r a t o a d i e m M ' , d o i xiifng vdi d i e m M q u a M' m a t p h a n g (P) 11; D O I V l D U d W G THi T i m h i n h c h i e u c i i a d i e m t r e n dii'ofng t h a n g De t i m hinh chieu ciia diem M t r e n diTofng t h a n g A, t a l a m nhii Viet phirong t r i n h sau: m a t p h ^ n g (P) d i qua M v a v u o n g goc v d i A Tim t o a giao d i e m H ciia di/dng t h a n g A va m a t phSng (P) T i m d i e m d o i x i i n g c i i a m p t d i e m q u a m p t ditofng thang De t i m d i e m M ' d o i xufng ciia d i e m M qua difofng t h i n g A, t a l a m n h i f sau: T i m t o a h i n h c h i e u H ciia d i e m M t r e n diTorng t h S n g A S L T d u n g he thOfc vecto M M ' ^ M H hoac c o n g thufc t o a t r u n g d i e m d e s u y r a t o a d i e m M T i m hinh chie'u song song vdi phu'dng ciia dvfcfng thang m p t dvfoTng thang A tren m p t mat phang ( P ) De t i m hinh chieu song song theo p h u o n g ciia diTofng t h a n g d , ciia d i / d n g t h a n g A t r e n m a t phSng (P), t a l a m nhiT sau: V i e t phifcfng t r i n h m a t p h a n g ( Q ) d i qua A v a s o n g s o n g vdfi diTofng t h a n g d V i e t phifcfng t r i n h difofng t h a n g A', giao t u y e n ciia h a i m a t p h a n g (Q) v a ( P ) Hoc va on luyen theo CTDT mon Toan THPT : /; 283 Tim hinh chieu (vuong goc) cua diiifng thang tren mat phang De t i m h i n h chieu cua diidng t h a n g A t r e n m a t p h a n g (P), t a l a m n h i / sau: V i e t phiTcfng t r i n h - V i e t phiicfng t r i n h m a t phSng (Q) d i qua A va vuong goc vdri m a t phang (P) - diicfng thSng A', giao t u y e n ciia h a i m a t phSng (P) va (Q) Chii y: Co t h e coi triiofng hgfp la m o t t r i / d n g h o p dac biet cua trU'dng h o p t r e n day Ili.CAC KHOANG CACH K h o a n g each gii?a h a i diem K h o a n g each giaa h a i d i e m A ( X A ; yA; ZA), B ( X B ; ys; ZB): A B = IABI = ^ix^-x^f+{y^-yj'+{z^-zj' K h o a n g each tijf mpt diem d e n mpt mat phang Khoang each tii diem Mo(xo; yo; Zo) den mp (P): Ax + By + Cz + D = la + By^ + CZQ + D AXQ d ( M , (P)) = VA^TB^TC^ K h o a n g each tijf mpt diem den mpt di^cfng thang K h o a n g each tii m o t d i e m M den diiofng t h a n g A, d i qua diem Mo va CO vectcf chi phifong u l a : d ( M , A) = ll^o-^' t r o n g [ M ( , M , u] la vecto t i c h c6 hLfdrng ciia h a i vectcf MQM va u K h o a n g each giffa h a i dvfcfng thang cheo n h a u Cho h a i dirofng t h a n g A, A' cheo A' d i qua d i e m M ' va c6 vectof chi phiTofng u' - A d i qua M va c6 vectcf chi phuong u - K h o a n g each giaa h a i dudng thSng A, A' l a : d(A, A') = [u, u ' ] M M ' [u, u'] T r o n g do: [ u , u'] l a vectcf t i c h c6 hiTdng ciia u, u' [ u , u ] M M ' la t i c h v6 hi/dfng ciia h a i vectcf [u, u'l va MM' [u, u'] la d a i eiia vector [u, u'] 284 TS Vu The' Huu - Nguyln VTnh Can [u, u ' ] M M ' l a gia t r i tuyet doi cua t i c l i v6 hifdng cua h a i vecto [u, u'] va M M ' Chii y: Ngoai r a t a can nh6 m o t so k i e n thufc ve k h o a n g each t r o n g m o n H i n h K h o n g gian Idp 11: Khoang each giOfa diibng thang A den m a t phang (P) song song vdfi A: (A // (P)) t h i bang khoang each tU mot diem M e A den m a t phang (P) Khoang each giOa hai mat phang song song (P) // (Q) t h i bkng khoang each t i i mot diem M thupc mat phang den mat phang Khoang each giOfa h a i diidng t h a n g song song t h i bang k h o a n g each tif m o t d i e m thuoc diTcrng t h a n g den diicfng t h a n g k i a BAI TAP 130 T i m h i n h ehieu cua d i e m M ( l ; - ; 2) t r e n m a t phang (P): 2x - y + 3z + = CHi DAN M a t phang (P) c6 vectof phap tuyen n = (2; - ; 3) Diiofng t h a n g A d i qua M ( l ; - ; 2) va vuong goe vdi (P) n h a n vecto n l a m m o t vectcf chi (x = l + 2t phifong, do, phifong t r i n h ciia A l a : A: d(M, A) = MHI = |z = V(2 + D ' + (-3 + 3)2 + (3 - 2f d ( M , A) = / l O Cdch A CO vector chi phiromg u = ( ; 2; - ) M a t phSng (P) d i qua M ( - l ; - ; 2) va vuong goc vdri A n h a n u l a m m o t vecto phap tuyen Phifcfng t r i n h m a t phSng (P) l a : l ( x + 1) + 2(y + 3) - 3(z - 2) = X + 2y - 3z + 13 = T h a m so t ufng vdri giao diem ciia A va (P) l a nghiem ciia phi/ong t r i n h (3 + t ) + 2(-l + t ) - ( - t ) + 13 = ^ t = - TCr day t a c6 H ( ; - ; 3) va M H = VlO 286 T S Vu The Huu - Nguyin Vinh CSn Chu y: De t i n h M H , ta c6 the suf dung cong thufc M H = d ( M , A) = [MoM, u] Vdi Mo(3; - ; 0), M ( - l ; - ; 2) ^ M ^ M = (-4; - ; 2) Ta t i n h di/oc [M ^M , u] = (2; - ; 6) |[MoM, ul! = Vlio , iui = V l =^ M H = d ( M , A) = Vl40 Vl4 = Vio b) H l a t r u n g d i e m cua M M ' cho ta M M ' = M H M ( - l ; - ; j , H ( ; - ; 3)=^ M H = (3; 0; 1) M'(x; y; z), M ( - l ; - ; 2) => M M ' = (x + 1; y + 3; z - 2) x + = 2.3 M M ' = M H o y + = 2.0 z - = 2.1 X = o y - - => M'(5; - ; 4) z = Chu y: Co the siTf dung cong thijfc cho toa dp trung diem doan thang x - 4t Cho diTong t h a n g A: y = 7t z = + 2t va m a t phSng (P): x - y + z + = 133 T i m phurong t r i n h h i n h chieu vuong goc ciia A tren mat phang (P) CHI DAN H i n h chieu A' ciia A t r e n m a t phang (P) l a giao t u y e n cua h a i m a t phang: m p (P) va m p (Q) d i qua A va vuong goc v(Ji (P) Difcfng t h a n g A c6 vectcf chi phiTcfng u = (4; 7; 2) M a t phang (P) c6 vecto phap tuyen n = ( ; - ; 1) M a t phang (Q) chufa A, d i qua d i e m Mo(0; 0; 3) va c6 vecto phap t u y e n q = [u,H] = ( 1 ; - ; - ) Phirong t r i n h m a t p h a n g (Q): l l x - 2y - 15z + 45 = Ducfng thang A' la giao tuyen ciia (P) va (Q) nen vector chi phuomg ciia A' l a u' = [u,q] Ta t i n h diiOc u' = (-28; 4; 20) = 4(-7; 1; 5) Ta chpn vectof (-7; 1; 5) l a m vecto chi phifcfng ciia A' Hoc va on luyen theo CTDT mon Toan THPT ' 11 287 De t h a y d i e m M 0;i5;r X = -7t thuQC A' Phiiorng trinh A': y = — + t ^ z = - + 5t Chiiy: B a i toan gom h a i b a i toan cor ban: - V i e t phi/dng t r i n h m a t phSng d i qua m o t difofng t h a n g va vuong goc vdri m o t m a t phang - V i e t phi/ong t r i n h giao t u y e n cua h a i m a t phang B a i toan v i e t phuong t r i n h h i n h chieu A' ciia diiorng thSng A t r e n mat p h a n g (P) theo phi/Ong ciia m o t difcfng thSng d cho triTdfc cung gom hai b a i toan cO b a n : - V i e t phiiong t r i n h m a t p h a n g (Q) chufa m o t di/ofng t h a n g A va song song vdfi dirofng t h i n g d ( A va d cheo nhau) - V i e t phuong t r i n h giao t u y e n A ciia (P) va (Q) CAC CAU HOI TlJ ON TAP VE CACH VIET PHlJCfNG TRINH MAT PHANG VA DlJClNG THANG H a y neu cac phifong phap g i a i cac dang bai toan sau : Cho ba d i e m A , B, C a) Chufng m i n h ba d i e m A, B, C k h o n g t h a n g hang b) V i e t phuong t r i n h m a t p h a n g (ABC) Cho d i e m A va dudrng t h a n g A a) Chufng m i n h A g A b) V i e t phiiong t r i n h m a t p h a n g (A, A), xac d i n h hdi A va A Cho h a i duc^ng t h a n g A, A' a) Chufng m i n h A, A' cat T i m toa giao diem b) V i e t phircfng t r i n h m a t p h a n g xac d i n h bdfi A, A' Chufng m i n h h a i difcfng t h a n g song song v d i va viet phiicfng t r i n h m a t p h a n g chufa h a i difcfng t h a n g song song Chufng m i n h h a i di/6ng t h a n g cimg thuoc m o t m a t phang H D : Co the chufng m i n h chung cat hoac song song vdri 288 H TS VD The Hyu - Nguyin Vinh CSn V i e t phLTorng t r i n h m a t p h d n g d i qua m o t d i e m va song song vdri m o t m a t p h a n g cho triTdfc Cho h a i diTorng thSng A va A' a) Chufng m i n h A va A' l a h a i duomg t h a n g cheo b) V i e t phirong t r i n h m a t p h a n g (P) t r o n g cac trirofng hgrp : - (P) chijfa mot di/cfng thang va song song vdi diromg thang lai - (P) di qua mot diem M cho tnidc va song song vdfi ca hai dUcfng thang Cho di/orng t h a n g A va m o t diem M V i e t phifofng t r i n h m a t phSng (P) qua M va vuong goc v6i A Cho dirdng thSng A va m a t phSng (P) V i e t phuong t r i n h m a t p h a n g (Q) d i qua A va vuong goc vdfi m a t p h a n g (P) 10 Cho h a i m a t phang (P), (Q) va m o t diem M Viet phifong t r i n h m a t phSng (R) di qua M va vuong goc vdi ca hai m a t phSng (P), (Q) H.D : Chia b a i toan t h a n h h a i bai toan cor b a n : + V i e t phiTcfng t r i n h giao tuyen A cua (P) va (Q) + V i e t phuofng t r i n h m a t p h a n g (R) d i qua M va vuong goc v6i A 11 Viet phiiomg t r i n h t h a m so va phiTcfng t r i n h chinh tac ciia diiotng t h a n g A di qua diem Mo(xo; yo; ZQ) va c6 vectcf chi phiforng u = (ai; a2; as) 12 V i e t phirorng t r i n h di/orng t h a n g d i qua m o t d i e m va song song v6i m o t diidng t h d n g cho trLfdc 13 V i e t phiicfn^ t r i n h diidng thSng d i qua m o t d i e m va vuong goc v d i m o t m a t phang «ho triidfc 14 V i e t phifong t r i n h diidng thSng d i qua m o t diem va vuong goc va cSt mot diicfng t h a n g cho trifdic 15 V i e t phirorng t r i n h giao tuyen ciia hai m a t phang 16 Chijfng m i n h diTofng thang va m a t phang cat T i m toa giao diem 17 T i m toa h i n h chieu cua m o t d i e m t r e n m o t m a t phang 18 T i m toa d i e m doi xufng v d i d i e m M qua m a t p h a n g (P) 19 T i m t o a h i n h chieu cua m o t d i e m t r e n m o t difcfng t h a n g T i n h khoang each tiT m o t d i e m den m o t difcfng t h a n g 20 T i m toa diem doi xufng v6i mot diem qua mot difofng t h a n g cho trifdfc Hpc va on luyen theo CTDT mon Toan THPT U: 289 C y E fE I MT ClVBMHTPHiG H Y Nl VI A BAITOAN18 PHl/dNG TRINH MAT CAU Phifcfng t r i n h m a t c a u a) M a t cau tarn I { a ; b; c), b a n k i n h r c6 phuofng t r i n h (x - a)^ + (y - b ) ' + (z - c)^ = r ' (1) b) Phi/ofng t r i n h bac h a i dang x' + Vdi + + + + 2Ax + 2By + 2Cz + D = (2) - D > la phuorng t r i n h ciia mot m a t cau t a m I ( - A ; - B; - C ) va b a n k i n h r = V A - ^ + B ' + C - D c) V i t r i tiforng doi giiia m a t cau va m a t phang Cho m a t cau t a m I , b a n k i n h r va m a t phang (P) * d ( I , (P)) > r => M a t p h a n g v a m a t cau k h o n g c&t * d ( I , (P)) = r (3) M a t p h a n g va m a t cau t i e p xiic v d i "Dicu khodng kien can vd dii de mat phang (P) tiep xuc vai mat cau (S) Id each tit tam I ciia mat cau den mat phang b&ng ban kinh ciia mat cau" !iiJcAC BAVTOAN c:d B A N ^ C a c d a n g t o a n Ve m a t cau va m a t phang t a thadrng gap cac b a i t o a n sau : a) Cho phuang trinh mat cau, t i m t a m v a b a n k i n h : co sd de g i a i loai t o a n n a y l a cac cong thiifc (1), (2) b) Viet phuang trinh mat cau De v i e t phi/cfng t r i n h m a t cau, t a can xac d i n h t a m I va b a n k i n h r ciia no, sau sii dung cong thufc (1) hoac (2) c) Toan lien quan den mat phang V i e t phiicfng t r i n h m a t cau t i e p xiic v d i m a t p h d n g : Siif dung cong - X e t v i t r i ti/Ong doi cua m a t cau vdfi m a t phdng - thufc (3) - Viet phuorng t r i n h m a t phang tiep xuc vdi mat cau va d i qua mot diem hoac tiep xuc t a i mot diem thuoc mat cau : Sijf dung cong thijfc (3) 290 III TS Vu The Hi^u - NguySn Vinh Can - T i m t a m va ban k i n h dtfcfng t r o n giao tuyen ciia mat phSng vdfi mat cau : + T a m K ciia di/ofng t r o n giao tuyen (C) ciia m a t phSng (P) v d i m a t cau (S) l a h i n h chieu vuong goc ciia t a m I ciia m a t cau t r e n m a t phang (P) + Ban k i n h r' ciia diTcfng tron giao tuyen d diicfc t i n h theo cong thijfc r'^ = r^ - d'^ t r o n g r l a ban k i n h m a t cau, d = d ( I , (P)) = I K Chu y : T r o n g k h i g i a i cac bai toan c6 l i e n quan den v a n de tiep xiic giCira m a t phSng vdii m a t cau hoac giOfa diiong t h i n g v d i m a t cau t a can luon luon n h d : " M a t phang (P) (hoac dircfng thang d) tiep xuc v6i mat cau t a m I t a i diem M t h i (P) (hoac d) vuong goc v6i ban k i n h I M ciia mat cau" hoac I M = r 2] B a i tap 134 Trong cac phi/ofng t r i n h sau day, phuang t r i n h nao l a phuorng t r i n h cua mot m a t cau va t r o n g t r i i d n g hop l a phiicfng t r i n h ciia m a t cau t h i hay t i m t a m va b a n k i n h ciia m a t cau ay a) (S) : x ' + y^ + - 4x + 6y - 2z + 10 = b) (S) : x ' + y^ + - 3x - y + 2z + 15 = c) (S) : X - + y^ + z^ + 4x - 2z - 2z - = CHI D A N a) Dua phircfng t r i n h ve dang : (x - 2f + (y + 3)^ + (z - i f = Ta di/tfc m a t cau t a m 1(2; - ; 1), ban k i n h r = Chil y : Co t h e g i a i nhif sau TCf viec ap dung phirorng t r i n h dang (2), => A = - ^ 2C = - C O 2A = - 2B = ta => C = - B= V i A^ + B^ + C^ - D = (-2)^ + 3^ + (-1)^ - (+10) = > => PhiiOng t r i n h x^ + y^ + z^ - 4x + 6y - 2z - 10 l a phi/Ong t r i n h ciia m a t cau t a m I ( - A ; - B ; - C ) = (2; - ; 1) Ban k i n h r = Vi = b) Khong phai phi/Ong t r i n h mat cau v i t a c o A = — ; B = — ; C = - ; D = 15 2 =^ A^ + B^ + C^ - D < c) M a t cau t a m I ( - ; 1; 1) va r = VlO HQC va on luyen theo CTOT mon Toan THPT 291 Viet phiicfng t r i n h mat cau cac trifdng hop : a) Co tam 1(1; 2; - ) , ban kinh r = 2V3 b) Co tam I ( - l ; 2; -3) va di qua diem M(0; 4; 2) c) Di qua hai diem A(2; - ; - ) , B(2; - ; 3) va c6 tam nam tren difcfng X = + 2t thang A : y = - i - t z = 3-2t d) Nhan doan thSng AB la dudng kinh, vdi A ( - l ; 2; 0), B(3; 4; -2) C H I a) D A N (S): (x - 1)^ + (y - 2? + (z + 3)^ = 12 o x'^ + y'"^ + b) (S): (x + if + (y - 2? + (z + f = 30 - 2x - 4y + 6z + = x^ + y^ + z^ - 2x - 4y + 6z - 24 = c) Tam I cua mat cau la giao diem ciia A vdi mat phSng trung trifc ( P ) cua AB Ta C O : ( P ) : 2x + y - z - = TCr day t i m dirge 1(3; - ; 1), ban kinh r = l A = Vs ^ (S): (x - 3)^ + (y + 2f + (z - 1)' = x^ + y^ + z^ - 6x + 4y - 2z + •= d) Tam mat cau I la trung diem ciia doan thSng AB : 1(1; 3; - ) Ban kinh r = l A = V G =^ ( S ) : (x - 1)^ + (y - 3)^ + (z + 1)^ = o x^ + y^ + z^ - 2x - 6y + 2z + = Chu y : Co the giai nhif sau : Goi M(x; y; z) la diem thupc mat cau t h i MA = (x + 1; y - 2; z) MB = (x - 3; y - 4; z + 2) M thupc mat cau dircfng kinh AB nen MA J_ MB < > MA.MB = = (x + l)(x - 3) + (y - 2)(y - 4) + z(z + 2) = x^ + y'^ + z- - 2x - 6y + 2z + = Chufng minh rSng mat phSng ( P ) : 2x + 2y - z - = va mat cau ( S ) : x"^ + y^ + z^ - 2x + 2y + = tiep xuc vdi Tim toa tiep diem C H I - D A N Mat cau (S) c6 tam 1(1; - ; 0) va ban kinh r = (1) Mat khac d(I, (P)) = 292 2+ ( - ) - - = => d(I, ( P ) ) = r => dpcm V2^72^+M7 TS, VO The Huu - Nguygn Vinti Can - Tiep d i e m M cua m a t p h a n g c h i n h l a h i n h chie'u cua d i e m I t r e n (P) ^5 1 ^ Ta t i m difofc M 3'3' 137 Cho m a t cau (S) : + + + 4x - 4y + 6z + 13 = v a diem M ( - ; 2; - ) a) Chufng m i n h d i e m M nam t r e n m a t cau (S) ( b) Viet phiforng t r i n h mat phang (P) tiep xiic vdi mat cau (S) t a i diem M CHi DAN a) The toa cua d i e m M vao ve t r a i ciia phifong t r i n h m a t cau (S) de thay toa d i e m M thoa m a n phiicfng t r i n h m a t cau b) M a t cau (S) c6 tarn I ( - ; 2; - ) va ban k i n h r = Ta c6 I M = (2; 0; 0) M a t phang (P) vuong goc vdi I M , n h a n I M l a m vecto phap t u y e n T a CO : (P) : x + = 138 Cho m a t cau + y^ + + 2x - 2y + 2z + = v a m a t p h a n g (P) : 2x - 3y + 6z + = V i e t phiiOng t r i n h m a t phang (Q) d i qua d i e m M ( l ; - ; - ) song song vo'i m a t p h a n g (P) va tie'p xiic vdri m a t cau (S) CHI D A N M a t cau (S) c6 t a m I ( - l ; 1; - ) v a b a n k i n h r = M a t phang (Q) song song vdi m a t p h a n g (P) c6 phtfOng t r i n h (Q) : 2x - 3y + 6z + D = 0, D ^ Ta CO : d ( I , (Q)) = - - - +D -12+ D V ' + (-3)' + ' (Q)tiep xiicvdri m a t c a u (S): d ( I , (Q))= r D-12 = 1=> D - * V d i D - 12 = => D = 19 ( t m ) va t a di/gc m a t p h a n g ( Q i ) : 2x - 3y + 6z + 19 = * Vdri D - = - = > D = ( t m ) v a t a diTOc m a t phang (Q2) : 2x - 3y + 6z + = 139 Cho diem M ( l ; 1; V ) va m a t p h a n g (P) : 6x + 2y - 3z - = V i e t phuong t r i n h m a t cau (S), t a m M va t i e p xuc v d i (P) Hoc va on luyen theo CTOT mon Toan THPT ij" 293 CHI DAN Ta CO : d(M, (P)) = 3V7 V7 Phi/ong t r i n h mat cau (S) : (x - 1)^ + (y - 1)^ + (z - V?)' = y 140 Cho mat phSng (P) : 2x - y - 2z + 10 = va mat cau (S) : + y^ + - 2x - 4y + 2z - 19 = a) Chufng minh rang mat cau (S) va mat phang (P) cSt b) Tim tam K va ban kinh r' ciia diTcfng tron giao tuyen CHI DAN a) Ta CO mat cau (S) c6 tam 1(1; 2; - ) va b a n kinh r = d(I, (P)) = 4; b) De thay r'^ = d(I, (P)) < r => (P) v a (S) cat n h a u - d^ = 5^ - 4^ = r' = Goi K la tam ciia difofng tron giao tuyen ciia (S) va (P) t h i K la hinh 16 5^ 3' ' j chieu vuong goc ciia I tren (P) Ta tinh K Cho mp (P) : 2x - 2y - z - = va diem 1(1; 2; 3) Viet phircmg trinh mat cau (S) CO tam la diem I , cat mat phang (P) theo mot difofng tron c6 ban kinh r' = Va tim tam ciia difcmg tron giao tuyen CHI DAN Ta CO : d(I, (P)) = 2-4-3-4 72^ + {-2? -9 + {-If Ban kinh mat cau r : r^ = d^ + r'^ -3 r^ = 3^ + 4^ = 25 => r = Phirorng t r i n h mat cau (S) : (S) : (x - if + (y - 2f + (z - 3)^ = 25 Tam K ciia dirofng tron giao tuyen la hinh chieu vuong goc ciia I tren mat phang (P) Ta tinh dirge : K(3; - ; 4) X Cho dtfdng t h i n g A : =t y =1 va diem M(0; 2; 4) Viet phiTcfng z=-l-2t t r i n h mat cau (S) tam M va tiep xuc vdi A CHIDAN Ta CO : r = d(M, A) Ta tinh duac : d(M, A) = S, suy (P) : x^ + (y - 2f + (z - 4)" = l:i T S Vu The HI^J - Nguygn Vinh Can MVCLUC DAI SO VA GIAI TICH « CHUYEN DE I DAI SO TO HOP VA XAC SUAT §1 Hoan vi, chinh hap, to hap §2 Nhi thCfc Niutan §3 Xac suat CHUYEN DE II PHUdNG TRINH VA BAT PHl/dNG TRINH DAI SO §1 Phuong trinh, bat phuong trinh bac nhat mot an §2 Phuong trinh, bat phuong trinh bac hai mot an §3 Bat phuong trinh bac hai mot an §3 Phuong trinh, bat phudng trinh chifa gia trj tuyet doi §4 Phuong trinh, bat phuong trinh chifa cSn thi/c §5 phuong trinh nhieu an CHUYEN DE III PHUdNG TRINH LI/ONG 5 13 17 23 23 28 31 36 39 48 60 GIAC CHUYEN DE IV PHUdNG TRINH, BAT PHUdNG TRINH M VA LOGARIT U §1 Luy thi/a va ham so luy thil/a §2 Logarit, ham so mu, ham so logarit §3 Phuong trinh, he phuong trinh mu va logarit §4 Bat phuong trinh, h? bat phuong trinh mu va logarit CHUYEN OE V Gidl HAN - OAO HAM - KHAO SAT HAM SO VA DAO HAM §1 Gidi ban cua day so §2 Gidi han cua ham so §3 Dao ham va quy tSc tinh dao ham §1 Md dau §2 Mien xac djnh, mien gia trj §3 Su dong bien, nghjch bien cua ham so §4 Ci^c tri cua ham so §5 Gia trj Idn nhat, gia trj nho nhat cua ham so §6 Dudng ti?m can cua thi ham so §7 Su loi Idm va diem ud'n §8 Khao sat su bien thien va ve thj mgt so ham so §9 MOt so bai toan thudng gap ve thj ham so §10 Sir dung djnh | lagrang va tinh don dieu cua ham so de chifng minh [ va giai bat ding thi/c, bat phuong trinh Hoc va on luy?n theo CTDT mon Toan THPT 82 82 83 88 95 100 100 105 107 110 113 117 121 125 129 132 132 146 153 •295 CHUYEN O E VI NGUYEN HAM VA Tl'CH PHAN 160 184 § Qng dung cua tfch phan 170 § Tich phan 160 § Nguyen ham H I N H HOC CHUYEN D E I DUCiNG THANG VA MAT PHANG TRONG KHONG GIAN BAI TOAN Quan he song song 193 193 BAI TOAN Quan h vuong goc nhOng bai toan ve khoang each O BAI TOAN Cac bai toan ve goc 196 206 C H U Y E N O E II T H E TICH CAC KHOI 209 221 BAI TOAN The tich cac khoi tron xoay 209 BAI TOAN Tinh the tich cac khoi da dien C H U Y E N D E III VECTd - VECTd VA TQA 00 (TRONG MAT PHANG) 226 BAI TOAN Xac dinh tpa dp cua diem - Tpa dp cija vecto - Vecto cung phi/ong 2 BAI TOAN Tich v6 hudng cua hai vecto C H U Y E N D E IV Dl/CJNG THANG TRONG MAT PHANG 229 232 254 BAI TOAN 10 Cac van de ve khoang each va goc 241 BAI TOAN : Ducing thing vuong goc 232 BAI TOAN Viet phuong trinh dudng thing C H U Y E N D E V Dl/dNG TRON VA E L I P BAI T O A N 1 Xac 260 dinh phuong trinh dUcJng tron xac dinh tarn va ban kinh ducing tron 260 BAI TOAN 1 Elip va phuong trinh chinh tic C H U Y E N D E VI VECTd VA TQA DQ TRONG KHONG GIAN 266 268 272 BAI TOAN 15 Viet phUOng trinh mat phing 271 BAI TOAN 14 Tpa dp vectO Tich c6 hu6ng cua hai vecto 269 BAI TOAN 13 Bieu thtfc tpa dp cua tich v6 hudng cua hai vecto 268 BAI TOAN Vecto bang - Vecto cung phuong TOAN 17 BAI TOAN 16 BAI Viet phUOng trinh ducing thing Mot so bai toan lien quan den tinh vuong goc va khoang each C H U Y E N OE VII MAT CAU VA MAT PHANG BAI 296 TOAN 18 Phuong trinh m^t cau 276 282 290 290 TS V ThS' Hiiu - Nguygn V n Can Q Th vv w w 1 1 i i ^ ; i c " 1111( ) n ^ i i n > n i v n Email: baolongco 18DNguyen Thj Minh DT: 38246706 (M&i^ N G A N H A N G N G A N - 08083021 ha