Phương pháp và kĩ thuật ôn nhanh thi đại học đạt điểm cao môn toán nguyễn phú khánh

510.76 PH561P n c u Y i n P I I U Kiirinii PHUONO PHAP on niinnii TIII oni HOC D A T DIEM CAO NGUYEN PHU KHANH PHUONG PHAP & KI THUAT on niinnii TIII oni HOC DAT DIEM CAO \SP/ NHA XUAT BAN DAI HOC SlT PHAM Phucmg phdp vd ki thudt on nhanh thi Dai hoc dat diem cao mon Todn giiip hoc sinh on tap, van dung sang tao kien thuc, tong hop, phan tich, kiem tra nang lye toan dien, hucfng dan chien thuat giai de thi dai hoc Sach trinh bay noi dung tam on thi dai hoc gbm: - Phucmg trinh, bat phuong trinh, he phucmg trinh dai so He phucmg trinh mil va Idgarit - Bai todn to'ng hap Bat dang thuc; cue tri cua bieu thuc dai so - phuang phdp giai phuang trinh lugng gidc nhanh nhat - Van de lien quan den sophiec, dai sotohffp, xdc sudi - Cdc bai todn lien quan den ung dung ciia dqo ham va thi cua ham so: chieu bieh thien cua ham so; cue tri; gid tri Ion nhat va nho nhat cua ham so; tiep tuyen, tiem can (dimg vd ngang) cua thi ham so; tint tren thi nhirng diem c6 tinh chat cho truac, luang giao giua hai thi (mot hai thi Id dubng thang) - Tim giai han - Tim nguyen ham, tinh tich phan Ung dung cua tich phan: Tinh dien tich hinh phang, the tich khdi trdn xoay - Hinh hoc khong gian (tong hap): quan he song song, quan he vuong goc cua dubng thang, mat phdng; dien tich xung quanh cua hinh non trdn xoay, hinh tru trdn xoay; the tich khdi lang tru, khdi chop, khdi non trdn xoay, khdi tru trdn xoay; tinh dien tich mat can vd the tich khdi cdu - Phuang phdp toa mat phang: Xdc dinh toa cua diem, vecta Duong trdn, ba dudng conic Viet phuang trinh dudng thdng Tinh gdc; tinh khodng cdch tit diem den dudng thang - Phuang phdp toa khdng gian: Xdc dinh toa cua diem, vecta Duong trdn, milt cdu Viet phuang trinh mat phdng, dudng thdng Tinh goc; tinh khoang cdch tie diem den dudng thang, mat phdng; khodng cdch giita hai dubng thdng; vi tri tuang ddi cua dubng thdng, mat phdng vd mat cdu Mac du tac gia da danh nhieu tam huyet cho cuon sach, nhung sai sot la dieu kho tranh khoi, rat mong nhan duac sy phan bien va gop y quy bau ciia ban doc de nhung Ian tai ban sau cuon sach dugc hoan thien hon Tac gia CAU T R U C D E T H I DAi HQC M O N TOAN' I P H A N C H U N G (7 d i i m ) Cau (2 diem): a) Khao sat su bien thien va ve thi cua ham so b) Cac bai toan lien quan den u n g dung cua dao ham va thi cua ham so: chieu bien thien ciia ham so; cue t r i ; gia trj idn nhat va nho nhat cua ham so; tiep tuyen, tiem can (dung va ngang) cua thi ham so; t i m tren thi n h i i n g diem c6 tinh chat cho truac, tuong giao giira hai thi (mot hai thi la d u o n g thang) Cau (1 diem): Cong thiic lugng giac, phuong trinh lugng giac Cau (1 di£m): Phuong trinh, bat p h u o n g trinh, he phuong trinh dai so Cau (1 diem): - T i m gioi han - Tim nguyen ham, tinh ti'ch phan - u n g dung ciia tich phan: tinh dien tich hinh phang, the tich khoi tron xoay Cau (1 d i i m ) : Hinh hoc khong gian (tong hop): quan he song song, quan he vuong goc cua d u o n g thang, mat phang; dien tich xung quanh cua hinh non tron xoay, h i n h t r u tron xoay; the ti'ch khoi lang t r u , khoi chop, khoi non tron xoay, khoi tru tron xoay; tinh dien tich mat cau va the tich khoi cau Cau (1 diem): Bai toan tong hop I I P H A N R I E N G (3 diem) Thi sink chi dugc lam mot hai phan (phan A hoac phan B) A Theo chucmg trinh Chuan: Cau 7a (1 diem): Phuong phap toa mat phang: - Xac d i n h tpa dp cua diem, vecto - Duong tron, elip - Viet phuong trinh d u o n g thSng - Tinh goc; tinh khoang each t u diem den d u o n g t h i n g Cau 8a (1 d i i m ) : Phuong phap toa khong gian: - Xac d j n h toa dp cua diem, vecto - Duong tron, mat cau - Tinh goc; tinh khoang each t u diem den d u o n g thang, mat phang; khoang each giiia hai duong thang; v i t r i tuong doi cua duong thang, mat phang va mat cau Cau 9a (1 diem): - So phuc - To hop, xac suat, thong ke - Bat dang thiic; eye t r i cua bieu thiic dai so B Theo chuang trinh Nang cao: C a u 7b (1 d i i m ) : Phuong phap tpa mat phSng: - Xac d i n h toa cua diem, vecto - D u o n g tron, ba d u o n g conic - Viet phuong trinh d u o n g thang - Tinh goc; tinh khoang each t u diem den d u o n g th5ng C a u 8b (1 d i i m ) : Phuong phap tpa khong gian: - Xac dinh tpa dp cua diem, vecto - D u o n g tron, mat cau - Viet phuong trinh mat phang, d u o n g thang - Tinh goc; tinh khoang each t u diem den duong thSng, mat p h i n g ; khoang each giua hai duong thang; v i t r i tuong d o i cua d u o n g thang, mat phang va mat cau C a u 9b (1 diem): - So phuc - Do thi ham phan thue h i i u t i dang y = + bx + e px + q quan - Su tiep xiie cua hai d u o n g cong - He p h u o n g trinh m u va iogarit - To hop, xac suat, thong ke - Bat dang thiie Cue t r i eiia bieu thiic dai so PHlTdNG TRINH, BAT PHlTdNG TRINH, HE I PHlTOfNG TRINH Phuong trinh, bat phuong trinh, he phuong trinh dai so H( phuong trinh mil va Iogarit M O T SO H E P H U O N G T R I N H C O BAN 1) He bac nhat hai an, ba an 2) He gbm mot phuong trinh bac nhat va phuong trinh bac cao He ba p h u o n g trinh - He hai p h u o n g trinh - Phuong phap ehung: Su d u n g p h u o n g phap the • 3) He doi xung loai Phuong phap chung: Dat an phu a = x + y; b = xy 4) He doi xung loai Phuong phap chung: T r u tung vehai phuong trinh da cho ta dugc: (x - y).f(x; y) = 5) He phuong trinh dang cap bac hai Xet truong hop y = V o l y ?t 0, ta c6 the tien hanh theo cac each sau: - Dat an p h u y = t.x - Chia ca hai ve'cho y ^ , va dat t = — y MOT SO P H U O N G PHAP G I A I H E P H U O N G T R I N H 1) Phuong phap the • Phuong phap: Ta riit mot an (hay mot bieu thuc) t u mot p h u o n g trinh he va the vao phuong trinh lai • Nhqn dang: Phuong phap thuong hay su d u n g k h i he c6 mot phuong trinh la bac nha't doi voi mot an nao 2) Phuong phap cong dai so 3) Phuong phap bien doi tich 4) Phuong phap dat an p h u 5) Phuong phap ham so 6) Phuong phap su d u n g bat dang thuc C/iM y: ting dung dao ham gidi todn phuong trinh, he phuong trinh Su dung cac tinh chat ciia ham so de giai phuong trinh la dang toan kha quen thupc Ta thuong c6 ba huong ap dung sau day: Huang 1: Thuc hien theo cac buoc: Buac 1: Chuyen phuong t r i n h ve dang: f(x) = k Buac 2: Xet ham so y = f ( x ) Buac 3: Nhan xet: • V o i x = Xg f(x) - f(xg) = k, do X g la nghiem • Voi X > Xg o f(x) > f(xQ) = k, do phuong trinh v6 nghiem • Voi X < Xg f(x) < f(xQ) = k, do phuong trinh v6 nghiem • Vay X g la nghiem nhat a i a phuong trinh Huong 2: Thuc hien theo cac buoc: Suae 1: Chuyen phuong trinh ve dang: f(x) = g(x) Buoc 2: Dung lap luan khang d i n h rang f(x) va g(x) c6 n h u n g tinh chat trai ngugc va xac d i n h X g cho f ( X g ) = g(xg) Buac 3: Vay Xg la nghiem d u y nhat ciia phuong trinh Huang 3: Thuc hien theo cac buac: Buac 1: Chuyen p h u o n g trinh ve dang f(u) = f ( v ) Bwac 2: Xet ham so y = f ( x ) , dung lap luan khang d i n h ham so don dieu Buac 3: K h i f(u) = f(v) o u = v Cac vi du| V i du Giai cac p h u o n g trinh sau tren tap so thuc: V x - l - x + 13 ~ + X X V =x+ 2x V +-=-2x-4 + X x ' " ' + ^ ' ' - ' ' ' + = 2^''*'+2''"+9.2^*" X Lcri giai Dieu kien: x > — Phuong trinh da cho tuong d u o n g v o i VSx - - V3x+ 13 = -if VSx - - V3x +13)(VSx - + V3x + ) Truang hap 1: V S x - l - %/3x + 13 = o V s x - l = V s x T l S , binh phuong hai ve roi riit gon ta dugc x = (thoa man) Trucrng hap 2: V S x - l + V3x + 13 = (1) Neu x > l thi V T ( l ) > >/4 + Vl6 = , neu x < l thi VT ( l ) < ^/4 + Vl6 = De thay x = la nghiem p h u o n g trinh (1) Vay p h u o n g trinh ban dau c6 hai nghiem x = , x = 2, Dieu kien: + - ^ o x X > o x < - hoac x > K h i p h u o n g trinh da cho viet lai: x ^ l + - = -2x^ - 4x + + Vdi X > , phuong trinh o > y x + x = - ( x + x ) + Dat: t = V x ^ + x , t > , ta c6 2t^ + - = o t = hoac t = - - , doi chieu dieu kien ta dugc t = l , tuc la phai c6 Vx^ + x = o x^ + 2x = , phuong trinh c6 nghiem x = -l + \f2 thoa man dieu kien + V d i X ^ - , p h u o n g t r i n h o - V x ^ +2x = -2(x^ + 2x) + Dat t = > / x ^ + x , t > , ta c6: 2t^ - - = o t = | hoac t = - l , d o i chieu dieu kien ta dugc t = ^ , tuc la phai c6 4(x^ + 2x) = 4x^ + 8x - = , phuong trinh c6 nghiem x = ~ ^ '^'^''^ man dieu kien Vay phuong trinh da cho c6 nghiem x = -4-V52 ; X = -1 + r/2 3.Dieu kien: x # , x - i > , x - - > X X Phuong trinh da cho dugc bien doi ve dang: x - — = Ix - — - p x - — (1) X V X V X A p dung cong thuc: — - j = , v d i a > 0; b ^ 0; a ^ b va Va + Vb > Va-vb - X Khi ( l ) o x — = X ^ (2) x-UJ2x-^ * Neu X — > thi — x < 0, t u (2) ta c6 ve trai Ion hon 0, ve phai be hon 0, v6 li X x 4 * Neu X — < thi — x > 0, t u (2) ta c6 ve trai be hon 0, ve phai Ion hon 0, v6 li X X Vay X — = o x - = = > x = (thoa man dieu kien) X Phuong trinh da cho c6 nghiem nhat x = Phuong trinh da cho tuong d u o n g v6i (2'" +2 o - ( " +2 (2"+2")'-2 , phuong trinh da cho tro thanh: t^ - 6t^ - 72t +189 = c = > ( t - ) ' ( t ' + t + l ) = o t = 3, v i t ' + t + 21 > , V t > Voi t = o " + - = « " = ^ ^ « x = log/^*^^ / /— 3±V5 Vay p h u o n g trinh da cho c6 nghiem nhat x = log^ V i d u Giai cac phuong trinh sau tren tap so thuc: log25(x^-8x + 15)^ = | l o g ^ ^ + log5|x-5 2 l o g , (4 - x ) ' + ^ log, (x + ) ' = + log, (x + 6y 4" (V2.6"-4" + V4.24" - " ) = 27" -12" + 2.8" LOT gidi Dieu kien: x > 1;x # 3;x # Phuong trinh da cho tuong duong v o i logs x'' - x + 15 = ' g ^ + l0g5 X - X-1 o l o g g x - +log5 X - = l o g - y - + l o g | x - o | x - = x - l < : = > x - = x - l hoac x - = l - x x = - (vi ^ Vay p h u o n g trinh c6 nghiem x = - x^5) ' Dieu kien: - < x < va x ;t - (*) I^huong trinh da cho o g ^ ( - x ) - o g ^ x + = 3-31og^ {x + 6) o log^ (4 ~ x) + iog^ (x + 6) = + log^ |x + 2| o (4 - x ) ( x + 6) = 4|x + 4(x + 2) = ( - x ) ( x + 6) hoac 4(x + 2) = - ( - x ) ( x + 6) (vi (*) nen ( - x ) ( x + ) > ) + V6i x - + x - = 0c:>x = 2,x = - + V6i x ' - x - = 0x = l + 733,x = l - V 3 Vay p h u o n g trinh c6 hai nghiem x = 2; x = - V33 Dieu kien: 2.6" - 4" > va 4.24" - 3.16" > « x > log, - Ta C O cac danh gia sau: 4"^2.6" - " = 2".2"^2.6" - " < ^ " (4" + 2.6" - " ) = 12" 4" V4.24"-3.16" = 4" ^^4" ( " - " ) < ^ " (2" + V " - " ) = - " + i \ " V " - " < - " + - " ( " +4.6" - " ) = 12" 2 ^ ' Do 4" (V2.6" - " + ^ " - " ) < 2.12", dSng thuc xay k h i x = " - " + " = 27" +8" -12" > 3N/27".8".8" -12" = 2.12", dau bang xay x = Vay p h u o n g t r i n h phai c6 nghiem x = V i d u Giai cac phuong trinh sau tren tap so thuc: x ^ + x ^ + x + l = 3| x + 2 log4 N/X^ - + 3^2 log4(x + 3)^ = 10 + log2(x - 3)^ Lai gidi l D a t 3|^i±?.=y + i c ^ ^ i ± i = y + y + y + i o x = y ^ + y ^ + y (1) 10 T h e o d e b a i ta CO p h u a n g t r i n h x +6x + x + l = y + y = 2x +6x +6x (2), Ttr (1) va (2) s u y 2x^ + 6x2 + 6x - (2y^ + 6y2 + y ) = y o X o 2(x^ - y^) + 6(x2 - y ^ ) + 7(x - y ) = (x - y ) ( x + 2y2 + x y + 6x + y + 7) = o x = y h o a c 2x^ + y + x y + 6x + y + = (3) P h u a n g t r i n h (3) v n g h i e m v i 2x^ +2y^ + x y + 6x + y + = ( x + y)2 + ( x + y ) + + (x + l ) +{y + = (x + y + ) + ( x + l ) + ( y + l ) + l > l lf +1 Vx,y ^i-#-Vay x = y , t h a y v a o (2) ta d u o c 2x'' + 6x2 + 6x = x o + 6x2 + 5x = o ^^2x2 + 6x + 5) = C5> x = V a y X = la n g h i e m p h u o n g t r i n h D i e u k i e n : x < - hoac x > 81og4 7x2 - + j o g ( x + 3)2 = 10 + \og^(x 2log2(x2 -sf - ) + j l o g ( x + 3)2 = 10 + l o g ( x - ) l o g ( x + 3)2 + 3^/log2(x + 3)2 - = D a t t = l o g ( x + 3)2 > Ta CO p h u a n g t r i n h : t^O t2+3t-10 = o t = o l o g ( x + 3)2 = o (x + 3)2 = 16 X = - h o a c x D o i c h i e i i d i e u k i e n , ta c6 n g h i e m p h u a n g t r i n h la x = - V i d u G i a i cac p h u a n g t r i n h sau t r e n t a p so t h u c : x^ + 3x2 + 9x + + ( x - i o ) V - x = V x ^ + yjx^ +x^ +x + \ + Vx'* - Lai giai D i e u k i e n : x < Bien d o i p h u a n g t r i n h ve: (x +1)"' + 6(x +1) = V - x ( - x + 6) Dat u = x + l , v = \ / - x ^ T a c p h u a n g t r i n h : u ' ' + u = v'' + 6v o ( u ' ' - v^) + ( u - v ) = u = V hoac u2 + u v + v2 + = Trudmg hQfp 1: u2 + u v + v2 + = ( u + —)2 + Trumtghgp 2: u = v V - x = x + + = , v6 nghiem x + 1^0 X = -3+N^ - x = x2+2x + l 11 Voi t, =~l=i> A(1;1;1) loai v i A = I Vai =2=>B(3;5;5) x = 3+ t Phuang trinh A qua M va B la: y = + t : Truong hop loai v i A khong cat A, z = 5-2t Voi t , = - = > B ( - ; - ; - ) x = - l + 5t f;2;0 Phuong trinh d u o n g A : y = - + 10t=>A cat A, tai A z = - + 6t Goi O la tarn mat cau theo gia thiet O thuoc (P) ( b ; 2b - 2; O) Ta c6: V i (S) tiep xuc v a i A, va Aj Ian luot la A va B Nen O A = OB ^ b = -11 80 40 80 Cdch khdc: De thay A,; A^; A dong phSng Do vay ta phai c6: B la trung diem A M Goi A (a; a - : - a ) => B ^ - ^ ; a + ; - a • V ' [ T u d o ta t i m A ( ; - ; ) ; B(2;3;3) Ta goi tarn l ( m ; n ; ) thi dua vao d ( l ; A , ) = d ( l ; A , ) , phan lai danh cho ban doc u^ = ( l ; l ; - ) la mot VTCP cua A H , H G A H = > H = (2 + a;3 + a ; - a ) • C H = (a - ; + a ; - a ) H e BC ^ CH.u = a = => H ( ; ; ) x = 3-t => BC CO phuong trinh: y = 2+t (te#) z =3 U j = ( l ; - ; l ) lampt VTCPcuaBD,B = B C n B D ^ B ( l ; ; ) I e BD zi 1(1 + b; - 2b; + b); I H BD : ^ I H i ^ = =^ I 2 H , la diem doi xiing ciia H qua BD => I la trung diem cua H H , => H , ( l ; 3; 4) ^ B H , = (O; - ; l ) , H , e AB => AB c6 phuong trinh: y = - s (s z = 3+s A = A B n A H => A = (l;2;5) => A B = A C = BC = 2V2 Cv^Bc = A B + A C + BC = 672 329 Ta c6: d x-3_y-2_z-l x-2_y-5_z+3 -; d , :4 ' -3 > d , va d j cheo M N = ( - ; - ; - l ) => D u o n g thang chiia M N c6 p h u a n g trinh: u = ( l ; - ; ) la mot VTCP cua d , ; M N u = x+1 y-1 z+1 M N d, M N c= ( ? ) = M N ± A B => M N ( Q ) ; ( Q ) la mat phSng chua d , va A B :=> ( Q ) : 5x + 3y + z - 2 = 0, ( Q ) n ( d , ) = B B ( ; ; - ) => M B = 76 A e d , =^ A = (3 +1; - 3t; + 4t) => M A = J(t - ) ' + (2 + t ) ' + ( l + t ) ' M G (P) => M A = M B A CO toa dp: ( ; ; l ) hoac 30,53,-23 J3'l3' 13 Gpi E la trung diem ciia A B ; (P) la mat phang trung true cua AB (P) qua E va nhan AB la mot VTPT Vdi A = ( ; ; l ) = i > E = ; AB = ( ; ; - ) = > ( P ) c6 phucmg t r i n h : y - z - = J Voi A = 13 13'13' 23 30 53 E = ^69 46 26 13 49' 26 ; AB = ^9 -14 ^13 13 -3^ 13^ => (P) CO phuong trinh: 9x - 14y - 3z + 20 = Vay p h u a n g trin h cua mat phang (P) la y - 3z - = hoac 9x - 14y - 3z + 20 = Bai tap : Gpi d la d u o n g thang bat k i d i qua B ( - ; ; ) va cat d u o n g th§ng A Giasir d cat A tai M ( l + t ; ; - t ) Khi d c o V T C P B M = (2 + t ; - ; - t ) Taco B A = ( ; - l ; - l ) , BM,BA BM,BA = (2-t;2-2t;4-t) ^(2-t)^(2-2t)^(4-tf d(A,d) = t ^ - t + 12 V t^+2t + BM f (t) = ^ ^ ^ ^ = o t = ±2 (t2^2t + f Xet ham so f ( t ) = t^+2t + Vay khoang each t u A t o i d Ion nhat bSng ^ / l l k h i t = - l i n g v a i M ( - ; ; ) x=- l D u o n g thang d can t i m c6 p h u o n g trinh: y = 2-t z=t 330 Ox CO VTCP i = (1;0;0) qua 0(0; 0; 0) va A B CO VTCP AB = ( - l ; l ; - ) va i.AB = - ^ => Ox va A B khong vuong goc Taco [i,AB].OA = + + = nen A B va O x c h e o n h a u x=t Phuang trinh tham so cua Ox : < y = 0, M e Ox => M ( t ; 0; 0) z=0 S = M A + M B = ,/(t ~3f+0 + 4+s]{t - i f + ^ =s]{t -3f+4+^J{t - i f + Ta phai t i m t cho S dat gia tri nho nhat Trong mat phang toa Oxy, xet cac diem M( (t; O) e Ox va hai diem A , ( ; ) , B ( ( ; l ) thi S = M , A , + M , B t Ta thay A^, Bj nSm cung phia v a i Ox nen ta lay A / ( ; - ) d o i x i i n g v d i Atqua Ox Phuong trinh d u o n g thSng A ( ' B j : x + y - = S= A j + M , B ( nho nhat M la giao diem ciia Ox va A^' Bj => 3t - = hay t = ^ • Vay M ( - ; ; ) la diem can t i m Cdch khdc: Ta cd the tim diem M bang phucmg phdp khdo sat ham so' T u bieu thuc S = ^(t - ) ^ + + ^/(t - ) ^ + T a x e t h a m s o f ( t ) = J(t - ) ^ + + ^/(t -if+1 t-3 t-2 Co dao ham f ' ( t ) = f'(t) = t-3)%4 t-3 ^/(T^ ^ { t - l f + t-2 :+ , (te#) = J(t-2f+1 ^ =0 (t-2) -(t-3) O ^ ' J(t-3)%4 J(t-2f+1 vai dieu kien < t < ta c6: (*) o (t - ^ [(t - i f +1] = (t - i f [(t - 3)^ + 4] «(t-3f=4(t-2f t - = 2(t-2) t - = -2(t-2) t = lg[2;3] Lap bang bien thien cua ham so f ( t ) Suy m i n f ( t ) = f /38+VlO 3, Vay M A + M B dat gia tri nho nhat bang ^ ' ^ ^ ^ ^ , dat duoc tai t = - , tiic la M ( - ; 0; ) 3 331 D u o n g thang d c6 p h u o n g trinh tham so x = l + 2t y = + 2t z =l+ t d qua diem N ( ; ; ) , C6 V T C P u = (2;2;l) va AB = ( ; ; - ) Ta CO u C D = + - = ^ =:>d khong vuong goc v6i A B va [ u , A B l N A = - = => d va A B cheonhau Chu v i tam giac M A B la 2p = ( M A + MB + A B ) , A B khong doi nen 2p dat gia tri nho nhat k h i M A + MB dat gia t r i nho nhat Xet diem M e d => M ( l + 2t; 2+2t; +1), ta phai t i m t de M A + M B dat gia trj nho nhat Xet f ( t ) = M A + MB =^|{2^ + 2f +{2t + lf +1^ + yjiltf +{2t-2f+{t + \f => f (t)=V9t^ + 12t + + V9t^ - 6t + = ^/(3t + 2)^ +1 + ^/(3t -1)^ + Co dao ham f'(t) = f'(t) = O o 3t + \/(3t + ) + 3t + 3t-l 7(3t + ) + o - 3t + j = 3t-l J(3t-l)2+4 =0 J(3t-l)2+4 -(3t-l) , = - = ^ = = = vol = V(3t + ) + l ^/(3t-l)2+4 — ^ t< - 3 cp^j^ t = - - , phuong t r i n h d u o n g thang can t i m la: x+l _ y _ z+ l ~4^~ a) d n d ' = M = ^ M ( - l + t ; t ; - t ) , t G i J ? ' , V T C P c u a d : U j = A M ( t - l ; t + l ; - t ) AB,Ud AB(2;2;-1); AB;Ud = ( l - t ; l ; - t ) = > d(B,d) = Xet ham so f(t) = I j ^ +24t + 54 2t^+4t + 12t^-18t + 18 V 6t^-2t + rnaxf(t) = f(0) = 18; minf(t) = f(2) = — 11 • ^ ^ d ( B , d ) < Vl8 x = 3t min(d(B,d)) = t = Phuong trinh d u o n g thang can t i m la: y - - l + 3t z = 2-2t x = -t max(d(B,d)) = v l t = Phuong trinh d u o n g thang can t i m la: y = - l+ t z = 2-t b) d n d ' = M = ^ M ( ^ l + t ; t ; - t ) , t e # , V T C P c u a d : U j = A M ( t - l ; t + l ; - t ) T u p h u o n g t r i n h A =o u^==(2;-2;l) va N = ( ; ; ) e A = (t l;4t-l;6t) AN(5;l;-2); AN - = 3., d(A,d) = (2 + t)^ (53t''"10t + = 3.Vfa) Xet ham so f(t) = — ^ ^ ^ ^ r:> maxf(t) = f ( — ) = - - , max(d(A,d)) = V26 53t^-10t+2 37 x = 29t Phuong t r i n h d u o n g thang can t i m la: y = -l-41t z = + 4t Goi VTCP ciia d u o n g thSng d la: u(a; b; c), a^ + + >0 d//(P)c=>Ud.nQ = c j c = a - b ; u ( l ; - ; ) Goi goc giua hai mat phSng la cp (0 < cp < — ) , 334 5a-4b cos((p) = V a ' - a b + 2b2 (5a-4br ^ \ a - a b + 2b^ Truong hap 1: Neu b = 0, coscp = -.^5 Truong hap 2: Neu b # , dat t = - , coscp = b Xet ham so f(t) = (5t-4)^ -, < cos(p < j-^^L_i)_ ^-.Mt) 3\5t2_4t + 5^3 5t^-4t + So sanh trudng hgp va t r u o n g hgp thi ^ cos(p < y min(cos{p) = => cp^^^ = ^O*^ J = ^ • b x-1 Phuong trinh d u o n g thang can t i m la: — ^ = max(cos{p) = a 5/3 y + ^ = z-2 Phuong trinh d u o n g thang can t i m la: X y + z-2 = ^—^ = — ^ Goi VTCP cua d u o n g thang d la: u(a;b;c),a^ + b"^ + c^ > dc(P)ouj.np =0 b = 0, chgn a = 1, c = X Phuong trinh d u o n g thang can t i m la: =l +t y =0 z =t max(d(B,d)) = \ / l o a = - b c h g n b = - l , a = l , c = - l 335 P h u a n g t r i n h d u o n g thSng can t i m la: x = ~ l - t y =t z =t 10 a) P h u a n g t r i n h m a t p h ^ n g (P) c h u a d c6 V T P T la n ( A ; B ; C ) , A ^ + B + C ^ > c6 d a n g A ( x - l ) + B ( y + 2) + C z = Taco: d c (a) « Uj".!!^ =0 o C = A + 2B G o i goc g i i i a h a i m a t p h S n g la (p, (0 < (p < ^ ) , coscp = A + 2B \/2A^ + A B + B ( A + 2B)^ Trudmg hap 1: N e u B = 0, coscp = 2 A + A B + 5B^ (1) A , cos(q)) = —(t + 2)^ Truong hap 2: N e u B ?t , d a t t = — B V2t2+4t + X e t h a m so f ( t ) = —l^t^) 2t2+4t + V a y m a x coscp = /30 , m a x f ( t ) = - tai t =1 h a y — = - • B (2) So sanh t r u o n g h o p v a t r u o n g h o p t h i (p^^^ coscp = — ^ v o i — = — B P h u o n g t r i n h m a t p h a n g c a n t i m l a : x + 2y + 5z + = b) P h u a n g t r i n h m a t p h S n g (P) c h u a d c6 V T P T : n ( A ; B ; C ) , A ^ + B^ > c6 d a n g : A ( x - l ) + B ( y + 2) + C z = Ta c6: d c ( a ) n^ = o C = A + 2B G o i goc g i i i a m a t p h a n g (P) v a d u o n g t h a n g d ' la T , ( ^ T < —) sin^ = A + 3B (4A + V A + A B + 5B2 3Br ' V A + A B + 5B Truatig hap 1: N e u B = 0, sin4^ = 2V2 N e u B *0,Aai t = —, s i n T = - j - ^ ^ ^ ' • B' 3V2t2+4t + Truanghapl: * 336 Xet ham so f(t) = 5V3 (4t + 3)^ 25 —, maxf(t) = — tai t =-7 hay 2t^+4t + sin \\i = So sanh truong hap va truong hap suy ra: ^ ^ ^ ^ sinT = voi = • Phuong trinh mat phang can tim la: 7x - y + 5z - = c) Phuong trinh mat phSng (P) chua d c6 VTPT n(A;B;C), > c6 dang A ( x - l ) + By + C(z-2) = Taco: d c (a) o u^.n^ = o B = - A - C A+C ' (A+C)^ = V A + A B + 5C2 " V A + A B + 5C2 Tru&ttghop 1: Neu C = 0, d(A,(a)) = d(A,(a)) = Trumg hap 2: Neu C # 0, dat t = —, d(A,(a)) = Xet ham so f(t) = (t+ir = 9^/f(t)• V5t^+8t + (t+ir va f'(t) = Oc=>t = ± l ; f ( - l ) = 0;f(l) = - , lim f(t) = - 5t^+8t + t^.±co Lap bang bien thien suy maxf(t) = - tai t =1 Vay maxd(A,(a)) = 3V2 ^ = So sanh truong hop va truong hop suy ra: A = C va B = -4C Phuong trinh mat phSng can tim la: x - 4y + z - = Bai tap 3: Lay tren duong thang d da cho hai diem M ( l ; 0; - l ) , N (O; 2; O) Mat phSng (p) can tim c6 phuong trinh dang: Ax + By + Cz + D = (A^ + B" + ^ ' [MG(P) [A-C +D = fC = [Ne(P) [2B + D = [ D= -2B > o) A-2B =>(p):Ax + By + ( A - B ) z - B = Hai mat phSng (P) va ( a ) : x + 2y - 2z - = c6 VTPT: n^ = ( A ; B ; A - B ) ; n ^ = ( l ; ; - ) Goi X la goc hop boi hai mat phang (p) va ( a ) : x + 2y - 2z - = thi cosx = - A + 6B —» np n 3V2A'-4AB + 5B' , (A-6B)' • cos X =9 ( A ' - A B + B ' ) ' 337 Xet B-O=>cosx = —- A -12 Xet B^O=>cos^x = 'A^ -4 Xet ham so f(t) = ^A^ vBy 36 +5 t ' - t + 36 ^ ^ A ( t ' - t + 5)' B t' -12t + 36 -; t e 2t' - t + -V 53 „, , 20t'-134t + 84 ^ t= l a co: f (t) = ——; ~ ; f (t) = 16 10 • \10, ~ ' ( t ' - t + 5y [ t = 6=:^f(6) = De thay goc x nho nhat cos x Idn nhat o f(t) Ion nhat A ( 7\ = — => — = — ; A = Lap bang bien thien cua ham so f(t) => max f(t) = f ,10, 10 B 10 B = 10 Phuong trinh mat phSng can tim: (P): 7x +10y - 13z - 20 = C M A T C A U Bai tap : Goi I la tam duong tron (C) cho nen I la hinh chieu cua A tren mat ph§ng (P) Goi A la duong thang qua A va vuong goc voi (P) nen A c6 phuong trinh tham so la x =l +t A:I la giao diem cua A va (P) va I e A i:> l ( l + 2t;2+ 2t;3 + 2t) z = + 2t Ie(P)=:>l + t + 2(2 + 2t) + 2(3 + t ) - = 0=>t = - l i z > l ( ; ; l ) Duong thSng d tiep xiic vdi duong tron (c) d(l;d) = R Duong th§ng d c6 VTCP la a = ( l ; - l ; - ) va di qua O(0;0;0) Taco IO = ( ; ; - l ) : IO;a = ( - l ; - l ; ) = ^ R = d ( l ; d ) = Ta laico l A = (l;2;2) ^ lA = O = AC n BD r = VlA'+R- = iO;a 28 (s): (x - l)^ +(y -2)^ +(z -3)^ = 28 Vay mat cau can tim la Goi Vi S A B C D la chop deu nen lO Suy I e ( S A C ) Ma B D ( S A C ) nen (a) chinh la ( S A C ) Nhu vay, yeu cau bai toan thuc chat la viet phuong trinh mat phang di qua ba diem: S, A , C 338 T a c : n ( ) = SA,SC = [ ( , , l ) , ( l , , - ) ] = (0,10,0) Vay ( a ) : y = Nhan thay: A, O thuoc mat cau Goi M la trung diem cua A O , suy M ( ; ; ) GoiB(a,b,c) Ta c6: O A ( ; ; ) ; M B ( a - ; b - ; c ) , tam giac A B O can tai B nen O A ± MB =^ a + b = (1) Taco: 4>/3 = S^„„ = ^ A O B M = | V B M =^ B M = ^6 = ^ ( a - ) ' + ( b - ) ' + c ' = (2) Do B e ( l ; R ) nen ( a - ) ' + ( b - ) ' + ( c - ) ' =12 (3) T u ( l ) , (2), ( ) s u y ra: B ; B 2- V Bai tap 2: Mot VTCP cua d u o n g thSng d la u ( ; l ; - l ) Mot VTPT cua mat phSng (P) la n ( l ; ; l ) Goi la goc giiia d u o n g thang d va mat phang (P) + 2-1 Ta c6: sin cp = cos u , n = - = > ( p = t M X = 30° Goi R la ban kinh mat cau (S) => l A = R Tam giac I M A vuong tai A Ta c6: T M X = 30° ^ A M = R>/3.S„,^ = 3V3 o |lA.A = M3V3 o R = N/6 Gia sir l ( l + t ; l + t ; ~ t ) , t < - - 3t-3 T u gia thiet ta CO khoang each d ( l , ( P ) ) = R c ^ — p - = V o t = - l =>l(-l;0;l) Phuong trinh mat cau (S): (x +1)" + y" + (z -1)^ = Bai tap 3: I , ( ^ l ; ; - ) ; R, = 2^2 Goi r la ban kinh d u o n g tron giao tuyen I , (d) o r = AR] - d ( l , ; ( P ) ) = I , ( t - ; t;2t - ) , d ( l , , { P ) ) = ^ R ^ - r ' = 7i '3V2I VTi ^ 4~2 t = 0:^(S'):(x + l ) ' + y ' + ( z + l ) ' =4 t = -2=i>(x + ) ' + ( y + ) ' + ( z + ) ' = 339 Bai tap 4: Goi I la trung diem cua AC, I 2 V\D la hinh chop deu nen ABCD la hinh vuong Ma AC = 3>/2 BI = ; B thuoc (Q) => B(a; b; - a - 2b) BI.AC = b= 24-5a ABCD la hinh vuong: BI = 3- — [2 a = 3=i.b = l 57 a=— 107 -a + J f5 — -h [2 (a e .(a.2b-5)^4 J 57 (loai) ==>B(3,1,-2) vol a = - 107 > ( A B C ) : x + 2y + z - = (cung chinh la phuong trinh ( A B C D ) ) x = - + 2t d y = - + 2t ^ z = -2 +1 Gpi d la duong thang qua I va vuong goc vdi ( A B C D ) Vi S A B C la hinh chop deu nen S thuoc d => S - + t ; - + 2t;-2 + t 2 Trong mat ph3ng (SAl) dung trung true cua SA cat SI tai O Vi O thuoc d la true va O thuoc trung true cua SA nen SO = OA = OB = OC = OD = R Vay O la tam mat cau ngoai tiep hinh chop S.ABC ban kinh R = SO Ma ^ R X ; I X cung nhin OA duoi mot goc vuong nen OHAI npi tiep duong tron duong kinh OA Vi S nSm ngoai duong tron • SH.SA = SI.SO SH.SA SO = R = SI SA' 2SI 2h Ma theo eia thiet — = - h = — ^ h h = SI = - S I = (-2; - t ; - t ) o S l ' = — =^t = - = > S 2-3-^ 2,3,^ ^ ' 16 hoac t = — => S l ; ; - ^ 4 Bai tap 5: A e d = > A(4 + 3t;4 + 2t;4 + t ) 340 Goi I la tam mat cau (S) =i> I = (l;2;3) => lA = (3 +3t;2 + 2t;l +1) =^ AI = |l +1| Vl4 (1) Dat AB = a; AI n(BCD) = H; (s) c6 ban kinh R = 14 AH±(BCD)=>AH = ^ ; =^AI = AH + IH = ^ IH = VR^-BH^ =J^^T^ + ^ | ^ ^ (2) Lai c6: Al' = R ' +a= r^ AI = VR' +a' (3) TLr(2)va(3) ^ V R ^ ^ = ^ + « a^ = 2R^ =^ A I = RVS = ^/l4 (4) Diem A C O cac toa do: (4;4;4); ( - ; ; ) Tir (1) va (4) + t = Bai tap 6: Ta CO tam mat cau (S) la l(l;l; -2), ban kinh la: R,s, = 4, d(l;(P)) = - Vi d(l;(P))[...]... bang bien thien Dua vao bang bien thien ta thay v o i x e (-co; +00) thi t G ( - 1 ; 45] Do do phuong trinh da cho c6 nghiem k h i va chi k h i phuong trinh ( t - 1 ) ^ 18 —- = m CO nghiem t e ( - 1 ; Vs] Xetham f(t) = ( t - l f + — tren ( - 1 ; S],iac6 t+1 ci> X = f'(t) = 2 ( t - l ) 2 Gidi han: l i m f(t) =+00; f(2) = 7;f(V5) = - + t~>(-i)+ 2 18 ^ va f'(t) = 0 (t + lf — 2 Lap bang bien thien, suy... a m so d o n g bien, VP (*) la h a m so nghjch bien ma V T ' 3^ ^ 4 ; = VP ( 3^ "4 nen tren ( - 1 ; - i ) thi x = - ^ la nghiem duy nhat thoa man ( * ) 24 Xetx 6 ( - — ; + o o ) , thi V T ( * ) la ham so dong bien, V P ( * ) la ham so nghich bien 6 ma V T ( I ) = V P ( l ) nen tren (-~;+co) 6 thi x = 1 la nghiem duy nhat thoa man ( * ) Vay phuang trinh da cho c6 3 nghiem: 2 f)ieu kien: x 1; x - ... so thuc: ( l + y 2 ) + x ( x - 2 y ) = 5x 1 2 ( l + y ^ ) ( x - 2 y - 2 ) = 2x 2>/x^+3y-Jy^+8x-l=0 x(x + 8) + y ( y + 3 ) - 1 3 = 0 18 Lai gidi 1 Neu X = 0 thi he da cho l + y2=0 , , he nay v6 nghiem (l + y 2 ) ( 2 y - 2 ) = 0 ^ ' Neu X ^ 0 thi chia ca hai ve ciia ca hai phuong trinh da cho x ta duac: 1 + y^ 1 + y^ + ( x - 2 y ) = 5^ (x-2y-2) = 2 l.y^ 1+ + (x-2y-2) = 3 ^(x-2y-2) =2 1+y X ' '^ =... 4 J Ma nen x = 1 la nghiem d u y nhat cua phuong trinh 2 Dat f ( x ) = Taco f ' ( x ) = + V 4 ^ ,v6i 2 < x < 4 ir-^=L==-—, Ht/O^ ^ - t/(4-xf Vxe(2;4):f'(x) = 0 o x - 2 = 4 - x o x = 3 Lap bang bien thien suy ra: f ( x ) > f (3) = 2 Vx G [2;4] => Phuong trinh f (x) = \/x-2 + N / 4 - X = 2 C6 nghiem d u y nhat x = 3 V i du 18 Giai cac phuong trinh sau tren tap so thuc: 47X+21 1.2x2 2 7 1 3 ' ^ ' -... nghiem x = 3; x = 8 3 Dieu kien: x>l 4X'N/>O^-12X' Bien doi phuong trinh da cho, ta dugc: +13x>/)rri-8x-4Vx^l - 3 = 0 x4x + -[{24x + l-3j +i\-2^Jx^J =0 x V ^ ( 2 V ^ - 3 ) ' = 0 De v e t r a i bang 0 thi ta phai c6: 5 x =• Vay X = ^ la nghiem cua p h u o n g trinh da cho V i du 6 Ciai cac he p h u o n g trinh sau tren tap so thuc: 1 2x + 3 y = 5 2 < 3x2-y2+2y = 4 x ^ + 2 x ^ y + x2y^ = 2 x + 9... 3 taco ( 2 ) c ^ ^ 7_ 7 y+1 x-1 Xet ham so g(a) = ^ ^ , a e (0;+oo), ta c6 g'(a) = — v a a a" Do X -1 6 g'(a) = 0 o a = e •g(x-l)>g(2) = ^ ; y + 1 ^ 4 ^ g ( y + l)^g(4) = ^ Tu do suy ra x = 3, y = 3 Thir lai x = 3, y = 3 thoa man he phuang trinh V i du 22 Giai cac he phuong trinh sau tren tap so thuc: 1 ( 2 x 2 - 3 x + 4 ) ( 2 y 2 - 3 y + 4) = 18 X +y + x y - 7 x - 6 y +14 = 0 2 x" - y'* + 28x + 31... f(^y - 1 ) => x = sjy - 1 Voi X = ^ y - 1 thay vao p h u a n g trinh (2) giai dugc x = 1; x = 2 2 x=l fx = 2 y = 2'ly = 5 \/x^Tl - 3 x ^ y + 2 Y^/iy^Tl + 1 I = 8x^y^ (l) A J x^y-x + 2 = 0 (2) Voi y < 0 thi V T ( I ) > 0 , V P ( l ) < 0 => He phuang trinh chi c6 nghiem (x, y ) voi y > 0 Vi y > 0 nen tix p h u a n g trinh (2) cua he suy ra x > 2 Khido: (l)c:>Vx^+l-3x^ + 2 = 2 x ^ y f j 4 y ^ + 1 - 1... p h u o n g trinh da cho tro thanh: a'' - lOa' + a + 20 > 0 (*) Nhdp: a' - 1 0 a ' + a + 20 = (a' + m ) ' - [ ( 2 m + 10)a' - a - 2 0 + m ' ' Ta chon m debie'u thuc tron^ dd'u phdn tick la 1 hhng d&ng thiec: A' = l - 4 ( 2 m + 1 0 ) ( m - - 2 0 ) = 0 Gidiradugc 'T>=Y- Bat phuong trinh (*) ^a" - a - 4 ) ( a " + a - 5 ^ ^ 0 , ta giai ra duoc: i ( 3 + VT7) hoac — ^ x ^ - ( 5 - V 2 T ) 3 Dieu kien:... l o g ^ ^(x^ - 1 ) ^ l o g ^ ^ j [2(2x - 1 ) ; x^ x^ - 1 ^ 4 x - 2 X>1 X>1 X>1 < = > l < x < 2 + V3 4x + l ^ 0 Vay bat p h u o n g trinh c6 tap nghiem S = (1;2 + \l3] Ldi giai Nhan xet rang, de x-~^ 0thi truoc het x > 0, k h i do x + — > 0 va ~ > 0 x^l x^ V i vay bat p h u o n g trinh o 2x + 1 2 i 2.|x2-±>A x ^ > ~ - x X x x^ x^ 34 — -x1 X x^~2>0 X>1 x^l 2 x3-2 ^ Ci> 3 - < X < ^... x^ + X + 1 = a.b Phuong trinh da cho tro thanh: a + b = l + a b < » ( a - l ) ( b - l ) = 0a-l = 0 hoac b-l=0a = l hoac b = l + Vdi a = l t h i V x - 1 = l < : i x - l = lx = 2 + V6i b = l thi Vx^+x^+x + 1 = l » x ' ' + x ^ + x = Oc=>x^x^+x + l j = 0x = 0(loai,do \2 1 + - > 0, vol X > 1) hoac x + X +1 = 0 (phuong trinh nay v6 nghiem vi x + x +1 = x+— 4 2, moi x ) V Vay phuong trinh da