Giới thiệu phương pháp và kĩ thuật ôn nhanh thi đại học đạt điểm cao môn Toán: Phần 1

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"Phương pháp và kĩ thuật ôn nhanh thi đại học đạt điểm cao môn Toán" cuốn sách hay dùng để ôn tập những kiến thức môn toán để đạt điểm cao. Bên cạnh đó còn hướng dẫn các em cách giải bài tập toán nhanh và hiệu quả, nhằm mục đích giúp các em đạt điểm cao nhất trong các bài kiểm tra, kì thi sắp tới.

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Nội dung Text: Giới thiệu phương pháp và kĩ thuật ôn nhanh thi đại học đạt điểm cao môn Toán: Phần 1

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 trong 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. - 8 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 do 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 do thi ham so; tint tren do thi nhirng diem c6 tinh chat cho truac, luang giao giua hai do thi (mot trong hai do 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 do trong mat phang: Xdc dinh toa do 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 do trong khdng gian: Xdc dinh toa do 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 3
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 1 (2 diem): a) Khao sat su bien thien va ve do thi cua ham so. b) Cac bai toan lien quan den u n g dung cua dao ham va do 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 do thi ham so; t i m tren do thi n h i i n g diem c6 tinh chat cho truac, tuong giao giira hai do thi (mot trong hai do thi la d u o n g thang)... Cau 2 (1 diem): Cong thiic lugng giac, phuong trinh lugng giac. Cau 3 (1 di£m): Phuong trinh, bat p h u o n g trinh, he phuong trinh dai so. Cau 4 (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 5 (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 6 (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 trong hai phan (phan A hoac phan B) A. Theo chucmg trinh Chuan: Cau 7a (1 diem): Phuong phap toa do trong 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 do trong 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. 5
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 do trong mat phSng: - Xac d i n h toa do 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 do trong 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 quan. px + q - 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 • Phuong phap ehung: Su d u n g p h u o n g phap the - He hai p h u o n g trinh - He ba p h u o n g trinh 6
3) He doi xung loai 1 Phuong phap chung: Dat an phu a = x + y; b = xy . 4) He doi xung loai 2 Phuong phap chung: T r u tung vehai phuong trinh da cho nhau ta dugc: (x - y).f(x; y) = 0 5) He phuong trinh dang cap bac hai Xet truong hop y = 0. 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 trong he va the vao phuong trinh con lai. • Nhqn dang: Phuong phap nay thuong hay su d u n g k h i trong he c6 mot phuong trinh la bac nha't doi voi mot an nao do. 2) Phuong phap cong dai so 3) Phuong phap bien doi thanh 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 do X g la nghiem. • Voi X > Xg o f(x) > f(xQ) = k, do do phuong trinh v6 nghiem. • Voi X < Xg f(x) < f(xQ) = k, do do phuong trinh v6 nghiem. • Vay X g la nghiem duy 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 nhau va xac d i n h X g sao cho f ( X g ) = g(xg). Buac 3: Vay Xg la nghiem d u y nhat ciia phuong trinh. 7
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 do f(u) = f(v) o u = v . Cac vi du| V i du 1. Giai cac p h u o n g trinh sau tren tap so thuc: 1. V 5 x - l - 7 3 x + 13 2. +-=-2x-4 + - 3 V X x 3. ~ + . X =x+. 2x-- 4. 2 ' " ' + 2 ^ ' ' - 9 . 2 ' ' ' + 1 8 3 = 2^''*'+2''"+9.2^*" X V X Lcri giai 1. Dieu kien: x > — . 5 Phuong trinh da cho tuong d u o n g v o i VSx - 1 - V3x+ 13 = -if VSx - 1 - V3x +13)(VSx - 1 + V3x + 1 3 ) . Truang hap 1: V S x - l - %/3x + 13 = 0 o V s x - l = V s x T l S , binh phuong hai ve roi riit gon ta dugc x = 7 (thoa man). Trucrng hap 2: V S x - l + V3x + 13 = 6 (1) Neu x > l thi V T ( l ) > >/4 + Vl6 = 6 , con neu x < l thi VT ( l ) < ^/4 + Vl6 = 6 . De thay x = 1 la nghiem p h u o n g trinh (1). Vay p h u o n g trinh ban dau c6 hai nghiem x = 1 , x = 7 . 2, Dieu kien: 1 + - ^ 0 o > 0 o x < - 2 hoac x > 0 . x X K h i do p h u o n g trinh da cho viet lai: x ^ l + - = -2x^ - 4x + 3 . + Vdi X > 0 , phuong trinh o > y x 2 + 2 x = - 2 ( x 2 + 2 x ) + 3 . Dat: t = V x ^ + 2 x , t > 0 , ta c6 2t^ + 1 - 3 = 0 o t = 1 hoac t = - - , doi chieu dieu kien ta dugc t = l , tuc la phai c6 Vx^ + 2 x = 1 o x^ + 2x = 1 , phuong trinh nay c6 nghiem x = -l + \f2 thoa man dieu kien. + V d i X ^ - 2 , p h u o n g t r i n h o - V x ^ +2x = -2(x^ + 2x) + 3 . Dat t = > / x ^ + 2 x , t > 0 , ta c6: 2t^ - 1 - 3 = 0 o t = | hoac t = - l , d o i chieu dieu kien ta dugc t = ^ , tuc la phai c6 4(x^ + 2x) = 9 4x^ + 8x - 9 = 0 , phuong trinh nay c6 nghiem x = ~ ^ '^'^''^ man dieu kien. 8
-4-V52 Vay phuong trinh da cho c6 nghiem x = ; X = -1 + r/2 . 3.Dieu kien: x # 0 , x - i > 0 , 2 x - - > 0 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 > 0 . Va-vb - X Khi do ( l ) o x — = (2) X x-UJ2x-^ ^ 4 4 * Neu X — > 0 thi — x < 0, khi do t u (2) ta c6 ve trai Ion hon 0, ve phai be hon 0, v6 li. X x 4 4 * Neu X — < 0 thi — x > 0, khi do t u (2) ta c6 ve trai be hon 0, ve phai Ion hon 0, v6 li. X X 4 2 Vay X — = 0 o x - 4 = 0 = > x = 2 (thoa man dieu kien). X Phuong trinh da cho c6 nghiem duy nhat x = 2 . 4. Phuong trinh da cho tuong d u o n g v6i (2'" +2 - 2 ( 2 " +2 - 7 2 ( 2 " +2 ") + 185 = 0 o (2"+2")'-2 - 2 ( 2 " + 2 - " ) ' - 7 2 ( 2 " + 2 ") + 185 = 0 2 , phuong trinh da cho tro thanh: t^ - 6t^ - 72t +189 = 0 c = > ( t - 3 ) ' ( t ' + 6 t + 2 l ) = 0 o t = 3, v i t ' + 6 t + 21 > 0 , V t > 0 . Voi t = 3 o 2 " + 2 - = 3 « 2 " = ^ ^ « x = log/^*^^ / /— 3±V5 Vay p h u o n g trinh da cho c6 nghiem duy nhat x = log^ V i d u 2. Giai cac phuong trinh sau tren tap so thuc: 1. log25(x^-8x + 15)^ = | l o g ^ ^ + log5|x-5 2 2. l o g , (4 - x ) ' + ^ log, (x + 2 ) ' = 3 + log, (x + 6y 3. 4" (V2.6"-4" + V4.24" - 3 . 1 6 " ) = 27" -12" + 2.8" 9
LOT gidi 1. Dieu kien: x > 1;x # 3;x # 5 . Phuong trinh da cho tuong duong v o i logs x'' - 8 x + 15 = ' 0 g 5 ^ + l0g5 X - 5 X-1 o l o g g x - 3 +log5 X - 5 = l o g 5 - y - + l o g 5 | x - 5 o 2 | x - 3 = x - l < : = > 2 x - 6 = x - l hoac 2 x - 6 = l - x x = - (vi x^5) 3 ^ ' 7 Vay p h u o n g trinh c6 nghiem x = - . 3 2. Dieu kien: - 6 < x < 4 va x ;t - 2 (*) I^huong trinh da cho 3 1 o g ^ ( 4 - x ) - 3 1 o g ^ x + 2 = 3-31og^ {x + 6) o log^ (4 ~ x) + iog^ (x + 6) = 1 + log^ |x + 2| o (4 - x ) ( x + 6) = 4|x + 2 4(x + 2) = ( 4 - x ) ( x + 6) hoac 4(x + 2) = - ( 4 - x ) ( x + 6) (vi (*) nen ( 4 - x ) ( x + 6 ) > 0 ) + V6i x - + 6 x - 1 6 = 0c:>x = 2,x = - 8 . + V6i x ' - 2 x - 3 2 = 0x = l + 733,x = l - V 3 3 . Vay p h u o n g trinh c6 hai nghiem x = 2; x = 1 - V33 . 3. Dieu kien: 2.6" - 4" > 0 va 4.24" - 3.16" > 0 « x > log, - Ta C O cac danh gia sau: 4"^2.6" - 4 " = 2".2"^2.6" - 4 " < ^ 2 " (4" + 2.6" - 4 " ) = 12" 4" V4.24"-3.16" = 4" ^^4" ( 4 . 6 " - 3 . 4 " ) < ^ 4 " (2" + V 4 . 6 " - 3 . 4 " ) = - 8 " + i 2 \ 2 " V 4 . 6 " - 3 . 4 " < - 8 " + - 2 " ( 4 " +4.6" - 3 . 4 " ) = 12". 2 2 2 4 ^ ' Do do 4" (V2.6" - 4 " + ^ 4 . 2 4 " - 3 . 1 6 " ) < 2.12", dSng thuc xay ra k h i x = 0. 2 7 " - 1 2 " + 2 . 8 " = 27" +8" -12" > 3N/27".8".8" -12" = 2.12", dau bang xay ra khi x = 0 Vay p h u o n g t r i n h nay phai c6 nghiem x = 0 V i d u 3. Giai cac phuong trinh sau tren tap so thuc: 1 . 2 x ^ + 6 x ^ + 6 x + l = 3| x + 2 2. 8 log4 N/X^ - 9 + 3^2 log4(x + 3)^ = 10 + log2(x - 3)^ Lai gidi l . D a t 3|^i±?.=y + i c ^ ^ i ± i = y 3 + 3 y 2 + 3 y + i o x = 2 y ^ + 6 y ^ + 6 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 2 x +6x + 6 x + l = y + 1 y = 2x +6x +6x (2), Ttr (1) va (2) s u y ra 2x^ + 6x2 + 6x - (2y^ + 6y2 + 6 y ) = y - X o 2(x^ - y^) + 6(x2 - y ^ ) + 7(x - y ) = 0 o (x - y ) ( 2 x 2 + 2y2 + 2 x y + 6x + 6 y + 7) = 0 o x = y h o a c 2x^ + 2 y 2 + 2 x y + 6x + 6 y + 7 = 0 (3) P h u a n g t r i n h (3) v 6 n g h i e m v i 2x^ +2y^ + 2 x y + 6x + 6 y + 7 = ( x + y)2 + 4 ( x + y ) + 4 + (x + l ) 2 +{y + lf +1 = (x + y + 2 ) 2 + ( x + l ) 2 + ( y + l ) 2 + l > l 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 = 0 o ^^2x2 + 6x + 5) = 0 C5> x = 0. V a y X = 0 la n g h i e m p h u o n g t r i n h . 2. D i e u k i e n : x < - 3 hoac x > 3 . 81og4 7x2 - 9 + 3 j 2 1 o g 4 ( x + 3)2 = 10 + \og^(x -sf 2log2(x2 - 9 ) + 3 j l o g 2 ( x + 3)2 = 10 + l o g 2 ( x - 3 ) 2 l o g 2 ( x + 3)2 + 3^/log2(x + 3)2 - 1 0 = 0 . D a t t = 7 l o g 2 ( x + 3)2 > 0 . Ta CO p h u a n g t r i n h : t^O o t = 2 o l o g 2 ( x + 3)2 = 4 o (x + 3)2 = 16 X = - 7 h o a c x 1. t2+3t-10 = 0 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 = - 7 . V i d u 4. 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 : 1. x^ + 3x2 + 9x + 7 + ( x - i o ) V 4 - x = 0 2. V x ^ + yjx^ +x^ +x + \ 1 + Vx'* - 1 Lai giai 1. D i e u k i e n : x < 4 . Bien d o i p h u a n g t r i n h ve: (x +1)"' + 6(x +1) = V 4 - x . ( 4 - x + 6). Dat u = x + l , v = \ / 4 - x ^ 0 . T a c 6 p h u a n g t r i n h : u ' ' + 6 u = v'' + 6v o ( u ' ' - v^) + 6 ( u - v ) = 0 u = V hoac u2 + u v + v2 + 6 = 0 . Trudmg hQfp 1: u2 + u v + v2 + 6 = 0 ( u + —)2 + + 6 = 0 , v6 nghiem. 2 4 x + 1^0 -3+N^ Trumtghgp 2: u = v V 4 - x = x + 1 X = 4 - x = x2+2x + l 11
2. Dieu kien: x > 1. Dat a = V x - l , b = Vx^ +x-^ + X + 1 , voi a > 0 , b > 0 , t a c 6 V x ' * - l = \ / ( x - l ) ( x ^ + x 2 + x + l ) = V ^ ^ . V 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 X > 1) hoac x + X +1 = 0 (phuong trinh nay v6 nghiem vi x + x +1 = 1 + - > 0, vol x+— 4 2, V moi x ) Vay phuong trinh da cho c6 nghiem duy nhat x = 2. Vi du 5. Giai cac phuong trinh sau tren tap so thuc: 1. X + V 4 " X ^ = 2 + 3xv'4-X- 2. 2 x ' - 6 x + 10-5(x-2)Vx + l = 0 3. 4 ( N / X + T - 3 ) X ' + ( I 3 V X + 1 - 8 ) X ~ 4 V X - 1 - - 3 -0 1. Dieu kien: -2 < x < 2. Dat t = x + \/4-x' => t" = 4 + 2x\l4~x- => wli-x' =~ ^, 2 Phuong trinh da cho tro thanh: t = 2 + 3 - — ^ c > 3 t ' - 2 t - 8 = 0ci>t = - -4h o a c t = 2. 3 2-x>0 x=0 + Voi t = 2, ta c6: x + V 4 - x ' = 2 ^Ji-x' = 2-x o 4~x^ = 4 - 4 x + x^ o x = 2' 4 / 4 / 4 + Voi t = — , ta CO x + v 4 - x ^ = — v 4 - x ^ = 3 x X = -2±Vl4 9x'+12x-10 = 0 x= -2-Vi4 Vay phuong trinh da cho c6 ba nghiem x = 0; x = 2; x 2. Dieu kien: x > - 1 . Phuong trinh da cho tuong duong 2(x ~ 2)" + 2(x +1) - 5(x - 2)>/x + l = 0 12
2(x - 2) - v/x + 11 ll^jx + l - (x - 2)1 = 0 « 2(x - 2) - Vx + 1 = 0 hoac 2>/x + l - ( x - 2 ) = 0 . x^2 x>2 + Voi V x + 1 =2(x-2) 50x = 3 4 x ' - 1 7 x + 15 = 0 X = 3; X = — 4 x>2 x>2 + Voi 2 V x + l = x - 2 „ , „ X = : [x'-8x = 0 [x=0;x = 8 Vay phuong trinh da cho c6 nghiem x = 3; x = 8 . 3. Dieu kien: x>l. Bien doi phuong trinh da cho, ta dugc: 4X'N/>O^-12X' +13x>/)rri-8x-4Vx^l - 3 = 0 x4x + -[{24x + l-3j +i\-2^Jx^J =0 x V ^ ( 2 V ^ - 3 ) ' = 0 5 De v e t r a i bang 0 thi ta phai c6: 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: 2x + 3 y = 5 x ^ + 2 x ^ y + x2y^ = 2 x + 9 3' +6.2' = 11 + 4'' 1. 2. < 3. < 3x2-y2+2y = 4 x^ + 2 x y = 6x + 6 2>'+4.3^ = 7 + 9" Lai gidi 1. 2x + 3 y = 5 (1) 3x^-y2+2y = 4 (2) 5-3y Tij phuong trinh (1) suy ra x = , khi do (2) tro thanh 5-3v -y2+2y-4 =0 o 3 ( 2 5 - 3 0 y + 9 y ' ) - 4 y ^ + 8 y - 1 6 = 0 o 2 3 y 2 - 8 2 y + 59 = 0 o y = l hoac y = — 23 31 59 Vay he phuong trinh c6 nghiem la ( x ; y ) = (l;l); '23'23 x ' * + 2 x ^ y + x^y^ = 2 x + 9(l) 2. X ' + 2 x y = 6x + 6 (2) 13
Trumig hap 1: x = 0 khong thoa man (2) 6x + 6 - x Trudng hap 2: x^O, (2) suy ra y = , the'vao (1) ta dugc , 2x ( 2 ^ ( 2 ^2 6x + 6 - x + x' 6x + 6 - x = 2x + 9 \.x ) V 2x J
Trumg hap 1: x = y , the vao phuong trinh (2) ta dugc + 4 x - 1 2 = 0 o X = - 6 hoac x = 2 . 2 2 V + xv + X Trumtg hop 2: = - 1 => xy < 0 . y Khi do (2) o 2x^ + 4y^ ~ 9xy + 4x - 16y = -36 o 2(x +1)^ + 4(y - 2f - 9 x y = -18 . Truong hop nay khong xay ra do xy < 0 => 2(x +1)^ + 4{y - 2)^ - 9xy > 0 . Vay nghiem cua he p h u o n g trinh la ( x ; y ) = (2; 2); ( - 6 ; - 6 ) . V i du 8. Giai cac he phuong trinh sau tren tap so thuc: 2- f x ^ + y ^ + ^ ' ^ y =16 y 1- 1. x+y ^ x + y = x^ - y 1=2 X Lai giai x ^ + y ^ + ^ = 16 (1) 1. x+y ,2 /x + y = x -y (2) Dieu kien: x + y > 0 . Phuong trinh (1) (x^+y^)(x + y) + 8xy = 16(x + y) (x + y ) ^ - 2 x y (x + y) + 8xy = 16(x + y) (x + y) (x + y ) ^ - 1 6 - 2 x y ( x + y - 4 ) = 0 (x + y - 4 ) [ ( x + y)(x + y + 4 ) - 2 x y ] = 0 . Truong hap 1: x + y - 4 = 0 the vao (2) ta dugc: x ^ + x - 6 = 0 c ^ x = - 3 = > y = 7 hoac x = 2 => y = 2 . Truong hap 2: (x + y)(x + y + 4) - 2xy = 0 x^ + y^ + 4(x + y) = 0 , v6 nghiem do dieu kien. Vay nghiem cua he la (x; y) = (-3; 7); (2; 2 ) . 2. Dieu kien: x^- ,y> - . 2^ 2 T r u ve hai phuong trinh ta duoc - + 2-1-.12-1=0 Vx Vy V y V x Vy->/x y-x y-x = 0 o = 0. Vxy(^/^ + ^/y) xy 2-UJ2-' 15
Trucntg hop 1: y - x = 0y = x the vao (1) ta duoc —j= +. 2 — =2. Vx V X Dat t = t > 0 ta duoc V2-t^ = 2-1 2 - t ^ O t X = l,y = 1 . Truang hap 2: = 0 ,v6 nghiem do dieu kien. 7^(V^ + 7 y ) xy 2-UJ2-^ Vay he c6 nghiem duy nhat (x; y) = (1; 1). Vi du 9. Giai cac he phuong trinh sau tren tap so thuc: 2x^ + xy = y^ - 3y + 2 (xy+ l)x^ +(x + l)^ = x^y + 5x 1. < 2.< x2-y2=3 4x^y + 7x^ +2x^^y + l =2x + l Lai giai 1. 2x^+xy = y ^ - 3 y + 2 c : . y ^ - ( x + 3)y + 2-2x^ = 0. Coi day la phuong trinh bac hai an y tham so x, ta c6: A = (x + 3 ) ^ - 4 ( 2 - 2 x ^ j = 9x^+6x + l = ( 3 x + l f , s u y r a : y = 2x + 2 hoac y = -x + l. y = 2x +2 Jy = 2x +2 Truang hap 1: x2-y^=3 [3X^+8X + 7 = 0(A' = -5
4x^ + 7 x 2 + 2 x ^ f ' - ^ + l =2x + l V X y x-1 o(x-l)^-2x^ = 0 x - 1 - 2 x(x-l) =0 X Tniang hap 1: x - 1 = O o x = l=>y = - l . 1 Trumghop2: x - l = 2 x o x = - l = > y = 3 ; x = — =>y = 3 . Vay nghiem cvia he da cho la: (x; y) = ( l ; - l ) ; ( - l ; 3 ) ; ( - ; 3 ) . 3 Vi du 10. Giai he p h u o n g trinh sau tren tap so thuc: ^xy + (x - y X T x y - 2 ) + V x = y + ^ (x + l ) ( y + .Jxy + X - x^) = 4 Lcri giai Dieu kien: x y + ( x - y ) ( , y x y - 2 ) ^ 0 va x ; y ^ O . Phuong trinh dau ^xy + (x - y)(^^/y^ 4 ^ Ta CO y + Jxy = x' - x + = (x -1)^ + x + 1 +• -2>2 x +1 x+ l y+ >0 7xy + ( x - y ) ( 7 ^ - 2 ) + y V ^ + N/Y P h u o n g t r i n h (*) x = y, thay vao p h u o n g t r i n h t h u hai ciia he, ta d u g c x^ - 2x^ - 3 x 4 - 4 = 0 x = l hoac x = ^ - ^ ^ ^ hgp v o i dieu kien ta dugc l + \/l7 x = 1, x = . 2 Vay he da cho c6 hai nghiem (x; y) la: (1; l};__Lt.^^^^y^ ^
V i d u 11. Giai cac he phuong trinh sau tren tap so thuc: + + x + y = 18 ^ x"^ + y^ + 2(x + y) = 7 1. i xy(x + l ) ( y + l ) = 72 y ( y - 2 x ) - 2 x = 10 Lai giai ( x 2 + x ) + { y 2 + y ) = 18 1. H § p h u o n g trinh da cho ( x 2 + x ) ( y 2 + y ) = 72 x ^ + x = a, a > - - Dat \ 4 y ^ + y = b, b ^ - i 4 a + b = 18 a = 6 , b = 12 Ta duoc ah = 72 a = 12, b = 6 a =6 X +x = 6 X = 2, X= -3 Trumghap 1: b = 12 y ^ + y = 12 [y = 3 , y = - 4 X = 3, X= -4 Truong hap 2: Doi vai tro cua a va b trong truong hop I, ta dugc y = 2,y = - 3 ' Vay nghiem cua he la (x;y) = (2; 3); {2;-4); (-3; 3); (-3;-4); (3; 2); (-4; 2); (3;-3); ( - 4 , - 3 ) . x^ +y^ +2(x + y) = 7 (x + l ) ^ + ( y + l)^ = 9 2. He phuong trinh da cho y ( y - 2 x ) - 2 x = 10 (y-xf-ix + lf =9' Dat a = x + l , b = y + l = > b - a = y - x , t a dugc he (b-a)^-a^ =9 = > a ^ + b ^ = ( b - a ) ^ - a ^ o a ^ = -2aba = 0 hoac a = - 2 b . + Vdi a = 0 => b = ±3 =:> X = - l , y = 2 hoac x = - l , y = - 4 + V 6 i a = - 2 b ^ 5 b 2 = 9 o b = 4-,a = — ^ hoac b = -4=,a = -^ 3 u - 1 6 3 hoac X = - 1 + - 7 = , y = - 1 - Vay nghiem cua h § la (x; y) = (-1;2); (-1; - 4 ) ; ( - 1 — ^ ; - 1 + A ) . (_i + ^ . _^ 3 V5 V5 V5 sl5 V i d u 12. Giai cac he phuong trinh sau tren tap so thuc: ( l + y 2 ) + x ( x - 2 y ) = 5x 2>/x^+3y-Jy^+8x-l=0 1. 2. ( l + y ^ ) ( x - 2 y - 2 ) = 2x x(x + 8) + y ( y + 3 ) - 1 3 = 0 18
Lai gidi l + y2=0 1. Neu X = 0 thi he da cho , , 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^ l.y^ + ( x - 2 y ) = 5^ + (x-2y-2) = 3 1 + y^ 1+ (x-2y-2) = 2 ^(x-2y-2) =2 1+y X ' '^ = ^ " 2 y ~ 2 , t a d u o c h e phuong trinh: u +v =3 [u = l u=2 uv = 2 •!v = 2 hoac v =l u=l l.y^ + Voi = 1 [x = l + y2 rx = 2y + 4 v =2 x-2y-2 =2 [^=2y + 4 [y^-2y-3 =0 x = 2y + 4 x =2 „ fx = 10 hoac y = - l ; y = 3 ^ ly = - l y =3 u =2 1+ + Voi ^ =2 _ J l + y2=2x [x = 2y + 3 v =l x-2y-2 =l >l^ = 2y + 3 [y2-4y-5 = 0 x = 2y + 3 fx = l x = 13 ^ o •{ hoac y =- l ; y=5 |y = - l " y =5 Vay he da cho c6 4 nghiem ( x ; y ) = ( 2 ; - l ) ; ( 1 0 ; 3 ) ; ( l ; - l ) ; ( 1 3 ; 5 ) . 2. Dieu kien: x^ + 3y ^ 0, y^ + 8x > 0 . Dat u = J x 2 + 3 y , v = J y 2 + 8 x ( u , v > 0 ) 2u-v = 1 v = 2u-l v = 2u-l He da cho tro thanh: u^+v2=13 |u2+v2=13 [ u 2 + ( 2 u - l ) 2 =13 fv=2u-l v =2u-l u =2 5u^-4u-12 = 0 u = 2;u = — lv = 3 5 4-x^ y=- u=2 V x ^ 3 y =2 x^ + 3 y = 4 Vol lv = 3 i/y^Tsx = 3 y^+8x = 9 4-x^ + 8x=9 19
4-x^ y = x^-8x^+72x-65 = 0 (x--l)(x + 5)(x^ - 4 x + 13) = 0 4-x^ x= l x = -5 < hoac y= l y = -7 X = -5; X = 1 Doi chieu v d i dieu kien ban rfau ta thu dugc nghiem ciia he p h u o n g trinh la: (x;y) = ( - 5 ; - 7 ) , ( l ; l ) . V i du 13. Giai cac he phuong trinh sau tren tap so thuc: 1 2x x+y ^Ux-y -y]y-x =1 -+- 1. 2. 3x 3y^ 2x^+y^ (x>0,y>0). 7yjy-x + 6y - 26x = 3 ^2y + 2 0 = x ^ + x - 5 Lai giai ^Ux-y -yjy-x =1 ^llx-y-^y-x =1 1. 7 ^ y - x +6y-26x = 3 7 ^ y - x - 2 ( l l x - y ) + 4(y-x) = 3 Dat u = ^ l l x - y , v = ^ y - x ; u , v > 0. u - V = 1 u = v+1 He p h u o n g trinh da cho tuong duong: 7v-2u^+4v^ =3 4v^-2(v + l)^+7v = 3 ' u = 2;v = l 1 u = v+1 V l l x - y =2 llx-y = 4 X = — i 3 5 2 u = —; V = — -X +y=1 2v^+3v-5 = 0 2 2 ^ = 2 Vay he phuang trinh c6 nghiem (x; y) = r i . 3 ^ 2'2 2. Phuong t r i n h t h u nhat ciia he tuong d u o n g v d i ^ ^ = ^ P 4 « ( 2 x ^ + y ^ ) ^ = 3 x y ^ ( x + y) 3xy^ 2x^ + y^ 2^2 2 2x2+y2^ 3xy x+ y y 2+ = 3 y 1+y \ Dat t = ^ , ta CO p h u a n g trinh: (2 + t2)2 = 3 t 2 ( l + t) c=> t ' * - 3 t ^ +1^+4 = 0 o (t -2)2(t2 +1 +1) = 0 o t = 2 » y = 2x. Thay y = 2x vao p h u a n g trinh t h u hai cua he ta dugc: 2Vx + 5 = x^ + x - 5. 20
Dat z = -J^^ > 0 . Ta CO he phuong trinh z = X + 5 z = x+ 5 z^ = x + 5 z^ = x + 5 X = z 2z = x^ + X - 5 x^ = 2 z - x + 5 x^ - z ^ = 2 z - 2 x x + z + 2 = 0. Truong hop x + z + 2 = 0 khong xay ra v i x, z > 0. Voi X = z > 0 ta duoc x = ^ ^ ^ ^ , y = 2x = 1 + V2I. Vay nghiem cua he da cho da cho la: (x;y) = ( ;2x = 1 + V i d u 14. Giai he phuong trinh sau tren tap so thuc: x^y^ + 4x^y - 3xy^ + x^ + y^ = 12xy + 3x - 4y +1 3x^ - 2 y ^ = 9 x + 8y + 3 Lai giai He phuong trinh da cho tuong duong: y^^x^ - 3 x j + 4y^x^ - 3 x j + ^x^ - 3xj + y^ + 4 y + 1 = 2 ( x ^ - 3 x + l)(y2+4y + l ) = 2 3(x^-3x)-2(y^+4y = 3 3 x^-3x y2+4y) = 3 Dat u = x^ - 3x; v = y^ + 4 y , he tro thanh: 3u~3 u = l;v = 0 (u + l ) ( v + l ) = 2 V =• 2 o 5 3u-2v = 3 u = —; -4 (u + l ) ( 3 u - l ) = 4 V = 3 x^-3x-l=0 3±Vi3 3±N/I3 X = X = + Voi u = 1; V = 0, ta c6: 2 hoac y^ + 4 y = 0 y = -4 y = 0 x^ - 3 X + - = 0 9±N/21 + Voi u = - —; V = - 4 , ta c6: 3 « x = y2 +4y + 4 = 0 [y = -2 Vay he da cho c6 6 nghiem (x; y) la: 3+ V13 ;0 3-N/T3 , r 3 + Vl3 ' 9 + V21 ( 9-V2I ( ^ ; 0 ) ; ;(^^^;-4); ;-4 ;-2 ;-2 V i d u 15. Giai cac phuong trinh sau tren tap so thuc: 1. l o g 3 ( x - 2 ) = l o g 4 ( x ' - 4 x + 3) 2. log3 |x^ + X + 1 j - log3 X = 2x - x^ Lai giai 1. Dieu kien: x > 3 . Dat l o g 3 ( x - 2 ) = t > 0 C5> x = 3 * + 2 . 21