Tài liệu Chuyên đề bồi dưỡng học sinh giỏi-giá trị lớn nhất, giá trị nhỏ nhất phan huy khải (phần 5)

0 • cua (3) bo t^+t + l (3) m-3 ok \6i t G R. Lam tiWng tif nhir tren, ta c6: c6 nghipm t > 0. Dieu do xay ra khi DoP = .c Dc'n day gpi m la gia tri luy y cua ham so: f(t) = 2. Neu m ^ 3, khi do (2) c6 nghiem khi va chi khi phiTcfng trinh (m - 3)t^ + 2(m - 2)t + m - 3 = 0 ro X N2 /g / up t^-t-3 om " ' - " y - y , X,2• + xy + y 2 (2) « ( m - 3)x* + 2(m - 2)x^ + m - 3 = 0. X Tac6: ' Ta P^3^ (1) (1) o 3 + 4x^ + Sx" = m + 2mx^ + mx"* p ^ ^ x ^ j c j ^ J j ^ khi x 2 - x y - 3 y 2 > 0 X +xy+ y Do X + x y + y < 3 , nen -, X G . (iH-x^)' iL ie uO nT hi Da iH oc 01 / min (x;y)eD| 3 + 4x^+3x^ Khim = 3,thi(l)c6dang khix'-xy-3y'<0. 5 o-/3. (6) (x; y)€D2 (x; y)6D2 TO (1), (2). (3), (4). (5). (6) di den: max P = -3 + 4v'3 va (x;y)€D min P = - 3 - 4 N / 3 . (x;y)eD K h i m = - , t h i ( l ) c 6 dang 3 + 4x^+3x'^ = - 0 x " - 2 x ^ + 1 = 0 o x = ± 1 . 2 l + 2x2 + x^ 2 Vay maxf(x) = 3 o x = 0, xeR 1 '' min f(x) = - o x = ± l xeR 2 237 ChuySn dg BDHSG Join gia tr| I6n nhSt Nhdn g\i tr| nh6 nha't - Phan Huy Khai Cty TIMHH MTV DWH Khang Vi^t xet: f^h^n xet: Me'u ta CO bai toan sau: 1. Ta da suf dung cac dinh l i ve dau tarn thuTc bac hai de t i m dieu kien sao cho (3) Co nghiem khong am. < .. i , Chtfng minh rang v d i m o i x > 0, thi f ( x ) = x + \\ + — > 2 . ' " ' 2. Ta CO the giai bai toan tren bang nhieu each khac - V Bang phifdng phap chieu bie'n thien ham so' (xem bai 11, b ^ i 1, chiTdng 4 X K h i do bai toan ddn gian hPn han. That vay: cuo'n sach nay) • Ne'u X > 2, thi ro rang f(x) > 2 yj ,, Bang phtfcfng phap lu'dng giac (xem b a i 5, chtfdng 3 cuon sach nay) • X e t k h i 0 < X < 2, thi ITx) > 2 ! : o J x ^ + - > 2 - x o x ^ + - > 4 - 4 x + x^ ( d o 2 - x > 0 ) X x iL ie uO nT hi Da iH oc 01 / - Bang phiTdng phap bat dang thtfc (xem bai 5 phan nhan xet, chtfong 3 cuo'n sdch nay). B a i 7. T i m gia t r i nho nha't cua ham so: f ( x ) = x + (!) x^ + x O 4x^ - 4x + 1 > 0 <=> (2x -1)^ > 0 . vdix>0. la&o suy ra: f(x) > 2 V x > 0 va f(x) = 2 o x = 2 Tuy nhien v d i bai toan da cho, thi lam sao ma tim ra hang so' 2. Ci'i kho cua HUdngddngidi G p i m la gia tri tuy cua ham so': f(x) = x + ^/x^ + — , x > 0. bai toan la d cho do. V i vay dung phi/dng phap mien gia t r i ham so' giai bai todn nay la each giai tif nhien va cung kha ddn gian! + Jx^ + — = m (1) up s/ x>0 Ta K h i do he sau day (an x): X 0 0 o m > 2 (chu y i\i (2) c6 m > 0). bo 2m ce 2m 2 2 4 Do do m > 2 chinh la dieu k i e n de he (2), (3) c6 nghiem Tiif do suy ra v d i m p i X > 0, ta c6: f(x) 0 <> x2. <2 K h i m = 2, t h i he (2), (3) trd thanh 4x2-4x +l =0 Tilfd6suyra: minf(x) = 2 o x = - . x>o 2 1 , '^''^ir-^-^^' n Ham so f(x) xac dinh k h i x - x^ > 0 o 0 < x < 1. 1 ^ Gpi m la gia trj tiiy y eua ham so f(x). K h i do phiTdng tiVnh sau day (an x) ssi! ( I ) e d nghiem. ( l ) o Vx-x^ = m - x (2) , ' : (2) CO nghiem k h i va chi k h i hai di/dng y = V x - x ^ va > = m - x c^t nhau. y >0 Ta c6 y = V x - x ^ <=> • o x2+y^-x=0 w. X2 < X| < m. / ww m V i X) + X2 = — fa Do do k h i m > 2 thi (3) c6 hai nghiem du'cfng Xj, X2 v d i X| > X2. • HUctng ddn giai Tacd K h i m > 2, t h i (3) c6 nghiem. M a t khac ta c6: P = — > 0; S - — = — > 0 ' T i m gia tri Idn nha't va nho nha't cua ham so tren mien xac dinh cua no. x + Vx-x^ = m .c X Bai 8. Cho ham so f(x) = x + V x - x ^ . ' { 1 X 2 2 1 =4- I (1 Vay y = Vx - x^ la nuTa dtfdng tron tam tai d i e m I - ; 0 \2 ( p h i n n k m tren true hoanh). ^^rf va ban kinh R = - . . y = m - x o x + y = m. Tir do de dang suy ra hai dudng noi tren cat nhau k h i va chi k h i nam giiTa hai difdng x + y = ()vax + y = •• x + y = in Vay (2) nghiem CO Cty TNHH MTV D W H Khang V i f gia trj nh6 nhat - Phan Huy Khii Chuyen dg BDHSG To^n gia trj I6n nha't f - Vay y m i n f ( x ) = min{f(()); 1(1)) = min{(); 0} = 0 , 0^ ^ V2 ^ V 2 + I = 4 do suy ra min f ( x ) = 0 I max f ( x ) = I 2 D o ( ) < x < 1 = > d a t x = sin'(p ( 0 < (p < ^ ). T f f d 6 x + V x - x ^ =sin^(p + 7sin^(p-sin'*(p = sin^(p + -y/sin^ (p(l-cos^ (p) 1-cos 2(0 sin 2(0 1 yjl 7l' - + = - + —cos 2(p2 2 2 2 ^ 1. Cach giai tren la sif phoi hitp kheo leo giffa hai phffdng phap m i e n gia iri s/ V 1 0 < ( p < ^ =^ _ Z ^ < 2 ( p - i ^ < ^ ^ 2 4 ^ 4 4 up 2. X e t each giai khac sau day (bang phiTdng phap chieu bien thien ham so) iyi 'am x^ + 1 - 2 X l-2x Ta CO f ' ( x ) = 1 + zVx-x^ 2Vx-x^ Ta ham so \k phifc^ng phap do t h j . • 1 , , 2Vx-x^ - ( 2 x - l ) X c l khi - < X < 1, luc do Vict l a i f ' ( x ) = 2V X - X ok .c om /g V d i O < x < - = ^ l - 2 x > 0 = > f'(X) > 0 . fa 4(x-x^)-(2x-l)^ w. ww + 2x -1) D'AU cua f ' ( x ) la diiu cua 8x - 8x^ - 1. f * Tff do ta CO bang bien thien sau: f'(x) f(x) 0 y ^ 1^ + 0 2 x - 1 > 0 , lufdo t a c o <2 Bai 9. Cho x va y la cac so thiTc thoa man dieu k i e n (x^ - y ' + 1)^ + 4x^y^ - x^ - bo • J: -:^ 0; cos(p > 0) = sin^cp + sin(pcos(p x+y=0 Nhqn xet: f(x) = 2 ' |. L a i xct each giai sau (bang phu'ctng phap Iffdng giac hoa) va a ^ 4 ~ Ta thu l a i ket qua tren! () - CO nghiem. x^ + y ^ = m (x^ + y^)2 -3(\ + y^) + 1 + 4x2 = 0 j^v^, fx2+y2=m m 2-3m +l + 4x2=0 (3) (4) Cty TNHH IVITV DVVH Khang Vi$t Chuy§n dS BDHSG Toan gia trj Idn nhat va gia tri nhd nha't - Phan Huy Kh^i 3 + >/5 m" - 3m + 1 < 0 <=> 2 Nhirvay(5)la dieukicnde (4)c6nghicm. Ta CO Qlum^^ PHIfONG PHAP SO DUNG DO THIHOAC HlNH HOC (5) ^ *' (3) <=> 4x^+ 4y^ = 4m <=> -m^ + 3m - 1 + 4y^ = 4m <=> 4y^ = m^ + m + Dt TIM GIA TRj LON NHAT VA NHO NHAT CUA HAM SO i phiTrtng phap nay dSc biet thich help vdi nhffng bai toan tim gia trj Idn nhat, Do m^ + m + 1 > 0 V m , nen lif do suy ra he (3) (4) c6 nghiem khi va chi khi 'i nho nhat cua ham so, trong do cac bicu ihifc va dieu kien cua bai toan ban !• V a y (5) chinh la dieu kien de he (1) (2) c6 nghiem. I 1( dau da tiem an nhffng ycu to hinh hoc ma thoal lien ta chffa nhin ra no. iL ie uO nT hi Da iH oc 01 / thoa man (5). . i f Vdi Idp bai loan nay, ta lhff(:(ng sff dung cac tinh chat hinh hoc sd cii'p sau „ 3 + 75 . . „ 3-75 Tif do suy ra maxP = — - — va mmP = — - — . day: - Ml • Nhan xet: B a y gid xet each giai sau day: Theo iren dieu kien da cho v(3i (x; y) c6 dang y^f - + 1 + 4x^ = 0 3(X^ + y 2 ) x^+y^ O ' x^1+y^2 —3V: = 5- - 4 X ^ 4 2, -2J (7) 75 _ 75 3+ ^ 2 +y 2 3+ 75^ = — — X Cho diem M d ngoai dffdng thang d cho Irffdc. Khi do do dai dffdng vuong 3+ .c ok 75 It!} 2 +y = giac CO chu vi vii dicn tich Idn nhat. Ngoai ra ngffdi la cung dung cac kicn ihffc sau day: i ' Dung tinh chat ciia do dai vecld, hoac tinh cha't cua ti'ch v6 hffdng cua hai w-vecld. . S«^V*1 , . :, Dfing phffdng Irinh dffdng va mtit va cac cong Ihffc tinh khoang each cua hinh hoc phSng va hinh hoc khong gian. , (U . Tim gia tri nho nhat ciia bieu ihffc p P= . n X" r +-r + r n r • ' 1< HUiiiii' dan viai i€x cac vecld sau: li = x ; - 3-75« ertH>'('^"M*-MiM. ^ Cho X, y, z la cac so ihffc dffdng va thda man dieu kien x + y + z < 1. 1* t |/] .IS ' .|J H ' Trong cac lam giac cimg noi ticp mot dffdng Iron, thi lam giac deu la tarn " a i 1. Be thi tuyen sinh Dai hoc, Cao ddnf' kho'i A) x=0 x =0 2 /g x =0 x =0 * (8) w. 3+ Trong mot tarn giac. long hai canh luon luon Idn hdn canh Ihff 3. Ta 5 <-. 4 3 ^ntlc'le- dffdng thang d. up 2 '• - - 2 ' goc kc tff M xuong d ngan hdn moi dffdng xien ke iff M xuo'ng ciing (6) 4 Tur (6) suy ra B la dffdng c6 do dai nho nhat. s/ (X^ + Trong ta'l ca cac dffdng noi hai diem A, B cho trffdc Ihi dffdng thing noi A, V 3-75 V Fa CO u + V + w = , V = f , f 1' /; Xy 1 1 1 N 'i n /J t i i d " , } UtKi.l «J!i!'ft !f''.« , • x + y+z; — + — + X y z VaymaxP= Theo tinh chat ve do dai cua vecld long, la c6 u T a thu lai ket qua tren. Phep giai nay c6n ddn gian hdfn ph6p sur dung mi^" Tff(l)c6: + V + w > U+ V + w .(1) gia tri ham so. O/l 1 Cty TNHH MTV DWH Khang Vi?t Chuyen ai BDHSG To^n gia tr| Ifln nhft v i g\i tr| nh6 nhflt - Phan Huy Khii (dal t = ( X + y + z)'. Do 0 < x + y + z < 1 => 0 < I < 1). 'I 1 (x + y + z) + — + — + X v = k2W,k2 > 0 . '1-.' Ta c6: f'(t) = 1 - — , va bang bien thien sau: y u, v, w la cac vectd cung phiTdng, cung chieu ' (3) :\iu>,ii int't ri^'^rhb ;:5iim) ,0,': 'mM' uw-i 1 1 = 81(x + y + z)^ + - ri 1 n y zj x fi',; Theo bat dang thuTc Cosi, ta c6 VAb 1 '1 1 ^1 iV —+ —+ - X z) y 1 .1 h'un ipj'O - 1 1^ - >9. HUifiig dan giai ro Cho ham so f(x) = x + yfi-x^. nay trcn mien xac dinh cua no. om .c Hiiifng dan giai ok bo ce fa 4. Tim gia tri nho nhat cua ham so f(x) = Vx^ - x + 1 + Vx^ - V 3 x + 1 , vdi w. e R. ww X Hiidng dan giai f(x)=. 1 1 1 9 Ap dung bat dang thuTc Cosi cd ban, la c 6 : - + - + - > X y z x+y+z 81 (x + y + z) 2 • ^ X 1 2J N2 (10) j. Viet lai ham so' f(x) diTdi dang sau day . j > (x + y + z)' + Tim gia trj Idn nhat va nho nhat cua ham so sach nay. bien thien ham so nhi/ sau: ZJ . mien gia trj ham so de giai bai toan trcn trong bai toan 1, §1, chU'dng I cuon 2. Thay cho phiTdng phap bat dang thtfc ta c6 the sOr dung phiTdng phap chieU - „ Bai 3. (De thi tuyen sink Dai hoc, Cao dang khoi B) phi/dng phap siir dung ba't dang thtfc de giai bai toan dat ra. —+ — + X y , > Xem Idi giai ket hdp giiJa phifdng phap do thi, hmh hoc va phifdng phap 1. Trong bai trcn ta da kct hdp phiTdng phap suf dung vectd trong hinh hoc phang V i the (x + y + z) + i tr . ; (8) (9) Nhirthe minP = >/82 o x = y = z = ^ . 1 + |y - 2| . •' bai loan nay. Xem Idi giai trong bai 13, §2, chi/dng 1 cuon sach nay. <=> dong thcfi co dau bang trong (3), (5) (6) 1 h': Ta kct hdp phi^dng phap suT dung vectd va chieu bicn ihien ham so' de giai (7) > 162. o x = y = z= - . (l .^^.si^'K^'^Ai^-piih thi tuyen sinh Dai hoc, Caoddnfi khoi B) V; 1 ZJ - Nh4n xet: ^. P = V ( X - l ) ^ +y^ + ^{X + lf+y^ /g i I'' 1 Cho x. y la cac so thifc tiiy y. Tim gia tri nho nhat cua bicu thtfc (6) , TCr gia thiet 0 < x + y + z < I = > 80(x + y + zf < 80. T i r ( 2 ) ( 4 ) ( 7 ) ( 8 ) t a c 6 P > >/82. fiiii2.(De (5) TJ —+ — + X y 1 Ta thu lai kct qua trcn. up Ttr (5) (6) suy ra 81 ( X + y + /.Y + 1 —+ —+ X y Thco bat dang thuTc Cosi cd ban thi (x + y + z) A Dau bang trong (9) xay ra . 1 'i0M> (ii 81(x + y + /.)' + [ - + - + - >18(x + y + z) U y 7.) 1 Vaytuf(lO) suyraminP= ^^min f(t) = - 8 0 ( x + y + z)^ (4) — + — + - , Ta 1 —+ — + s/ (x + y + /.)' + I'd) 1(1) Dc thay , 0 t iL ie uO nT hi Da iH oc 01 / Dau bang trong (2) xay ra o X.,... ,1 t, ^ I (2) + X - + — Xet he true toa do Oxy, va tren do xet cac diem A (1) va C(x;0),xe K. 81 Xet ham so f(l) = t^ + y, Khi do tir (1), ta c6 f(x) = CA + CB. d day 0 < t < 1. 245 Chuyen dg BDHSG ToAn gia tri I6n nhat va gia tri nh6 nhS't - Phan Huy Khai Cty TNHH MTV D W H Khang Vi§t R6 rang ta co C A + CB > A B , trong do -> O M „ = OC + AB = 2 1 V^^' siin45|^^N/2 I sin75" • 2 1 +cos 30 ,0 73 V2 + 73 V 2 Nhir the ta C O f(x) > N / ^ Vx G E . Goi C„ = A B n Ox. Ta c6 C„A + C B = 74-273 (2) AB. =J(73-1)" Ti( do suy ra minICx) = 72 o Nhir vay ncu dat Xo = OC,, , thi l'{xn) = >/2 = 73-1. rt>;yrtq o.^{^U-wwrt,:-; x = 73 - 1 . Ta thu l a i kc't quii trcn. ' xeR iL ie uO nT hi Da iH oc 01 / Nhif vay la eo min = 1+ <=:> x = Xd = 73 - 1. 4. Cac ban Ihijr x c m eo the giai bai toan trcn bang each khac ma l a i dOn giiin hdn hai each tren khong? Nhqn xet: \ ' 1. X|) C O the linh nhif sau: ,c. Bai 5. Cho x, y la cac so thiTc thoa man dieu k i c n < x + y + 2 > 0 DifcJng thang qua A B eo phlfdng trinh y+ ] X - r^! , 73 - i n ! m-^X .^m , • nhui 2. R6 rang phu'rtng phap do ihj to ro day uy life eiia no qua RJi giai trcn. < 0, k h i do ta c6 om X ••:(*>Y i " ( - x ) - V x ^ + X + 1 + Vx^ + N/3X + I. " (**) bo V i the do chi quan l a m den m i n f ( x ) , ta chi can xet khi x > 0. I S noi tren chinh la giac A B C j>f K h i do ta eo: I'M ^ MA+ MC > AC = sfl ^ Thco dinh ! i ham so sin trong A M O C , ta c6 ba max ' ^ i ! n.f OM, sin 75" sin 45" max OM' MeAABC 2 = max O A ^ ; OB^; OC^} = max {20; 16; 4} = 20 = O A ^ , minP = min O M ^ = O H ^ (d day ta ke O H 1 B C ) " '" -'t' " MeAABC Theo cong thi?e linh khoang each tif O den difdng thang y - 2x - 4 = 0, ta c6 oh j i o i :)i.n,} M . OC Ro rang (x + y ) = (x;y)eAABC X c l diem M sao cho M O C = 60"; M O A = 30", v i O M = x. f(x) = N/2 <=> M = M„ e A C . ca maxP= Do do ap dung djnh l i ham .so cosin, xet h m h vuong O A B C vd'\A = OC = ' : r diem Ta c(') x ' + y ' = OM^. T a c o r(x) = J x - - - x . l . c o s 6 0 " + l - + J x 2 - - x . c o s 3 0 " + l ^ . \„i "'^ A = ( - 4 ; 2), B = (0; 4) va C = ( - 2 ; 0). ^ -'-a w. ^ ''''Wb ' I \ t . : ^ canh), trong do ww I ^' tap y) (kc ce fa Do X < 0, ncn lit (*) va (**) suy ra IXx) > f ( - x ) . thay M(x; lam ok | - { x ) - V x ^ - x + 1 + V x ^ V 3 x + l , •.tjiri ' ^• - - - ''^ kien da cho. hdp cac .c Ncu ^ c' - phang toa do thoa m a n he dieu De /g 3. X c l C c i c h gi£u bang phifdng phap hinh hoc sau day Ta x„ + — = ~ =>xn= yfi 2 2(1+ V3) s/ Cho y = 0 + y^ Hu0 vJv»-. T i m gia t r i Idn nhat va nho nhat cua bieu thiJe P = ' "yf" 2 -x + 2y-8<() 0H = ^^.(OP^ K ! i 247 Chuy6n BDHSG Toan giA trj I6n nhal yia In nh6 nhift - Phan Huy Khii Cty TNHH MTV DWH Khang Vigt 8 V a y maxP = 20 <» M = A Binh luqn: x = -4 o y = 2 16 m i n P = = — < = > M = H<=> ; . 5 ; X = Khi (x; y ) thoa man (1), ta viet lai bieu thi?c P drfdi dang — 5 4 8 x ^ ^ x _ ^ y _ ^ ^ 3 ^ l - 2 2 , , 2 ^ rkng neu ve he true tpa do Oxy, K h o n g the c6 phiTdng phap khac hieu qua hdn phu'dng phap do thi (2) ^ va hinh hoc gian bai toan nay ; , ,» 2x + y > 2 : d day M ( x ; y) thoa m a n (1). • iL ie uO nT hi Da iH oc 01 / A^^rtjcef: X e t bai toan tiTdng ti/sau: . TCf (2) suy ra maxP = 8 + - m a x O M ^ , x + 3y < 9 Cho X , y la hai so' thifc thoa man he dieu k i e n : x > 0; y > 0. { r > > t . . . . ^ f < - . i . . 3,.... t minP = 8 + - m i n O M ^ T i m gia t r i Idn nha't va nho nhat cua b i c u thtfc P = x^ + y^ - 4x - 8y. ^ t*M'*'-,:xl'- 2 MeQ Hudiigddngiai a day Q 1^ difdng tron t a m 1(4; 3) Cac d i e m M ( x ; y) thoa man he da cho la toan bo ti? giac A B C D v d i A = (1; 9).B = (0;2), • va ban kinh R = 3. C = (0; 3), D = (9; 0). V i e t lai P diTdi dang: P = (x - 2)^ + (y - Af - De thay m a x O M ^ = 0 M ^ ; m i n O M ^ = 0 M ? 20. Men up '1 1 1 « om fa 1 M = H 3 5 x = — ; y = —. 2 2 = 32 24 y =- minP = 1 0 o M s M , = o 5 \ 1 (*) - (**) ' = -59 Cac ban c6 the tU" nghiem lai cac ket qua (*) va (**) mot each de dang. ^Hn xet: Ta co the SIJT dung phiTdng phap lifdng giac hoa de giai bai toan tren " h l 'sau: Tsan- Tiif ( I ) suy ra '''^i X - 4 = sin 9 , • vdi 0 < (p < 271 y - 3 = 3cos(p T i m gia trj Idn nha't va nho nha't ciia bieu thuTc P = 4x + 3y. Tiif do P = 4x + 3y = 4(4 + 3sincp) + 3(3 + 3 coscp) = 25 + 12sin(p + 9cos9 Htidiig ddn giai V i e t l a i dieu k i e n da cho diTdi dang (x - 4)^ + (y - 3)^ = 9. 2=10 8 B a i 6. Cho cac so thiTc x, y thoa man dieu k i e n : x^ + y^ + 16 = 8x + 6y * M = M2 o • X bo B2 = 9 ; y = 0. minP = - — Nhir the maxP = 40 o ce M = D I w. o x -A C3 T i r d o suy ra maxP = 45 o 4 X .c cua I tren C D . Tir do suy ra maxP = 8^+ 32 = 4 0 ; min P = 8 + ok ^ d day H la hinh chic'u /g m i n M I ^ = H I ^ -' 2 ^' ww i (xem hinh ve) Do 0 1 = 5 => OM2 = 0 1 + 3 = 8; O M , = 0 1 - 3 = 2. ro De thay m a x M I ' = D I ' = 65; : d day M | , M2 Ian li/dt la cac giao diem cua 0 1 va difdng tron s/ ABCD. Ta Gpi I la d i e m I = (2; 4), k h i do P = M l ' - 20, d day M ( x ; y) thuoc tiir giac • Men C) (1) (3) Ap dung ba't d^ng thufc quen bie't: V a , ta c6: Tilf d6 suy ra cac d i e m M ( x ; y) thoa man (1) n ^ m tren diTdng tron tam ta' ~-\/a^ + b^ < a s i n a + b c o s a < Va^ + b^ . d i e m 1(4; 3) va ban k i n h R = 3. T a c 6 : - 1 5 < 12sin(p + 9cos(p< 15. * ^"^^ Cliuyen BDHSG To^n gia tri I6n nha't va gia trj nhd nha't - Phan Huy KhSi Cty TMHH MTV DWH Khang Vlgt Bay gic( liT (3) (4) suy ra maxP = 40; minP = 10. xet: Ta c6 the giai bai toan tren bang phiTdng phap mien gia trj ham so Ta thu lai ket qua Ucn (vdi phep giai raft gpn gang). X + 3y-10>0 Bai 7. Cho x va y la cac so' thifc thoa man dieu kien • X +y- 6<0 X - y + 2 > 0. nhu^sau: : ,•„ ,.v,.;;,r \ iL ie uO nT hi Da iH oc 01 / ' \. ,^ ' : \• / [x-y+2> CO nghiem. ' JM.J '•• \-- • • !' -y+ 2 = 0 X • f^, j j ^ , ^ . j j ~ ' ^ V i l M ' ( l ) ( 2 ) (3) ( 4 ) « , y+ ra-10>0 -y + m - 6 < 0 o y>10--m (6) y>m-6 (7) ''M- y ^ i ^ . (8) «^^-^.'^ Ta , ' Ne'um>8thi 1 0 - m < 6 - m = : > ( 6 ) ( 7 ) o y > 6 - m y>10-m ro i , /g , ' • y<- A ^ • ' r^ m + 2 khim<8 . , Vay ( 5 ) ( 6 ) ( 7 ) ( 8 ) o y >m- 6 om lit; ' Neu m < 8 thi 10 - m > m - 6 => (6) (7) o y > 10 - m .c i' \r + IS \ • up s/ / 1^-t) (4) x = m - 2 y (5) = m-2y Xet he (6) (7) ta CO / (1) T u f ( l ) s u y r a x = m - 2y. • -3y + m + 2 > 0 / fX + 2y = m ^ ^'.^^^'.1 i Goi m la gia tri tiiy y cua P, khi do he ,sau day (an x; y) • ^ ^ "^^ 10 > 0 (2) x +y-6<0 (3) Tim gia tri Idn nha't cua bieu thiJc P = x + 2y. .\ dhn giai ; rn + 2 k h i m > 8 ok Gia su" M(x; y) la diem, trong do x, y thoa man he dieu kien da cho. bo De dang tha'y rang khi do mien D cac diem M(x; y) nhu" vay du"dc bieu dicn ce tren mat phang toa do xOy la toan tarn giac ABC, vdi cac dinh A(4; 2); w. - Tim gia trj Idn nha't va nho nha't cua tham so m, sao cho diTdng thang x + 2y = m cat tam giac ABC. Bang each tinh tien difdng thang x + 2y = 0 (theo chieumuiten) •' m +2 ^10-mo8>m>7 '•, ww Bai loan da cho trcf thanh: fa B(2;4)vaC(l;3) filf d6 suy ra khi m < 8, he (5) (6) (7) (8) c6 nghiem khi fm«'l---x1 I C6n khi m > 8, he (5) (6) (7) (8) c6 nghiem khi Nhir vay he (5) (6) (7) (8) c6 nghiem (ttJc Ih he (1) (2) (3) (4) c6 nghiem) khi chi khi 7 < m < 10. Dieu dau tien ma difrfng thang x + 2y = m gap tam giac ABC chinh la die'm >m-6ol0>m>8 / w ^ ; ' Tir do suy ra maxP = 10 o x = 2; y = 4; minP = 7 o x = : l ; y = 3 ^^nh luqn: R6 rang phuTcfng phap giai bang do thi va hinh hoc to ro hieu qua Khi do m = 1 + 2.3 = 7. Diem cuo'i ciing ma du'dng x + 2y = m gap chinh la diem B(4; 2). Khi do m = 2 + 2.4 = 10. Vay maxP =10<=>x = 2;y = 4; min P = 7 < = > x = l ; y = 3. AAB*- hcfn h^n so vdi phifdng phap mien gia tri ham so. j ^ ^ i 8. Cho X, y la cac so' thiTc thoa man dieu kien sinx + siny = ^ . Tim gia tri Idn nha't va nho nha't cua bieu thtfc P = cos2x + cos2y. ai BDHSG Toan gia tr| I6n nhS't vi g\& tr| nh6 nhSit - Phan Huy Hiidng ddn gidi Khai Cty TNHH MTV DWH Khang Vi§t xet: Ta c6 the suT dung phuTdng phap "chicu bien Ihien ham so " de giai bai toan tren nhU'sau: Dat u = sinx; v = siny. Khi do ta c6 cos2x + cos2y = 1 - 2sin^x + 1 -2sinV = 2 - 2(u^ + v^). Bai loan da cho trd thanh: \'y A-.:.,-':! \ . Tim gia tri Idn nha't va nho nha't cua bieu thiJc Q = + v - — u (I) 2 ; Do do fa ce 4 (d day H 1^ hinh chicu cua O ircn AB). Tiirdd suy ra: 1 1 u = — sinx = — max? = 2 - - = - o M s H < : : > 4 4 :i lit 4 4 1 1 smy = V = — 4 u = —;v = l 4sinx = -—;siny = l 2 M s A m m ? = 2 - - = — <=> 2 2 1 MsB 1 u = 1; V = — sinx = l;siny = -— 2 w. 8 ww MeAB s/ Ta /g ok bo min OM^ - O H ^ 4= -1 B om 2 MeAB Ta CO ngay max OM^ - O A ^ = 0 B ^ = 1 + - = - , MeAB u up "\ 1 O 1 .c MeAB u ^ T -^ i • -1 7 Vay maxP = f v4. 4 'il'::i 0 • ^ » 1P u=— 4 ^=4 ' u =— ; v =l 2 = — <=> minP = min< f . 2 ;f(l) 2 21 21 u~l;v = - - . 2 - min< : —gang va ddn gian! Ta thu lai ke't qua trcn. Phep giai cung—gpn fiaf 9. Cho bon so thi/c x, y, z. t thoa man dieu kien x + y = 6; + t^ = 1. ""im g'la tri nho nha't cua bieu thtfc P = x' + y^ - 2xz - 2yt Htidngddngiai t, , , Xet he true toa do Oxy. Khi do diem M(x; y) vdi x + y = 6 n^m trcn dir5ng i h ^ n p X + y = 6; con diem N(z; t) vdi r + t^ = 1 nlm trcn diTiIng tron ddn vi A + y^ =; 1 tam tai O va ban kinh bhng I .J Chuyen dg BDHSG T(rfn gi^ tr| I6n nhtft g\i trj nh6 nhift - Phan Huy Khi\ Cty TNHH MTV DWH Khang Vi^t d day H la hinh chie'u cua O Vie'l l a i bieu tMc P diTdi dang P = (X - zf + (y - t ) ' - (z' + r ) = (X - zf + (y - t ) ' - 1 (1) „ tren dtfdng thang X + y = 1. /. , 72 ' ^'^A Ro rang do O H = — , nen ta c6 t i r ( l ) Tir(l)suyra P = MN^-1. (2) V a y minP = ( m i n M N ^ ) - 1, P= Vx^+z2+Vy' + t 2 > ^ . M ( x ; y) Ihuoc di/cJng thang X+ y t= 6, con N(z; c> O M , M N cung phu-png, cung chicu va N = H . . iron ddn v i . K e , . 0M„ 1 di^ng \g X + y = ^'^'I'a?' f;Tac6: ; ' OM = (x;y); MN=:(y;t). Do vay O M , M N cung phiTPng, ciing chieu va N s H k h i va chi k h i ''^ " ' (3) y = t. va gia suf , . Mat khac N(x + y ; z + t) = H OMo cat dirc:fng s/ N„. up K h i do CO ngay (3) /g M„N„ = 3 ^ - 1 => MoNf, = 19 - 6^2 . _ ^ i ^ ^ ok o M = M„;NsNo<=> r X = z= a Tir (3) (4) suy ra 2 X = z= a [z = l = 3 Vay minP = ^ ce Nhgn xet: LcJi gitii bkng phifdng phap do tfii that trong sang va gon gang! fa ww each giffa hai d i e m tren mat phang tpa do. w. + t^ = 1 vii bieu thtfc P gdi ta ngliT den khoiing T i m gia tri nho nhat cua bieu thiJc P = 7x^ + z ^ + • • - VK!; ... f^^^,^g giai 2 X, y, z la ba so thifc dtfdng va thoa man dieu k i e n : xyz(x + y + z) = 1. Tim gia trj nho nhat cua bieu thiJc: P = (x + y)(x + z). * .t s^' ^-^i" Hudng ddn giai T r 6 n mat phang toa dp xOy xet hai diem M ( x ; z) va N(x + y ; z + t) ^ t n tam giac A B C c6 3 canh Do (x + y ) + (z + t) = 1, nen diem N nhm tren diTdng thang x + y = 1 '^B = X + y ; A C = X + z; BC = y + z. T a c d O M = Vx^+z^ , M N = ^(x + y - x ) ' + ( z + t - z ) ^ =^|y^+l^ Ta CO O M + M N > ON > OH (1) I c6 the giai each khac l a i hay hdn each giai tren khong? '^^J11. Cho • ' 1 1 vdi 0 < a < y = t = - - a . 2 ^Hn xet: Phu-png phap do thj to ro hieu qua trong viec giai bai l o a n tren. Cac hn 10. Cho bon so thiTc x, y, z, t thoa man dieu k i e n x + y + z + t = 1. « • 2 Scr d i CO I d i g i a i nhif vay, v i trong bai toan cai hon " h i n h h o c " da the hicn 10 qua cdc dieu k i c n x + y = 6; y = t = - - a vdi 0 < a < - . 2 bo Tilfdo suy r a m i n P = 1 8 - 6^2 .c L nen ta CO x + y = z + t = - . (4) >I nhih itim 1 2 ^ •••• M•• ~ 2 (4) om Do OMo = 3V2 ; ONo = 1 ro min(MN-) = M o N f , . - Bai U'2j Ta tron ddn v j t a i - * Pafu b^ng trong (2) xay ra t) thuoc dirdng 6, (2) iL ie uO nT hi Da iH oc 01 / vdi • ' ^6 rang A A B C ton t a i v i thoa " l a n cac tien dc vc dp d a i canh ^"Ja mot tam giac. A OB, Chuy6n ai BDHSG Toan gia tri lan nhS't va gia tri nhd nhat - Phan Huy KhSi Cty TNHH MTV DVVH Khang Vi§t K h i do ne'u goi p la niira c h u v i cua tam giac thi p = x + y + z. IhuTc M A + A B + B N > M N = V ( 6 - 2 ) ^ + ( 4 + 4 ) ' =AS Heron, la c6 SABC = V P ( P - B C ) ( P - A C ) ( P - A B ) Gia sur M N cat x + 2y = 9 tai M,, va cat 2x + y = 4 lai N,,. = 7 ( x + y + z)xyz . Dd thay M„ = (5; 2); N„ = (4; 0). T i r g i a l h i e t s u y r a S A B c = 1- (1) M a t k h a c SABC = - A B . A C . s i n A = j ( x + y)(x + z ) s i n A . ' ' ' in do suy ra m i n ? = AS <=> (2) HUdng ddn giai c=> (y + z)^ = (X + y)'+ (X + z)^ V i e t l a i P diTdi dang (4) <=> yz = xy + xz + x^. P= V(x-3)^+(y-5)2 +V(x-5)2+(y-7)2 . xyz(x + y + z) = 1 Txi do ta c6 minP = 2 <=> < yz = xy + xz + x^ Ta toa dp xOy. s/ up T i m gia t r i nho nhat cua bieu thiJc ro P = Jx^ - 12x + y^ - 8 y + 52 + Jx^ + z^ + y^ +1^ - 2 x z - 2 y t + Vz^ + - 4 z + 8l + 2(1 om /g Hudng ddn giai .c V i e t lai P difdi dang sau ok P = ^ ( x - 6 ) 2 + ( y - 4 ) 2 + V ( x - z ) 2 + ( y - t ) 2 + V ( z - 2 ) ^ + ( t + 4)^ . Gia sur M ( x ; y ) v d i x, y thoa m a n X - 2y + 2 = 0, turc la M ( x ; y) nSm tren diTdng thang x - 2y + 2 = 0. R6 rang tuT (1) suy ra P = M A + M B Theo m o t b a i loan (1) quen bie't ciia h m h hoc phang, t h i neu X€i hai d i e m gpi A ' la d i e m d o i y) va B( z; t) ce bo T r e n mat phang toa do Oxy ve hai diTdng thang x + 2y = 9; x + 2y = 4. w. fa + 2y = 9 ww Trong d6 A ( x ; xi^ng cua A qua dirdng thang x - 2y + 2 = 0 va gia sur A A ' ca't dirdng t h i n g X 2y + 2 = 0 tai M„, tifdng tfng n ^ m thi M A + M B > M„ + M„B, V M e (d), cl day (d): x - 2y + 2 = 0. , , tren hai duf5ng Co the tinh diTdc ngay Mo = (5; 3,5) => MoA + MoB = A ' B = 6. thang X + 2y = Tird6tac6P>6. (2) 9 v a x + 2y = 4. H i e n nhien ta c6 256 (1) Xet dirdng thang x - 2y + 2 = 0 va hai d i e m A ( 3 ; 5), B ( 5 ; 7) tren mat phang B a i 12. Cho bon so x, y, z, t thoa man dieu k i e n x + 2y = 9; z + 2t = 4. P = M A + AB + BN z = 4 ; l = () P = -Jx^ + y^ - 6 x - lOy + 34 + ^ x ^ + y^ - l O x - 14y + 74 o B C ^ = AB^ + AC^ Tir(l)tathay x = 5;y-2 T i m gia trj nho nhat cua bieu thiJc a : s r : x } . . i v ' (3) \ ! ' i ! i it ' pai 13. Cho X, y la hai so thi/c thoa man dieu kien: x - 2y + 2 = 0. Dau bang trong (3) xay ra o sinA = 1 <=> A = 90" M ( 6 ; 4) va N ( 2 ; - 4 ) . M ginh lu4n: K h o n g the c6 IcJi giai nao lai ngan gon va hay hOn lc:fi giai tren. Do s i n A < l , n e n t i r ( l ) ( 2 ) suy ra ^ ( x + y)(x + z) > 1 => (x + y)(x + z) > 2 VayP>2 (3) iL ie uO nT hi Da iH oc 01 / Thco cong Dau b ^ n g trong (2) xay ra o M s Mo o fx = 5 \y; =i 3. , 5^. v, . V§y minP = 6 c : > x = 5 ; y = 3,5. 257 Chuyfin dg BDHSG Toan g\A trj Idn nhS't va g\i trj nh6 nha't - Phan Huy KhAi Cty TNHH MTV D W H Khang Vi$t Nhdnxet: 1. T i r x - 2 y + 2 = ( ) = ^ x : = 2 y - 2 . B ,5^..,,,^,, Tur do ta c6: 1 2 ' ~2, I vii T(x; x). 2 • 2J Khi do t i r ( l ) , ta c6 f(x) = TA + TB + T C ; P = 7(2y - 2 ) ^ + y^ - 6 ( 2 y - 2 ) - lOy + 34 + V(2y - 2)' + y^ - 1 ()(2y - 2) - 14y774 Dc thay ABC la tam giac deu = Vsy^ - 3 0 y + 50 + ^/5y^ - 4 2 y + 98 . , . • ,., ,: . iL ie uO nT hi Da iH oc 01 / vdi tam la go'c toa do O. *' Bai loan trd thanh: \ rskmisl Theo ket qua quen bict trong hinh hoc phang: Tim gia tri nho nhat cua ham so f(y) = 75y^-30y + 50 + 7 5 y ^ - 4 2 y + 98 „ 'JUtU UM.i ftiJj. Do ABC la da giac deu, ncn ! vdi y e R. vdi Cac ban hay thuT dung phifcfng phap chicu bien thicn ham so de giai tiep! "•• Ta ^ , p = Jx^ + y ^ + z 2 + 2 x - 6 y + 4z + 14 + V x ^ + y ^ + z ^ + 1 8 x - 8 y - 1 8 z + 178. ' HU('/ng ddii: Xet mat phang (P): 2x - y + z + 1 = 0 , ro up Tim gia In nho nhai eija bieu thiJc: s/ Vay min f(x) = 3 <=> x = 0. » .r (D Da'u bang trong (1) xay ra khi va chi khi trong cac tam giac AMP, BMN, CNP cd w. tam giac trting vdi AABC. Dieu dd xay minP = 3x/26 o M = M„ o x = - 2 ; y = 2; z = 3. ra khi va chi khi Irong ba so' x, y, z cd ww Bai 14. Tim giii tri nho nha't cua ham so xy - yz - jzx, HiUing d&n giai Ta cd: fa la diem doi xufng cua A qua (P). Tim gia tri Idn nhat ciia bieu thiye: P = x + y + z cac diem M, N, P sao cho A M = x; BN = y; CP = z. ce dday A ' = ( 3 ; 1;0) Bai 15. Cho x, y, z la cac so thiTc thupe doan [0; 1 ]. Vc lam giac deu ABC canh bang 1. Tren cac canh AB, BC, CA Ian li/dt lay .c Khi do P = M A + M B , d day M(x; y; z) e (P). Dap so: minP = A ' B = 3>y26 , xeK ^ttmh ^ va hai diem A ( - l ; 3; - 2 ) va B(-9; 4; 9). M:+ A M = A Co the thay AB n (P) = M„(-1; 2; 3). vdi Do OA = OB = OC = 1, nen liT (2) (3) cd f(x) > 3 V x e R, >< /;? f(x) = 3c=> T = O <=> X = 0. nfi^h'iit OA + OB + OC. 2. Ta CO bai toan tiTdng ti/ (giai bang each sijf dung phi/dng trinh difdng thang , ^ trcn mat mpi diem T(x; x), ta cd a r i e b ' i b l J b H'l^si va mat phdng trong khong gian) ^ ^ ^ ^ moi diem phang toa dp - ndi Sau do hay so sanh vdi each giai bang phi/dng phap suT dung do thi va hinh hoe c( trcn! 1 •c mpt so bang 1, mot so bang 0 va mpl so f(x) = ^ / 2 ? - 2x +1 + sjlx^ - ( V 3 - l)x +1 + sjlx^ + (>/3 + l)x + 1 . ui iii.ii:) ,j:ii..> tuyy e [0; l l . T i r ( l ) c d ' HUofng dan giai ^ [ x ( l - z ) + y(l-x) + z ( l - y ) ] < ^ o x - x z +y - y x + z - z y < l Vict hji l(x) di/di dang \ f(x) = Vx- + ( x - l ) ^ + X - - if X + — 2J Tren mSt phing toa do xet cac diem A(0; 1); + x+- X 2 J + — 1) (1! Da'u bang trong (2) xay ra <=> cd da'u bang trong (1). r,r: Vay maxP = 1 <=> trong ba so x, y, z cd mpl so' bang 1, mpl so' bang 0 va mpt so'"yye|o;i|. Chuy6n 6i BDHSG Join gii t r i Idn nha't va gia t r i nh6 nhS'l - Phan Huy KhSi Cty TNHH MTV DWH Khang Vigl Nhqnxet: V 1. Ta CO the sijf dung phiTdng phap "ba't dang thuTc" de giai bai toan trcn nhu sau: .j.^^ Do X, y, z € [0; \ } , nen ta c6 ( x - 1)(1- y ) ( l - z) < 0 => X + y + z - xy - yz - zx + xyz - 1 < 0 => X + y + z - xy - yz - zx < 1 - xyz. , Do X, y, z dcu khong am, ncn xyz > 0. Tird6kethdpvdi(='0, t a c 6 P < 1. i Dau bang irong {**) xay ra <=> Cong tufng ve' ba bat dang ihtfc tren ;!':••• '.i , , , j, , (*) va P = 1 khi va chi khi trong ba so' x, y, z c6 mot so' bang 1, mot so' bang 0, so con lai tiay y e [0; 1] (ban doc co the nghiem lai dieu nay). Vay maxP = 1 . pal 16. Cho X, y, z la ba so' thifc thoa man dieu kien x + 2y + 3z = 4. (**) (x-l)(l-y)(l-z)-() /.j^j, xy/ = o ,„.,.., Tim gia tri nho nhat cua bieu thuTc: P = Vl6 + x^ + 2-^16 +y^ + 3 \ / l 6 + ? . Hudng dan giai vdi X + 2y + 3z = 4, tren mat phang tpa dp Oxy xet cac diem A(4; x); B(12; x + 2y); C(24; x + 2y + 3z). if. r K)A o trong ba so x, y, z c6 mot so' bang 1, mot so bang 0 va so' con lai luy y e[();l| TCf do suy ra maxP = I C5> x. y, z ihoa man (**"•) Do Ta thu lai kcl qua trcn. ,( up ro /g Trcn bon canh cua hinh vuong: dal AM = x; DN = y; CP = z; BQ = I (xcni 2 2 K'-^-)^j 2 ' + y + z + t - (xy + yz + zt + tx) < 2 Cac ban hay tif lam hai dicu sau day: ww NhiT vay maxP = 2. X - Giai lai bai loan trcn bang phu'dng phap "bat dang thu'c"! Tim gia trj Idn nhat ciia bicu thiJc P = x^ + y"' + z^ - x'y - y^z - z^x. Lcfi giai nhUsau: x{ 1 - y) > x'( 1 - y) ^ X _ x + 2y 12 =y=z= _ X+ 2y + 3z 24 2 1 +2+3 3 mien xac dinh cua no. R6 rang ta c6 x^ - 2x + 2 = (x - 1)^ + 1 > 0 Vx, + 4x + 8 = (X + 2)2 +4 > 0 Vx. Vay f(x) xdc dinh tren R. < I => X > x' 1- y >0 o O , A , B , Clhdng hang Hudng ddn giai Cho X, y, z la ba so ihiTc ihuoc doan fO: 1 ]. CO Da'u bang trong (1) xay ra 17. Tim gia Iri nho nhat cua ham so f(x) = Vx^ - 2 x + 2 + Vx^ + 4x + 8 trcn 3. Ta CO bai loan sau di/a vao bai loan da cho: Lai P > O C = 724^+4^ =4N/37 (1) M Khi nao ihi maxP = 2. X do do hien nhicn la c6: Vay minP = 4V37 <=>x = y = z = —. A - Do 0 < dinh. Vay P = OA + AB + BC, <=> X w. =^P<2 dicii BC = A / I ? + 9 ? = 3\/l6 + z^ 4 fa => X 2 ^ABCD ok ^ x(I-t) ^ y(l-x) ^ z(l-y) ^ - .c ^H\'Q bo CO S^MQ + SDMN + ^CNP + ce Ta om hinh vc). C(24; 4) la Ta s/ \ y + z + i - xy - yz - zt - tx. Hifi'liifi dan fiidi Y^' 2y + 3z = 4 - /i AB = V8^+4y^ = ijie + y^ 2. Ta CO bai loan lifting lif sau: Cho \ y. /. I ia bon so thifc ihuoc doan [ 0 ; 1 [ Ta giai bhng phu'dng phap sif dung hinh hoc nhU' sau: X + d i De thay OA = ^4^ +x^ ; Cach giai niiy ciing ra'l gon gang va di P TiJf do di/a vao bai tren ta c6 P < 1, iL ie uO nT hi Da iH oc 01 / ';>.. TiTdng tir CO y - yz > y^ - y^z, z-zx>z^-z^x. X - xy > x^ - x^y ,f \ Viet lai f(x) durdi dang sau day 261 Chuygn dg BUHSG Toan gia trj I6n nha't va gia Iri nhd nhfl't - Phan Huy Khai f(x) = V(x-l)^+l^ +7l'^-(-2)P+(-2)- . Cty TNHH MTV DVVH Khang Vi$t T r c n mat phang toa do Oxy x c t a i c diem A ( x - 1; 1) va B(x + 2; -2) R6 rang ta luon c6 O A + OB > A B = sjy + 3 ' = 3 V2 . ^ :l,:o;> oofo m^v (3). A, O, B thang hang. o O A va OB cungphiTdng, VA BAT PHUONG TBlNH CO THAM SO , dung gia tri Idn nhat va nho nhaft cua ham so' la mot trong nhiyng phu'dng - V M J u.i ( v ' v Dan hang trong (3) xay ra o , i S «grtt cnrfi ni>f 3V2 NHD NHAT TRDNG BAI TDAN GIAI PHU0NG TBINH y ':<^^^^^; : •^^^'^''^'^-^^J'^^'^i^^*""'" T i r ( i ) c 6 f ( x ) = OA + OB. nay cung thi/dng xuyen xua't hicn trong cac dc thi Toan cua cac de thi tuyen ,x osf - ft{ x sinh vao D a i hoc, Cao dang trong nhijfng nam gan day. :J § 1 . M O I L I E N H E G I Q A G I A TR! L 6 N N H A T , N H O N H A T C U A ngiTcfc chicu. o(x<=> H A M SO V A Si; C O N G H I E M C U A P H U O N G TRINH V A BA'T 1; l ) = k(x + 2 ; - 2 ) v d i k < ( ) x-1 liNG DUNG CUA GIA TBI LlN NHAT VA GIA TRj P^^Mif^ (1)'-" PHUdNGTRiNH = k(x + 2) Ta thu'dng xuyen sii" dung ket qua sau day: l = -2k Gia < x > x - 1 = - ^ ( x + 2) . . f su" f(x) la ham so lien tuc trcn mien D , va gia suf ton tai M - m a x f ( x ) ; m = min f(x). x=0 . ' • • --r-- - ' - i x = 0. iuI ilnjm^kisii'j •. CO nghiera k h i va chi k h i m < a < M . chi k h i m > a. om ce fa 5. Bat phu'dng Irinh r(x) < a diing v d i m o i x e D khi va chi k h i M < a. w. Chiing minh: , !• Gia suf he phufdng trinh da cho co nghiem, tufc la ton tai Xd e D sao cho f(X(i) = a. Theo dinh nghla ta co ^ m i n f ( x ) < f(X(,) < max f(x), "eD H : 4V (MUV' fl"(x) a dung v d i m o i x e D khi ok „ $ 2. He baft phuTdng trinh \ nghiem k h i va chi k h i M > a. [xeD .c ...1 t r(x) = a ro /g va 17. Cac ban c6 each giai nao khac ngan gon hdn khong? ... 3-i;sjq|> ^a.ji,} ylvtfriH , f 1 . He phiTdng trinh Nhgn xet: Ro rang phi/dng phap do thi to ro qua hicu qua trong hai bai toiin 16 ^ _. . ^, , Kin do ta co: up Vay m m l ( x ) = 3V2 o ' • xeD Ta 2x - 2 = - x - 2 o s/ o u ^'^-^ xeD \r =; ,! minf(x) dpcm. - • 263 Cty TNHH MTV DWH Khang Vigt Chuy6n dg BOHSG Toati gia tii Idn nha't va gia trj nho nha't - Phan Huy KhJIi 2. Gia sur he da cho Ro rang c6 n g h i c m , ttfc la ion tai x„ e D sao cho f(X(,) > X a. m a x f ( x ) > IXx,,) > a r'(x) xeD Dao l a i gia sur m a x f ( x ) > a i xeD Gia thie't phan chi?ng he da cho vo nghiem, ttfc la f(x) < a V x e D •a' TO T i j rdo d o suy s u y rra a ', maxf(x) dpcm. 3. Gia sur m > a • Lay X(, luy y e D . ,J'AHfl M O J M A W A i i l © tSH H : M 'tOM Ta c6 f(x„) > min f ( x ) = m > a -Q ^4 Dao '' HWmi Ta 1= I up s/ :o'} iiJ dh ' TCfdo o " D i e u k i e n de phiTcfng trinh c6 nghla la <^ bo w. •• ' ^ ^ f4-x>0 ww i» » ' o 1 < x < 4. ^ • sfe Tiir 264 do suy ra bang bien thien sau: I + - ^ 0 ^ he (1) (2) c6 nghiem g(t)oO 2 A / 4 ^ = V 2 X - 2 O 0 ciia bien mc'Ji t, ta da giai mot bai toan t i m gia tri Idn nha't, nho nha't ciia ham D e tim m i e n \ic dinh cfia t, x e t h a m so f(x) = V 4 - x + V 2 x - 2 v d i 1 < x < 2V4-X 1 1 1- K h i doi bien ta phai t i m m i e n xac dinh cua bien m d i . De t i m m i e n xac djnh Datt= V4-X+V2X-2 Tac6f'(x)= 2 Nhaiixet: fa HUdngddngiai CO nghiem. 7-473 min g ( t ) < m < max v'5 t ' = X + 2 + 2 V ( 4 - x ) ( 2 x - 2 ) i g(t) VA BAT PHOCfNG TRINH CO THAM SO CO MAT TRONG . N/6 ^' 73 g'(t) „...U>ljiwii §2. CAC BAI TOAN BIEN LUAN PHOONG TRINH V " ^ " ^ ^ ^ do suy ra k h i : 1 < x < 4, thi 73 < t < 3 . t xeD 4,5. Chu-ng minh tu-cfng tu" nhu" 2,3. , ' N/3 Ta CO g'(t) = 2l - 4, va c6 biing bien thien sau: K h i do do m = m i n f ( x ) , nen theo djnh nghTa ta c6 ton t a i Xo e D ma m = f(x ,.....,.4 3 l a => m > a. Nhu^vay ta CO dpcm. - V i the bai toan da cho t r d thanh: miOim- l a i gia sur f(x) > a Vx G D. I a Vx e D. 0 iL ie uO nT hi Da iH oc 01 / Ttf Tiir (1) (2) suy ra v6 l i , vay gia t h i c l phan chtfng la sai, ttfc la he da cho c 4 m i n 1"(X) = N/3 va max f ( x ) = 3. Vay xeD + i l(x) (1) 3 1 , . HUifng ddn giai I ' D a t l X x ) = 72x + 2 7 6 ^ ; g ( x ) = 72^ + 276^ Luc nay phiTdng trinh da cho co dang h(x) = f(x) + g(x) = m 265 Cty TMHH MTV DWH Khang Vigt BDHSG Toan gia tri Idn nhit va gia lii nh6 nhS't - Phan Huy Khii Chuygn PhiTdng trinh (1) xac dinh trong m i e n 0 < x < 6. Ta CO f (X) = _76^-N/2^ 2 2 2>/2x 276-X 72x(6-x) 0 X + 0 '">'^" ^"'^ ^^"^ f i m m de phiTOng trinh S V x - l + mVx + 1 = 4\/x^ - 1 c6 nghic icm. , nen C O bang bie'n thien sau- HUdng dan gidi 6 2 f'(x) P»i 3- phiTdng trinh 3Vx - 1 + mVx + 1 = 2 V x ^ ^ - p i c u k i e n de (1) c6 nghla la x > 1. Do x > 1, nen V x T T > 0, vay f(x) (1) <^ 3 J — + x +1 iL ie uO nT hi Da iH oc 01 / 4/(5 - ^(2x)'' TuTdng tif ta c6 g'(x) = ^^^^—, = - , nen cijng c6 bang bie'n thien sau 2^(2x0-^6-x)-^ /X —1 x —1 E)3t t = ^ . Do =1 Vx+I x+1 r a v d i t l a O < t < I) '•"•2 0 g'(x) 6 0 - ^ 0 ^ ^ max h(x) = h(2) = ^ ^ m + V 4 + 2 ^ + 2V4 = 6 + 3>/2 , ok () 1 nen > 0 nen dieu k i e n dat x+1 i x+1 ,f £ ^ m nghiem. 0 f'(x) ^•""•? .if :ji> 0. X m = 24) + 2^3 < m < 6 + 3V2 . 2. T u y nhien d day b a i toan d o i hoi: T i m m de phiTOng trinh c6 hai nghic''^ Cac ban nen lifu y rSng khi suf dung dieu k i e n c6 nghiem cua he phifdng trinh • " t a can lu\ y gia thie't phai ton tai mini"(x), m a x f ( x ) XeD • xeD xeD CJ day v i xet khi 0 < t < 1, nen khong Ion tai m i n f ( t ) nhu^g ton tai () min 1(1) thanh m > l i m 1(1)) 3 " S K I _ , • d ji; , ,, 267 Cty TNHH MTV DWH Khang Vi^t Chuy6n 6i BDHSG To^n gia tri Icin nhSt va gia tri nh6 nhS't - Phan Huy Khii f^' tuyen sink Dai hoc Cao danf^ khdi B) p^i 4. Cac ban can lifu y d i c u do (d day ta cung dfi ket hOp sit dung chieu b i f thien cua ham so' da xet trong bang bicn thien ham so' l'(t) trong bai). Tim m de phirong trinh Vx^ + m x + 2 = 2x +1 c6 hai nghiem thiTc phan biet. 2. Ta CO each giai khiic sau day: - 3 l - + 2 t - m = 0 (3) 0 0 De he (3) (4) c6 nghiem, triTdc h e t (3) can c6 nghiem, tiJc la phai x 2 + m x + 2 = (2x + l)^ Co <=> s !•;')'>; - x> — 2 •' • • <• , " x^ + m x + 2 = 4x^ + 4x + l iL ie uO nT hi Da iH oc 01 / X e t he HU ( ) < ^ m < ^ . K h i do (3) <^ t = ^ " ^ ^ ^ ' ^ ^ ' 3 x ^ + 4 x - l = mx (1) TiTdo suy ra he (3) (4) CO nghiem k h i va chi k h i ^, „ • i . , 1 2 3 0-l ^ . f(x) .c b. hoac (3) c6 nghiem, nhifng m o i nghiem deu khong thoa man (4) bo ok 2 K h i (3) CO nghiem t i , t2 thi tj + t2 = - , do do m o i nghiem t i , I2 khong thoa ce man (4) k h i va chi k h i ti < 0 < 1 < t2 V a y he (3) (4) v6 nghiem k h i fa w. De tha'y (5) xay ra k h i va chi k h i ww A'>0 (5) m<0 f(l)<0 m +l<0 , va CO bang bien thien sau: 1 0 2 i 1 + + +00 4 =1 i':' . +00 9 2-" -00 Tiir do suy ra he (3) (4) cd hai nghiem phan biet <=> m > —. ^hqnxet: 3 f(0)<0o n • (4) X /g D i e u nay xay ra trong hai IriTdng hdp sau: up ' Ta CO f ' ( x ) = ro Tru'dc hot ta tim dieu kien de he (3) (4) v6 nghiem s/ 3. L a i con each giai khac sau day ' = m (3) x> — 2 Ta Vl-3m y i 11 Do X = 0 khong phai la nghiem cua ( I ) V m , ncn (1) (2) 3 o i (2) X> — o m < - l . 1 m >3 1- Ro rang trong b a i nay ta c6 m i n f ( x ) = — , v a da k e t hdp v d i chieu bien -'- — 2 (6) I n g h i e m phan biet. ham so v d i cac phi/dng phap vijfa trinh bay! Cac ban thich each giai nao? 268 269 cnuyen de B D H S G loan gia tr| lan nhat va gia trj nh6 nhat - Phan Huy'Khai Cty TNHH MTV DWH Khang Vigt D i e u d o x a y r a k h i v a c h i k h i ( 5 ) c 6 h a i n g h i e m p h a n b i e t X j , X 2 sao c h o TCf do suy ra m a x ^ f ( t ) = 1(0) = 1, 0X, f ( t ) = f(N/2) = V 2 - 1 . min 1 R6 rang (5) c6 hai nghiem phan biet ( v i - = — < 0 ) a 3 C . i „ X | + - [ X | + X2 >0 X-, + • > -1 U X|X2+-(x,+X2) + - > 0 2 3 m-4 m > - >-l 2 <=>m> —. m >1 : yjfKy, ^! Cho Tim J ' f . i ^ " * n n o c y Q ' * ^ ,, Ta t h u l a i kc't q u a t r e n . Bai l o a n da c h o In't t h a n h : 0, v a y (1) o Ta s/ /g t+2 w. Ta CO f'(t) = ,? . . t f'(t) f(t) —t 0 0 < m < 2, x^/: X e t each giai khac nhau: Bai loan da cho C O dang: T i m m de he sau 0 1 + 1 'nfrfBimbt, CO nghiem. t +2 nghicMii 2 (1) ww Timmdehe: to 2 no fa B a i loan da cho trd thanh: ce t+ 2 (2) I bo Luc d o la di/a phuTdng u-inh v e d a n g m(t + 2) = 2 - t^ - t . Do 0 < t < 1<1<2 .c 2 - 2V1 - x"* < 2, v a y 0 < t < V2 . (1) Ta C O f ' ( t ) = 2t + 1 va c6 biing bien thien sau: om , ok CO t ' = _ up c6 nghiem. . r(t) = t ^ + t - 2 = 2m ro + 2 ) = 2 V l - x ' * + V l + x2 - V l - x ^ Hudng dan gidi Ta Tim m de he • f y V I 6:. r,^nti\M <;<'»^>^ h-i .Khi d6t>0. I s i y Liic nay t h e o b i e n m d i t , phU'dng t r i n h da c h o c6 d a n g I " + t - 2 = 2m. S y. B a i 5. (De thi tuyen sink Dai hoc Cao dang kho'i B) D a t t = V l + x^ - V l - x ^ /n)e\y'' D a t l = , / l o g ] T + 7 . K h i 1 < x < 3"^ => 1 < t < 2 . B a n doc h a y tuT b i n h l u a n v e t i n h h i e u q u a cua m o i phiTcfng phdp. m ( V l + x^ - V l - x ^ I—7— phiTdng t r i n h l o g ^ x + J l o g ^ x + l - 2 m - l = 0 m d e p h i / d n g t r i n h c6 i t n h a t m o t n g h i e m t h u p c d o a n [ I ; 3 ^ 1 HU(ing dan gidi " Tim m de phiWng Irinh :.' Bai 6. (De thi tu\en sinh Dai hoc Cao dami kho'i A) y 9 -.Z'::: nhat c u a h a m so' de g i a i b a i t o a n n a y . .. in = i --.J 1 1 ^ - -6+ -4> 0 % h a i de tha'y ro t i n h h i e u q u a c u a phU'dng p h a p suf d u n g g i a I n I d n n h a t va n h o (9) l^x, + X 2 > - l ( '' glith luan: Cac b a n h a y g i a i l a i b a i l o a n t r c n b a n g phu'dng p h a p t a m thtfc bcic (8) 4 A p dung djnh H V i e t vdi (5), ta c6 (8) (9) m-4 Vayhc(2)(3)c6nghiem<=> V 2 - l < m < l . iL ie uO nT hi Da iH oc 01 / V i vay (7) <=> < ()/2 s,h cmi,} • 1 v, t = t= Jl^ + l - 2 - 2 m = 0 l Do do he (2) (3) c6 nghiem <=> min f(t) < m < max <=> - — < m < - 2 . • () <=> 0 8 < m < 3 < V8m + 9 < 5 E)6 chinh 1^ cac gia trj can tim cua tham so'm. 2. plnh luan: , CHI n;/.:,<,; -* ...^ J R6 rang suf dung phiTctng phap gia tri Idn nhat Ta ihu lai ke'l qua trcn. > ' ,:>•.. nh6 nhat ciaa ham so de gi^i b^i toan nay day hieu qua: Ldi giai gon gang, trong sang. Ban tha'y the nao vc tinh hieu qua cua hai each giai vijfa tnnh bay! iL ie uO nT hi Da iH oc 01 / 2. Ta hay x6t them cdc each giai khac dc nhan thay cdch giai tren la thich hdp nhS't. § 3 . sCr D g N G G I A TR! LCJN NHAT V A NHO NHAT C U A HAM SO 2: Bai toan da cho CO dang: D E BIEN LUAN PHUCING TRJNH V A BAT PHl/ONG TRJNH C O Tim m de he 3t - 2 t - 3 - m = 0 (4) 00 t| +t2 <2 ro trinh da t,i2>0 Dieu nay xay ra khi va chi khi -{t, +12 > 0 up Hitting ddn giai Vie't lai phiTdng fA'>0 '--y'iU. mot n g h i c m thupc d o a n Ta T i m m de a. (4) CO nghiem t|, t2 m^ 0 < ti < t2 < I Bai l o a n da cho trd thanh: w. ww Ta CO f'(0 = 6l - 2 va CO bang b i e n thien sau: 0 3 • .c 3m + 10>0 : Tur do ta c6 ce f(t) = 3 t ^ - 2 t - 3 = m (2) c6 nghiem. 0 0 < 2 x < 7 t = : > 0 < sin2x < 1. 2 Dira vao djnh Ii Viet, thi t, + ta = | ; t,t2 = om /g 1 ^ -> l - - s i n ^ 2 x + l - 2 s i n ' ^ 2 x + 2sin2x + m o 3 s i n ^ 2 x - 2 s i n 2 x - 3 = m ( . 2 ) -3-m>0 o 10 -Y0 3 b. (4) CO nghiem t i , t2 trong do mot nghiem 6 [0; 1] \h mpt nghiem khong thupc [0; 1] Dieu do xay ra o f(0)f(l) < 0, d day f(t) = 3t^ - 2t - 3 - m o (-3 - m)(-2 - m) < 0 o (3 + m)(2 + m) < 0 -33 U) 3m + 10 < 0 <=> m < - a. Ban than (4) v6 nghiem, tiJc la can c6 A ' < 0 x b. (4) CO nghiem nhiTng ca hai nghiem deu « |(); 1 ]. ;; ' 2 Khi (4) CO nghiem l , , la c6 t, + = - , nen dieu nay xay ra k h i v a chi X>3 <:> (2) (3) I f ( ( ) ) < 0 <=> m > - 2 . C:> i 1(1) < ( ) -2-m<() Vay he (4) (5) v6 nghiem o m < 10 3 m >-2 Do do he (4) (5) c6 nghiem o - 10 y < m < - 2 . T a thu lai ket qua Iren. Cddi 4: Trirdc h e l (4) c6 nghiem » 0, nhU" vay (1) • < v>. . i ^ ; ; , ^ . 7 X 1 f'(x) 0 1 oA'>0o3m+10>0c:>m>-^. Tir do suy ra 1 - V 3 m + 10 /g Vay he (4) (5) c6 nghiem Nhanxet: Tim m de he sau . s • f ;;(.,, l i m f(x) = + « . ok bo fa - 2 ( m + l)x + 5m + l = 0 (6) , C O nghiem x>3 • ' (7) X ww He (6) (7) v6 nghiem trong hai tru"dng help sau: a. (6) v6 nghiem <=> A ' < 0 o m ' - 3m < 0 o O < m < 3 B a i 2. T i m m de phiTdng trinh 72x^ - 2 ( m + 4)x + 5m + 1 0 + 3 - x = 0 c6 nghiem b. (6) C O nghiem X i , X2 va X| < X2 < 3. >f. ,^ * <, Trirdc h e l t i m m de he (6) (7) v6 nghiem. nhat va nho nhat trong tri/cfng hdp nay la help l i nhat. TCr d o s u y r a ( l ) o lim f(x) = + c » ma ta van sir dung) Bay gicf cha'c cac ban dong y vdi chiing t o i : Suf dung phtfcfng phap gia t n U^" V i e t lai phi/dng trinh da cho di/(3i dang sau: • 2. Xct each giai khac sau day (de cac ban thay no kem hieu qua hcJn phiftJng phap Bai loan da cho C O dang: Hitffng dan gidi :A, , • • 10 ^, <=> min r(x) < m <=> m > 3. ce - 2 < V3m + 10 < 1 om 10 2 / f x>3 Ta ro up s/ l + N/3m + 10 + m i n f ( x ) = 1(4) = 3 (con m a x f ( x ) = +oo hieu theo nghla 2 "•' +C0 0 1 f(x) : 4 1 x>3 K h i do (4) C O nghiem (4) , X -2x + ! l(x) = — - — = m (5) 2x-5 , ^ 2x^ - l O x + 8 . , ^, Ta C O I (x) = — — — va co bang bien thien sau: (2x-5)^ JiSji;" ••s/ . khil| ^ 0 3, nen 2x - 5 - 2 x + l = m ( 2 x - 5 ) (3) iL ie uO nT hi Da iH oc 01 / 'fij 6".. f ' i u . « po X > (2) , ^ ,, [A'>0 (1) X|+X2<6 x-3>0 Do X | + X2 = 2(m + 1); X|X2 " (8) (9) (8)xayrac::> j ( x , - 3 ) ( X 2 - 3 ) > 0 " (10) lit hi t \ ,1 (11). = 5m + 1, nen 2x^ - 2(m + 4)x + 5m + 1 0 = (x - 3)^ 4 275