Phitdng trinh, he phudng trinh, bat phudng trinh va bat ding thiic c6 moi lien he chat che vdi nhau. Chang han khi chiing minh mot bat dang thiJc, ta can dit doan dau bkng xay ra khi nao, dicu nay dan t6i vice tim mot nghiem nao do ciia phifdng trinh, he phitdng trinh. Qua day ta thay rSng viec giai phitdng trinh, he phitdng trinh la thi.tc sit c6 y nghla. B6i vay trong qua trinh sang tao va giai toan bat dang thiic se nay sinh ra nhu cau tim nghiem ciia phitdng trinh, h$ phitdng trinh, bat phitdng trinh. Nhieu bai toan ve phifdng trinh, ho phitdng trinh lai la sit che dau mot bat dftng thitc nao do. Dau hieu d i nhan ra dang toan nay la so phitdng trinh it hdn so an, phifdng trinh rat phiic tap, khong mau mitc, mang bong dang cua mot bat ding thiic nao do Vi bat ding thitc la mot llnh vifc r i t phat trien ciia Toan Sd Cap nen theo do, bang con ditcing bat dang thiic ta se sang tao ra rat nhieu phifdng trinh, ho phitdng trinh. Mot dieu dac biet lun y doi vdi phitdng phap nay la doan dUdc nghiem se gop phan rat Idn vao thanh cong ciia Idi giai.
V i du 1. Ta xtil hai bid dCing thiic c6 dau bang cung xdy ra khi x = - 1 :
^3000 ^ 2999 > 3000 = 3000 |x| > -3000x (i)
^3000 ^ > 1000 ' " V ^ ^ = 1000 |x^| > -lOOOx^. (ii) d ( i ) , (ii) cimg xdy ra khi va chi khi x = -1. Cong (i) vd (ii) ducfc
If 2^3000 ^ 3998 > -SOOOx - lOOOx^ <t=> x^""" + 500x^ + 1500x + 1999 > 0. (iii) p^y^ ằ=" d {Hi) xdy ra khi vd chi khi x = -I. Ta c6 bai toan sau.
gai toan 121 (De nghi Olympic 30/04/2011). Giai phiCc/ng trinh
x^"™ + 500x^ + 1500x + 1999 = 0. , (1) Giai- A p dung bat ding thiic Cauchy cho x^"™ va 2999 so 1, ta dUdc
^3000 ^ 2999 > 3000 '"7^3000 ^ 3000 |x| > -SOOOx. I n.ho (2) Dau bang trong (2) xay ra khi va chi khi x = -1. Tifdng t i i :
^.3000 ^ > 1000 '""V^ = 1000 jx^l > -lOOOx^ (3) n i u bang trong (3) xay ra khi va chi khi x = - 1 . Trt (2) va (3), ta ditdc
ft. 2x3""" ^ 399g > _3000x - lOOOx^
B <^x3""" +500x3+ 1500X + 1999 >0. ) \) Ma (1) nghia la dau bang d (4) xay ra, tiic la dau 6 (2) va (3) dong thdi xay ra. Vay (1) <^ x = -1. Phifdng trinh (1) c6 nghiem duy nhat x = -1.
Vf du 2. Ta se xet hai bat ddng thiic c6 dau b&ng cung xdy ra khi x = 3, chang han vdi x € +00), ta c6
S n ^ i = y24.24.24.(4x + 4) < + + ^^"^ = X + 13. (i)
- 3x2 ^ 9^. + 27 = (x - 3)2 (x + 3) > 0. (ii) Vdi X > -1 thi dau bang d ( i ) vd {ii) cung xdy ra khi vd chi khi x = 3. Tii
ii), {ii) vd x^ - 3x2 _ 9^ + 27 - (x + 13) = x^ - Sx^ - 8x + 40, ta dUdc bai iodn sau.
Bai toan 122 (HSG Qu6c gia nam 1995, bang A ) . Giai phudng trinh
x 3- 3 x 2- 8 x + 40-8^y4FT4 = 0. (1) Giai. Dieu kien x > -1. Xet hai ham so tren [ - 1 ; +00) la
/(x) - - 3x2 _ ^ ^^^-^ ^ sV^^^n.
83
Theo bat dang thiic Cauchy, ta c6
. ( X ) = V2^.24.2M4x-f4) < 2- + 2^ + 2^+(4x + 4) ^ ^ ^ Dau bang d (2) xay ra khi va chi khi x = 3. Mat khac
fix) - (x + 13) =x^ - 3x^ - 9x + 27 = (x - 3)^ (x + 3) > 0, Vx > - 1 . (3) Dau b^ng 6 (3) xay ra khi va chi khi x = 3. Ta
( l ) ^ / ( x ) - 5 ( x ) . (4) Vay (4) CO nghia la dau bang d (2) va (3) dong tlidi xay ra, hay x = 3 (thoa
man dieu kien). Phildng trinh (1) c6 nghiem duy nhat x = 3.
Bai toan 123 (De nghi Olympic 30/04/2010). Giai phuang trinh x^ + x^ - 15x + 30 = 4 V27(x + 1).
x + 2
V i du 3. Xet ham so f{x) = '.' Ta tun dudc gid tri Idn nhdt cua ham
\Jx^ + 8
so nay Id —, dn.t diccic khi va chi khi x — A. Bdi vay vdi moi x G R, ta c6
Vay
Cong lai ta diMc
Vx^ + 8+ ^y2 + 8 + y/z^ + 8>-={x + y + z + 6),
Dau hhng xay ra khi va cM khi x = y = z = 4. Ta thu diMc bdi todn sau.
Bai toan 124. Giii H [ ^ ^ " ^ + y ? T 8 = 6^/6.
Hifdng dan. Thco trcu ta chiing ininh dUdc
<' 7x2 + 8 + V y M ^ + y j + 8 > ^ ( x + y + 2 + 6 ) , Dau b^ng x^y ra khi va chi khi x = ly = 2 = 4. Bdi vay
( X -\- y + z = 12
I y ^ n ^ + v 9 T 8 + v/?T8 = 6v/6. ^(^;y;^) = (4;4;4).
pai toan 125. Gidi he | ^^2 ^_V+ i + ^y2 _y+i + - 2 + 1 = 3.
Vi du 4. Xet phuang trinh
^x^ + ax^ + bx + c = d\/x2 - 2x - 1 + x + e (rf > 0).
pfi^ang trinh nay cd dieu kien x^ - 2x - 1 > 0. Do dy/x^ - 2x - 1 > 0 nen
\J x^ -\r ax2 + 6x + c > x + e
ô^x^ + ax2 + 6x + c > x^ + 3ex2 + 3e2x +
^{a- 3e) x2 + (6 - 36^) x + c - > 0.
Ta can chon a,b,c,e sao cho
a - 3e = - 1 f a = 3e - 1 6 - 3e2 = 2 <^ 6 = 3e2 + 2 c-e^ = 1 I c = + 1.
Chang han chon c = 2, d> 0 tuy y, chdng han d = 2. Khi do a = 5, b = 14, c = 9. Ta dudc phuang trinh gidi bang phUdng phdp ddnh gid nhu sau.
Bai toan 126. Gidi phudng trinh
\/x3 + 5x2 + 14x + 9 - X = 2 ^^a;2 _ 2a; - 1 + l ) . (1) Giai. Dieu kien x2 - 2x - 1 > 0. Ta c6
\/x3 + 5x2 + 14^. + 9 = 2v/x2 - 2x - 1 + 2 + X. (2) Do ^12 - 2x - 1 > 0, nen tif (2) ta c6
v/x3 + 5x2 + I4x + 9 > 2 + X
<^x^ + 5x2 + 14x + 9 > 8 + 12x + 6 x 2+ x ^ ^
^ x 2 - 2x - 1 < 0.
Tit day ket hdp vdi dieu kien x^ - 2x - 1 > 0, suy ra (1) tifdng dudng vdi ( - 2x - 1 = 0 4ằ x 2- 2 x - l = 0 4 = > a ; = l± V 2 . I v^x3 + 5x2 + I4x + 9 = 2 + X
Bai toan 127. Gidi phUdng trinh v^l4 - x^ + x = 2 ( l + v/x2 - 2x - l ) • du 5. Tic mot phUdng trinh bdc ba ndo do, ch&ng han 4i^ + 3i = 2. Ddt
* = 2x, ta dudc 32x^ + 6x = 2 16x^ + 3x - 1 = 0. Xeô phudng trinh
\/ax3 + ax2 + 6x + c = d\J\Qx^ + 3x - 1 + x + e (d > 0).
Dieu kien IQx^ + 3x - 1 > 0. £>o ds/lQx^ + 3x - 1 > 0 nen
V Q X ^ + ax^ + bx + c>x + e i > i ',
, <=ằax^ + ax^ + 6x + c > 4-3ex2 + 3e^x + .1'uh
• ^ ( a - l) x 3 + ( o - 3 e ) x ^ + (6 - Se^) x + c - > 0.
Ta can c/ipn a, a, 6, c, e sao cho , a - 3e =
Chdng han chon e = 1, = 71, ta dudc phicang trinh
v / - 1 5 x 3 + 3x2 + 2 71 \/l6x3 + 3x - 1 + x + 1.
Th CO 6di iodn sau.
B a i t o a n 128. Gidi phuang trinh
\ / - 1 5 x 3 + 3x2 + 2 - X = 7 i \ / l 6 x 3 + 3x - 1 + 1. (1) G i a i . Dion kion IGx^ + 3x - 1 > 0. Ta c6
(1) \ / - 1 5 x 3 + 3x2 + 2 = 71^16x3 + 3x - 1 + X + 1. (2) Do 71>/16x3 + 3 x - 1 > 0 nen tir (1) ta suy ra
\ / - 1 5 x 3+ 3 x 2 + 2 > X + 1
<^ - 15x^ + 3x^ + 2 > x^ + 3x^ + 3x + 1
<!=^16x^ + 3x - 1 < 0.
Tit day ket hdp v6i dieu kien, ta ditdc 16x^ + 3x - 1 = 0. Do do
Dat X = ^, thay vao (3), ta dUdc
i - \t^ + 3( = 2. (4)
V i ham so / ( f ) = \t^ + 3( c6 / ' ( x ) = 12^2 + 3 > 0, G E nen / dong bien, suy ra (4) c6 khong qua mot nghiem. Xet
2 V
Po do. neu dat a = ^ 2 + \/5 t h i 2 = i (^a^ - Ta c6
= 3 1 / a [2 V ' W J
+ 4 2
T3
1 ^ 1>
" 2
pfing la nghiem duy nhat ciia (4). PhUdng trinh da cho c6 nghiem duy nhat
V i d u 6. Xet mot ham so dong bien tren K ; f{t) = t^ + t^ + t. Vdi t>lc6
fit)>f(l)^t^ + t^ + t>3. (1)
Xet X > y > 0. t = - > 1. TH (1) ta CO I - ] + { - (2)
• y \yj \yj ^ 1 > 3 .
Do (2) nen cong them vdo ve trdi cua (2) mot biiu thiic theo z, khong dm vd y.
hhng khong khi vd chi khi z = 1, ta dvtcic
fxY fx\^ fx\ ,2
- + - + - + ^ - l ^ > 3 .
\vj \yj \yJ (3)
Ta se thiet lap mot phUdng trinh md tit dieu kien cua no dan tdi x > y :
x^z + yz^ + z^^/x -y = Vx + ^j. . , ., . (4) Tw: (3) vd (4), ta c6 bdi toan sau.
B a i t o a n 129 ( D e nghi O l y m p i c 3 0 / 0 4 / 2 0 1 1 ) . Gidi he phUdng trinh f x^ + x3j/2 ^ j . y i + y 5 - 2 ^ 2y^z + 2y^ (1) JHU I ...
\ + yz2 + ^ 3 ^ ^ _ y ^ ^ + (2) jutvm
G i a i . Dieu kien x > y > 0. Neu y = 0 t h i thay vao (1) ta dUdc x = 0, thay vao (2) dUdc z t u y y. Tiep theo xet y ^0. Chia (1) cho y^, ta dUdc
+ V?// V?y/ (3)
Xet ham so f{t) = + f3 + ^^ vt G R. Ta c6 f'{t) = + 3^^ + 1 > 0, suy ra ham / dong bien tren E. M a - > 1 nen
Bai the \ 2 / / \y) '^^^ ~ ^ ^^^^^^ ^^^^S. xay ra k h i va chi khi
2 = 1
Vay (3) <^ I f r f Thay vao (2) ta ditdc ii:
Nghiem ciia he la (x; y; z) = ( 1 ; 1; 1 ), (x; ?/; = ( 0 ; 0 ; c) e K , t u y y j .
V i d u 7. .Ywai ;;^d< tu bat ddng thiic '
rfau t a n ^ xd?y ro khi va chi khi x = y, ta c6 bai toan sau.
B a i t o a n 130 ( H S G Q u o c gia - 2009). Gidi he phudng trinh 1
: -r ZJ''^ V I + 2x2 ^ 1 + 2xy ^ + (1) v / x ( r - 2 x y + v / y ( i - 2 y ) = - . (2)
G i a i . D i e u kieri 0 < x < ^ , 0 < y < ^ . Vdi dieu kien n a y t a c6
\ / i + 2x2 yiTv ~ v T T 2 i ^ '
Daiig tht'rc xay ra khi va chi khi x = y.
Chting minh. Theo bat dang thii:c BunhiaCopxki, ta c6
( * )
1 1
+
Dau dang thiic xay ra k h i va chi k h i
1 + 2x2 1 + 2y2
+ 2 x 2 yr+v <^ x^ = < ^ X = y (do X > 0 , 2/ > 0 ) .
Lai C O
1 + 2x2 1+22/2 1 + 2 x y
2 + Axy + 2?y2 + 4 x ? / + 2 x 2 4,^3,^ - 2(1 + 2 x 2 ^ 27^2 + 4 x 2 ^ 2 )
^ '' ' ( l + 2 x 2 ) ( l + 27/2)(l + 2x2/) 4x7/ - 2?/2 + 4x7y^ - 2x2 _^ 4^3,^ _ g^2^y2
= • ( l + 2 x 2) ( l + 22/2)(i + 2x,y) ^^,„,. , 1 . 4x?y(?/2 - 2x?; + x 2 ) - 2(x - yf
' ( l+ 2 x 2) ( l + 2 2 / 2 ) ( l + 2xy) , ' .'^ ' 2 (T - ? / ) 2 ( 2 x i y -1 ) / , ^ ^ 1
= ( l + 2 x 2 ) ( l + 22/2)(l + 2 x 2 / ) ^ ° ( d o O < x , < -
Daug t l i i l c xay ra k l i i va c h i k l i i x = y. Vay (*) dUcJc cliitng uiiuli. Dau bang xay ra X = 2/. D o d o (1) 2/ — 2;. Thay vao (2) diTdc
y x( l - 2 x ) ^ 4^ 162x^ - 81x + 1 = 0 ^ x = t
9 2.162 Hf p h U d n g t r i n h d a cho c6 h a i n g h i e m
/ 9 + \/73 9 + y 7 3 \ (x; y) =
36 36 , (a;; y) = / 9 - 9 - \ / 7 3 \
36 36
J
Bai toan 131 ( D e du" bi t h i h o c s i n h gioi c a c trifclng C h u y e n k h u vufc D u y e n H a i v a D o n g B a n g B a c B o n a m 2010). Gidi he phiCcfng trinh
j 2^^+!/ + 2''+y^ = 8 1 v ^ + v / y = 2.
Giai. Ta CO X + 2/ > ^ ( v ^ + v ^ ) ' - 2;x2 + 2/' > ^ i^ + vf > 2- Suy ra x^ + y^ + x + y> 4. Theo bat d i n g thilc Cauchy ta c6 *
2''^'+y + 2''^+^ > 2 \ / 2^ ' + y ' + ^ ' + y > 2.V24 = 8.
Vay dan bftng xay ra. Tit do x = y = 1. He da cho c6 nghiem duy nhat (^;2/) = ( l ; l ) .
Bai toan 132 ( D e nghi thi hoc sinh gioi cac tru-^ng C h u y e n k h u vu"c
^ u y e n H a i v a D o n g B a n g B a c B o n a m 2010). Gidi h$ phuang trinh
\^ - 8j/3 = 2x2/(1 - 22/) (1) y ^ T T 4^ = l + ( ? ! ^ . (2)
'^iai. Dieu kien x^ + 4x > 0 x > 0. T i t (2), suy ra x > 0. T a c6 (1) ^ x ( x - 22/) = 42/2(22/ - x ) (x - 2y){x + Ay"^) = 0 <^ x = 2y.
Thay vao phUdng trinh (2) dildc
3\/x3 + 4x = + 2x + 4. , -^^^ J (3) Theo bat dang thutc Cauchy, ta c6
^ ^ > x ^ x 2 + 2x + 4 = ^ ^ +5(^:2 + 4 )+ 2x
> x + ^(x^ + 4) + 2x = ^ ( ^ ^ + 2x) > ^.2v'x3 + 4x = 3v^x3 + 4x.
Dau dang thulc xay ra khi x = 2. Nghiem ciia he la (x; y) = (2; 1).
Lvfu y. Cc'ich khac lc\h phvtdng 2 vc ciia (3) duoc (x - 2Y{x? - x + 4) = 0.
B a i toan 133 ( D l nghi O l y m p i c 3 0 / 0 4 / 2 0 1 1 ) . Gidi phudng trinh
13v/x2 - x^ + 9Vx^ + x4 = 16. (1) G i a i .
C a c h 1 . Dieu kien - 1 < x < 1. Ta c6
1 3 v/ x 2-x4 + 9V'x2 + x4 = 9 |x| v/l + + 13 |x| V l -
• = - . 3|x|.2\/l+ x 2 + ^ | x | . 2 v / l- x 2
< ^ [9x2 ^ 4 (1 ^ ^2)J + ^ [^2 ^ 4 (1 _ ^2)j ^ ig
Dau bang xay ra khi va chi khi
3|x| = 2 v T T ^ 9x2 = 4( l 4- x 2 )
|x| = 2 v / r ^ ^ 1 x2 = 4 (1 - x2) 4 (2 2 ]
Tap nghiem ciia phitdng trinh (1) la | -7=, - — I.
l\/5 V5 ) C a c h 2. Dieu kien - 1 < x < 1. Ta c6
(1) x2 ( l 3\ / l - x2 + 9v/l + x2)^ = 256. (2)
Thco bat dang thiic Bunhiacopxki :
/ i 3 v ^ l 3( l - x2) + 3V3\/3(l + x2)]' < (13 + 27) (13 - H x ^ + 3 + Sx^)
= 40 (16 - 10x2)
•ằx2 = - X- 5
0di v^y x2 (I 3^ 1 -X2 + 9N/ 1 +X2 ) < 40X2 (16 - 10x2) , (
dan hhng d (3) xay ra khi va chi khi f v/ i 3( r 3 - ^ ^ v g i T ^
v/l3 3^3 I^at khac theo bat dang thiic Cauchy ta c6
'l0x2 + (16 - 10.T2)
(3)
4.10X2 _ 10x2) < 4
2 <•
= 4.64 = 256, (4)
d i u bang 6 (4) xay ra khi va chi khi 10x2 = I6 - 10x2. Tif (3) va (4) suy ra
T9,p nghiem ciia phiMng trinh (1) la ~
Lifti y. Thco doi ItJi giai, d ca hai each ta deu thay rat ddii gian va dc liicu.
Tuy nhien thuc te day khong phai la bai toan de. Bai toan nay ra ddi di.ta tren bai toan: "Chiing minh rang v6i moi 0 < x < 1, ta deu c6
X ( 9 + x2 + 13v/l - x 2 ) < 16.
Tac gici bai toan bat dang thufc tren la Tien SI Tran Nam Dung, giang vien khoa toan, tntdng Dai hoc Khoa Hoc T u Nhien, D H Q G thanh pho Ho Chi Minh. Bai toan bat d i n g thiic nay da dudc ra trong de t h i Olympic 30/04/1996, trong ky thi do chi c6 mot em hoc sinh lam dudc bai nay. L d i giai dildc t i m ra bang phitdng phap can bKng ho s6. C h i n g lian, xet each 1:
Vai a > 0, /? > 0, ta c6
n / - 2 4 13 r-TTn ^ ^ 1 3 a V + ( l - x ^ ) 13 (rv^ - 1)-x^ + 13 13Vx2 - x^ = —.Ja^x^ (1 - x2) < — . =
a a 2 2a 9 9 / 3 V + (1 + x2) 9(/32 + l)x2 + 9
9Vx2 + x4 = ^.v//?2x2(l+x2) <
Suy ra
I 13v/x2 - x4 + 9v/x2 + x4 <
Dau bang xay ra khi va chi khi
13 (Q2 - 1) ^ 9 (^2 + 1)
2a 2/3
2/3
2 13 9
{ ? 2 ^ 2 Z j ; ^ 2 ^ ( a 2 + l) x 2 = ( / ^ 2_ i) , 2^ i .
Can chon a > 0, /3 > 0 sao cho
Q 2 + 1 = ^2 _ 1
13 (g^ - 1) 9 (/j^ + 1) ^ 2a ^ 2(3 ~^
Tir day t a c6 \bi giai nhir da t r i n h bay d each 1. ^ B a i t o a n 134. Giai he phuanq trinh ( 3 ^ 1 + 2x2 + 2^40 + V = 5v/li
\ + y=l.
B a i t o a n 135 ( D e d y b i k y t h i c h o n hoc s i n h g i o i k h u v y c D u y e n H a i va D o n g B a n g B a c B o n a m 2 0 1 0 ) . G'ldi he phudng trinh
a: — 3
G i a i . Dieu kien — ^ > 0. Theo bat dftng thiic Cauchy t a c6
+ + l + l + l > 4 y ( ^ T ^ 7 ^ l T l = 4|x + y| > 4 ( x + y ) . (1)
Dau bang a (1) xay ra k h i va chi k h i a: + y = 1. K i t hdp vdi dieu kien
X - 3
^ — ^ > 0 suy i& -2 < x,y < 3. PhUdng t r i n h thu: hai cua he tUdng dudng vdi
^4 g^2 - T j ,
Xet ham s6 f{x) = — + _ + _ + 3 in (3 - x) (vdi - 2 < x < 3). Ta c6
- - 1 , ^ Z . ^ - (^^ + 9^ + - 3) + 48
^ ^ ^ 16 16 ^ 8 ^ x - 3 1 6 ( x - 3 )
= - + 9x2 _ 133. ^ 6 ^ ( x - l) ^ ( x 2- x + 6)
i A. 16(x - 3) ~ 16(x - 3) - ^
fl-V (vi - 2 < X < 3).
Suy ham so / nghich bien tren ( - 2 ; 3), vay / ( x ) = f{y) ^ x = y. Ta c6 he
B a i t o a n 136 ( H S G t i n h B i n h D j n h , n a m hoc 2 0 1 0 - 2 0 1 1 ) . Giai he x^ + y^ + z^^<l (1)
Tif (1) t a C O- 1 < x,2/,z < 1. Tut (1) va (2) suy ra
^2007 ^ y2009 ^ ^2011 > ^6 ^ ^8 ^ ^10 j -
^ ^ 6 (1 _ ^2001) + y8 (1 _ y2001) + ^10 (1 _ ^2001) < Q
-Tii d i i u kien - 1 < x, y , 2 < 1, t a de dang thay rang ' - / I
x*^ (1 - x 2 ™ i ) , / (1 - y2ôoi) , (1 - . 2 ô ° i ) > 0. . ; , , . ^ ,
Do do, phai c6 dang thiic xay ra, tiic la
x 6( i - x 2 o i i ) = o r x - o v x = i
^10 ^1 _ 22011^ = 0 l 2 = 0 V 2 = l.
Ket hdp vdi dieu kion x*^ + y^ + < 1, t a thay he bat phudng t r i n h da cho CO cac nghiem la ( 1 ,0 , 0 ) , ( 0 ,1 ,0 ) , (0,0,1).
V i d u 8. Xuat phdt til phep bien doi tUdng duong '<
{ x2 + 4 x + 4 < 0 ,'
o2 + 4y + 4 < 0 "ôt\
? + 42 + 4 i O
(x + 2)2 + ( y + 2)2 + (2 + 2 ) 2< 0
""l 1 ?S}f 111 ?m n"^ r x2 + 2014X + 4 < 2010x
1 tx X ^^UXWll + 2)^ < 0 1 + 2014. + 4 < 2010., ta se xay dmg mot he da,ng lap doi vdi ba an (x; y; z), dUUc giai bang each cong ha bat phuong tnnh, sau do sit dung bat ddng thtic. Do la bai toan sau.
( x2 + 2014x + 4 < 2010y B a i t o a n 137. Giai he I y2 _^ 2014y + 4 < 2010z [ z2 + 2014z + 4 < 2010x.
G i a i . He da cho tUdng dildng vdi x2 + 2014x + 4 < 2010y y2 + 2014y + 4 < 2010.
.2 + 2014z + 4 < 2G10x
^ x2 + 4x + 4 + 4y + 4 + z2 + 42 + 4 < 0 x2 + 2 0 1 4 x 4 - 4 < 2010y
^ . y2 + 26i4y + 4 < 2010z f.ZZ^
^ 1 .2 + 20142 + 4 < 2 0 1 0 x ^ I 2 = - 2 . I (x + 2)^ + (y + 2 ) V2^ + 20142 + 4 < 2010x ( 2 + 2 ) 2 < 0
^ {x + 2f + {y + 2f + {z
da cho CO nghiem duy nhat (x; y; 2) = ( - 2 ; - 2 ; - 2 ) .
B a i t o a n 1 3 8 ( H S G Q u o c g i a - 2 0 1 3 ) . Gidi hf phuang trlnh s i i r x ' H \ T - + ./cos'-^y -f \r
s i i r . c V cos^
sin'-^f/ -) h \/cos^:r • 1 sill y
1
_20y_
G i a i . D i c u k i r n sin .r C O S T sin//cos/y ^ 0, .T// > 0. N h a n t h e o ve h a i i)hU(Jn<.
t r i n h c n a he, t a t h n (hi'dc
siii'^.r + , + \ + —\
Hurx V coH-^y J I
I ain^y H + \: H ^
siii^y V cos-^x
= 20 xy
{x + y) 2 • (1)
T h e o h a t Atxug thn'c B n n h i a c o p x k i , t a c6
cos .T H ~ > s i n . r c o s . T , c o s f r y V
sin X H TT- s m ./;/
s i i i 2 ; r l
/ | s i n 2 x | 1 3
+
sin:z;cosx[
2
> |sin2.r 1 + 3
\ 2 •2|sin2.T| 2|sin2.T|
T u r t i i g t i t , t a chi'tng m i n h rhroc
siii^?/ + I cos''?/ H ^ s i n •(/ / \ J
1 \
2|sin2x| 2 | s i n 2 x i
1 \5
> —4 .
N h i r vay. t h e o l i a t d a n g thi'rc Co-si, suy r a ve t r a i c u a (1) 16n h d n hoac bang
4W sin'^x -f- 1 \
s i i i ^ x / cos'^.r
cos-^x sin^y 4- cos'^J/ H ^
sm y/ V cos-^y 1 \
> 4 25^
42 = 10.
Mat k h a c , v d i xy > 0, t a c6
^2 ^
(.r + yY > 4.ry 1
x2 ^ 7 =^ 20
X + y)^ 4
xy < 10.
S u y r a ve p h a i ciia. (1) n h o h d n hoac b a n g 10. N h u vay cac d a u d a n g thi'tt c u n g x a y r a , d o (!6-
; :, { |sin2x| = l ^ x = 2 / = ^ + ^ , ; t e Z .
f h i t l a i : v d i X = y = - 4- —, fc e Z, t a C O
1 X
s i i i ^ x = cos^x = sin^y — cos^y = - , 2 ' x - f - y x - F y 2 '
[jlii do C H h a i ve c i i a i n o i p h i t d i i g t r h i h t r o u g he d c u b a n g \/To.
Y^y j;z=y = ^ + ^.keZ]h t a t ca cac n g h i e m c u a he d a cho.
I^IO P h i f d n g p h a p t h a m b i e n
J)6i v d i m o t so he b a t p h u d u g t r i n h d o i x i t n g d d n g i a n , c6 t h e a p d u n g p h U d n g p h a p d5 t h i de g i a i . T r o n g b a i n a y c h u n g t a se l a m q u e n v d i p h i t d n g p h a p b i l u d i e n n g h i e m t h o n g q u a t h a m so, dUdc goi l a p h U d n g p h a p t h a m b i e n . B a i t o a n 1 3 9 . Gidi he x'^ + y' + xy = 1.
G i a i . H e d a c h o v i e t l a i d i r d i d a n g 4- y = 1 - a, o > 0 2 - f y2 + x y = 1
x + y = I - a, a > 0 xy = (1 - a)^ - 1.
, ' 1 . ^
ifj .he/' \'. • Vay X va y l a n g h i e m c i i a p h i r o n g t n n h - (1 - ô ) i -I- (1 - a)^ - 1 = 0. (1) D i e u k i e u d o i vcJi a :
I A = ( l - a ) ' - ^ - 4
r a > 0
^ 1 ( - 1 ) ^ < ^ ^
(1 - a f - 1 ( a > 0
> 0
^ < a - \ < — ^ 0 < a < l + ^
Vdi d i g u k i e n (2) t h i (1)
a - 1 - \ / 4 - 3 ( l - a ) ^
a - l 4 - \ / 4 - 3 ( l - a ) ^ t = 1—
(2)
N g h i e m c i i a he :
a - 1 - J A - i { l - a f
" — ; 2 = =
ô - 1 + i74 - 3 ( 1 - a ) '
y = 2
X =
; <
a - 1 + ^ 4 - 3 (1 - _
a - 1 - A / 4 - 3 ( 1 - o)2 '
y = 2
/ 2 v d i 0 < a < 1 -I-
B a i t o a n 140. Gidi he I ^ItK^'^^'^^
G i a i . Viet he da cho dirdi dang
.T^ + ?/2 = xy + 1 + 0 , +y'^ = 4xy + b 3xy + b - l - a = 0
^\ "
2a - 26 + 2 -2 + y2 ^ 4 x y + b 2xy =
a;2 + y2 = ( x - y ) 2 =
4a - 6 + 4 2a + 6 + 2
I (x + y)^ = 2a - 6 + 2 Dieu kien doi vdi a, 6 :
(vai a < 0, ^ < o) ( v d i a < 0, 6 <
( v d i a < 0, 6 < o | (;
( v d i a < 0, 6 < o) (1)
| 2 a - - ° ^ + 1 > 0 ^ | ô < 0 , ^ < 0 ^ / 0 > a > - l - ^
K h i do tir (1), t a diwc I x-y = ±^ — [ x + y = ±y/2a'
/2a + 6 + 2 - 6 + 2, {x; y)
{x; y)
(x; y) ••
(x; 2/):
+
^ 2 a - 6 + 2
12
2a + 6 + 2 s/2a - 6 + 2 12 ' 2 2a + 6 + 2 V2a - 6 + 2
12 ' 2
12
+ 2 a + 6 +12 2
12 V2a - 6 + 2 / 2 a + 6+ 2 / 2 a + 6 + 2 v/2a - 6 + 2
12 12
/ h \ vdi - 2 < 6 < 0 v a - l - - < a < 0
V 2 y x + 2/ < 2
B a i t o a n 1 4 1 . Tim nghiem {x;y) cua he | ^2^^^^
thiic x2 + - xy rfai /^za ir? Idn nhat, nho nhat.
G i a i . Viet he ditdi dang
{
+ xy = 3 sao cho bieu
x + y = 2 - a , a > 0
(x + y)2 - xy = 3 ^
x + y = 2 - a , a > 0 xy - (2 - a)2 - 3.
x,y la nghiem ciia phiTdng t r i n h
^2 + (2 - a) t + (2 - a)2 - 3 = 0. (1) p i l u kien doi vdi a :
a > 0
A = (2 - a)2 - 4 f(2 - a)2 - 3 > 0 <^ 0 < a < 4. (2) Khi do
x^ + y^ - xy = (x^ + y^ + xy) - 2xy
= 3 - 2 (2 -ay-3 = 9 - 2 (a - 2 ) ^ Do dieu kien (2) nen
0 < (a - 2)^ < 4 - 8 < - 2 (a - 2)2 < 0 <^ 1 < 9 - 2 (a - 2)2 < 9.
Vay gia t r i Idn nhat cua + y^ - xy la 9, dat dildc k h i a = 2 va x + y = 0
xy = - 3 ^ X = = - V 3
X = - \ / 3 , y = \/3.
i l l .l,ô5''')
Gia t r i nho nhat cua x^ + y2 - xy la 1, dat diWc k h i a = 0 hoac a = 4.
• Vdi a = 0 t h i
• V6i a = 4 t h i
r X = 1
i y = i- f x = - l
I 2/= - 1 -
B a i t o a n 1 4 2 . Xdc dinh cdc gid tri cila m de he sau c6 nghiem duy nhat x + y = 2
xy = 1 ^ x + y = - 2
\y = 1 ^
{I X + y < m
+ y* < m + x2y2.
G i a i . Do vai t r 6 cua x, y la binh dang nen neu (x; y) = (a; /3) \h nghiem cua he t h i (x; y) = (/3; a) cung la nghiem. Vay dieu kien can de he c6 nghiem duy
f < -
nhat la a = /3. The vao he, ta dUdc <^ " - 2
• Neu m < 0 t h i khong ton tai a.
• Neu m > 0 t h i t o n t a i v6 so a sao cho - - ^ < a < m i n | y , .
• Neu m = 0 t h i a = 0. He c6 dang x + y < 0
\ ^0
^ ^ t luan : He c6 nghiem duy nhat k h i va chi k h i m = 0.
r x = p
1 y = 0.
B a i t o a n 143. Xdc dinh m de' he sau c6 nqhiem duv nhdt I x^ + 2y <m
\ 2x < m.
G i a i . Do vai tro ciia x, y la binh dang nOn ncu {x; y) = {a; ft) la nghiem ci'ui he t h i (x; y) — {ft; a) cung la nghiem. Vay diSu kien can de he c6 nghiem duv nhat la a = /3. The vao he ta diidc -\- 2a — m < 0. PhiTdng t r i n h nay co nghiem duy nhat <i^A' = H - m = 0 < ^ m = - 1 . Thay vao he da cho
f x^ + 2y<-i ^ f itly^-]
C^.2 + 2 y < - l
^ 2/^ + 2 x < - l ^ { . I (x + l ) ^ + ( t/ + l) 2 < 0 ^ ^
Kgt luan : He da cho c6 nghiem duy nhat khi va chi k h i m = - 1 .
\ + 3.T + 1 < ?/
B a i t o a n 144. Gidi he \ ->r Zy-\-\ z
G i a i . He da cho titdng dudng v6i x2 + 3x + 1 < y y2 + 3j/ + 1 < 2 2^ + 32 + 1 < X
x^ + 2x + 1 + J/2 + 2y + 1 + 2^ + 22 + 1 < 0 x^ + 3x + 1 < 2/
V ^2 + 32 + l < x ^ 1 ' ^ - Z } I (x + l f + ( y + l) 2 + ( 2+ l) 2< 0
Ho da cho c6 nghiem duy nhat (x; y\z) = ( - 1 ; - 1 ; - 1 ) . B a i t o a n 145. Gidi he { ^2 "^_^|^T< ^
G i a i . Ta CO
r x + 2y = 2 ( x^2-2y 1 x 2 - 2 . / < l ^1 ( 2 - 2 y) 2- 2 y 2 < l
N g h i e m c u a h e d a c h o l a j ^ = ?-2< vdi ^ ~ ^ < f < ^ + B a i t o a n 146. Gidz /.e { i = 3..
Qiai. Viet he dvfdi dang
f x- J/ = l+ a , a > 0
t x'-xy-2y'=3 ,
y = X - (1 + a ) , a > 0 ' * - x [x - (1 + a)] - 2 [x - (1 + a)f = 3
y = X - (1 + a ) , a > 0
2 x 2 - 5 ( l + a)x + 2(l + a)2 + 3 = 0. (1) pjgu kien cua a :
f fl > 0
\ = 25(1 + a)2 - 8 [2(1 + a)^ + 3] > 0 r a > 0
1 A = 9(1 + a)2 - 24 > 0 a > 2\/6 - 1.
V|y he CO nghiem^
X _ 5(1 + a) ± y/Hl + ay - 24 l + a ± v / 9 ( l + a)2 - 2 4
4
2\/6 , vdi a > — 1.
Bai toan 147. Gidi he { ^ - f y ^ < ' L
G i a i . Dat x + y = uvhxy = v, dieu kien - 4?; > 0. Ta c6 he "
( V > u (v = u + a, a>0
\ ^ \2 - 2 (u + a) < 1
(v — u + a, a>0 ( V = u + a, a > 0
^ \u - 1 )2 < 2 + 2a ^ 1 1 - y2 ^ r +2 < u < 1 + 1 + 2
n > 2 + 2\/rTS
u<2- 2x71 + a .
Dieu kien - 4(; > 0 <^ - 4 (u + a) > 0 <^
• Neu
1 - v /2^rT2 < 2 - 2vT+^ <^ (2 - \/2) V l + a < 1
<^l + a < 1 ^ l + a < i _ ^ 0 < a < i + v/2
~ 6 - 4 \ / 2 2 ( 2 - ^ / 2 ) ^
thi / ^ = u + g , I 1 - v/2^r+2 < u < 2 - 2 \/rT^.
• Ngu 1 - v'2a + 2 > 2 - 2 ^ 1 + o t h i he v6 nghiem.
Ket luan : He c6 nghiem x^h, y = t2 vdi
tl,2 =
1 - v/2a + 2 < u < 2 - 2 v / l + a
= u + o, 0 < a < - + \/2.
B a i t o a n 148 . G^d^he | + - y ) ! < 4 (1)
f/2 + ( x - y)2 > 4. (2)
G i a i . Tu: (1) c6 < 4, {x - yf < 4, suy ra |x| < 2, \x ~ y\ 2. He viet l a i + {x - yf = a, 0 < a < 4
y 2 + (x - y)2 > 4.
Do + (x - yf = u, 0 < a < 4 nen ton tai a e [0; 27r] sao cho X = -^/asina, x - y = ^/acosa.
. N i u a = t h i { ^ l O ^ _ ^ ^ { y = ^ . Thay vao (2), t a d a + a>4-^a>2. Nghiem ciia he da cho la | ^ ^ ^ 2 < a < 4
UV)(.
Neu Q 7^ TT, t h i dat i = t a n ^ , khi do sin a = 7 - ^ , cos a = - — ~ . Thav vao (2) ta diiOc
> 4.
> - .
a 1) Neu 0 = 0 t h i he v6 nghiem.
2) Xet 0 < a < 4. K h i do (3) <^ i 1 i ^ ^
dang thitc ( ^ ^ - ^ J + (^Y:p72 J = 1' va (4) suy ra
(3)
(4)
> -
a
2t \ 2 ^ / l - i 2 x 2 1 + ^2
^ - ^ ) ^2^)' + ( 2 - ^ ) (1 - ^ ' ) ' - 4* (1 - ^ 0. (5) i) Neu a = 4 t h i (5) c6 dang
(1 - i ^ ) ' - 4t (1 - > 0 ^ (J _ ^2) ( ^ 2 ^ 4 ^ _ < Q
e ( - 0 0 ; - 2 - ^ ] U [ - 1 ; - 2 + ^ 5 ] U [1; + 0 0 ) . (G) ii) Neu 0 < a < 4 t h i < = ± 1 khong la nghiem ciia (5). Chia ca hai vg ciia (5)
cho (1 - ^2)2 > 0, dat = z, ta duoc / 4 \
-2z+ 2 - - > 0, vdi 1 - - < 0.
\ a (7)
pigu kien de (7) c6 nghiem la
<^a2 - 12a + 16 < 0 6 - \/20 < a < 4.
Vay neu 0 < a < 6 - t h i (7) vo nghiem. Neu 6 - \/20 < a < 4 t h i a a 2 -
(7) <^ 21 < Z < 22 ; 21,2 = a a'' a
va 2t = Z ^ 2*2 ^ 2t - Z 3= 0 <!=>
l- < 2
K i t luan : Cac nghiem cua he la f x = 0^
{ y = s/E, 2 < a < 4 4t
t =
1 -
x =
2(^2 + 2t - 1) ^ (^))
neu 2 7^ 0 neu 2 = 0.
l + i 2 2\/at l+ ( 2
^ ~ 1 + ^2
(8)
^ = ; . neu 2^ 0 2 t h o a( 8 )
V [ f = 0, neu 2 = 0, /