. x^, 4x +3 y, Ta thay phuong trinh c6 nghỉm khi va chi khi — va
HOAN V! V6 NG QUANH
?ing tong quat ciia h^:
f( X l ) = g( X 2 ) f( X 2 ) = g( X 3 ) «K y < /. •1.) V »: -"ii i f K ) = g(xi)
. i n h chat 1: Neu f , g la cac ham cung don difu tang , hogc cung don dỉu
giam tren tap xac djnh va h? c6 nghi^m (xi;x2;....;x^) thi X j = X j =... = x„ -hung minh:
Gia su f,g la cac ham cung don di^u tang tren TXD
Khong mat h'nh tong quat ta gia su: X j = min{xi, X 2,...,x„} ;i iẸ!;.: •<
li do ta c6: x, < X 2 = ^ ^{•>'\)^^{^2)=> z{^2)^z['>^i)^ ^2 ^ X3.... x„ < X j
Tai lifu on thi daihoc ^ng tao vd gidi Fl, mxf i , ftp f i , um u j JIIny^ggin 1 1 W A .
Tinh chat 2: Neu f,g la cac ham khac tinh dan di§u va h? c6 nghi^m
I r ( x i ; x 2 ; . . . . ; x „ ) thi X j = X 2 = . . . = x„ neu n le va n chan. M O T S O V I D U X i = X 3 = Xg = ... = X n _ i ^^2 ~ '^3 ~ ^^6 = — ~ " n néu 8X3: = 60y2 -150y + 125 V i dv 1. Giai h? phucmg trinh 8y3 = 60z2 -150Z + 125 = 60x2--150X + 125 2 x - ^ , 2 ; Giai N2 T a c o 60x2-150x + 125 = 15 5 ^ ^ 5 ^ Tuang ty x > — — , y > — — 4 4
Xet ham so f (t) = 601^ - 150t +125 vol t > ^ 5 ^ 125 125 . + > nen z > — — 5 ^ ' Vs. Vi f'(t) = 1 2 0 t - 1 5 0 > 0 V t € 5 ^
-:+co , nen ham so f(t) dong bien tren
+) Néu x > y thi Sx^ >Qy^ nen f (y) > f ( z ) , hay y > z d o d o f ( z ) > f ( x ) nen
z > x . Vi the x > y > z > X (v6 ly). +) Tuong tu cung xay ra truong hop x < y . +) Tuong tu cung xay ra truong hop x < y .
V^y X = y va tu do CO x = y = z . Phuong trinh thu nhat ciia h# thanh Sx^ - 60x2+150x-125 = 0 hay (2x - 3 f = 0 nen x = —. 5
2 Vay nghif m can tim ciia h^ la (x;y;z) = f 5 . 5 Vay nghif m can tim ciia h^ la (x;y;z) = f 5 . 5
. 2 ' 2 5^ 5^ ' 2 . 3x3+3x = + 1 V i dy 2. Giai h? phuong trinh: • 3y3 + 3y = 3z2 + 1 3z3+3z = = 3x2 + 1
G i i i
Vi 3y2 +1 = 3x^x2 + i j nen x > 0 . Tuang ty y , z > 0 . > Tren khoang (0;+oo) cac ham so f (t) = + 3t,g(t) = 31^+1 deu dong bien.
neu x > y thi f ( x ) > f ( y ) nen g ( y ) > g ( z ) suy ra y > z . K h i do Igi c6 f ( y ) > f ( z ) nen g ( z ) > g ( x ) hay x > z , tuc la x > y > z > x do do phai c6 x = y = z .
Phuong trinh thu nhat ciia h? thanh 'ix^ - 3x2 + 3x - 1 = 0 (1)
Phuong trinh (1) tuong duong vox 2x^ + (x -1)^ = 0 nen x = — ^ . ^
1 + ^ Vay nghiem can tim ciia h^ la (x; y; z) = Vay nghiem can tim ciia h^ la (x; y; z) =
, l + ^ ' l + ^ ' l + ^
2x(y2+7| = y(y2+63) V i dy 3. Giai hf phuong trinh: • 2y(z2+7j = z(z2+63)
2z(x2+7J = x(x2+63)
Viet lai h? duoi dang:
G i i i y y2+63) y2+7 z z2+63 z2+7 |x2+63 x2+7 t]'\O r: • tft2+63)
Xet ham so f (t) = '- voi t e K . t2+7
( t 2 - 2 l f ( t 2. 7 f
Ta CO f'(t) >0 voimpi t € l
Suy ra ham so dong bien tren M . Vi the neu x > y thi f (y) > f (z) nen y > z , suy ra f ( z ) > f ( x ) dodo z > x , t u c l a x > y > z > x .
Vay X = y = z. Khi do phuong trinh thu nhát aia h? thanh - 49x = 0 nen
X = 0 hoac X = ±7.
Nghi^m aia h# can tim la (O; 0; O), (7; 7; 7;), (-7; -7; -7 ).
V i 4. Giai h^ phuong trinh:
Vx2 + 5 + V y ^ = y^
\/z2+5 + V x ^ = x2
Dieu kif n x>l,y>l,z>l.
Khi do y^ = 7x^ +5 + ^ y - l > ^6 => y > . Tuong ty x,z > . V x 2+ 5= y 2 - 7 y ^
Vy^+5 = z 2 - V ^
V z^ + S^ x ^ - V x ^ Viet h^ da cho ve dang:
Xet ham so f (t) = Vt^+S va g(t) = t^ - vol t - ; . o o 3
Ta CO f'(t)= , * >0,g'(t) = 2t 1 ^ 4 t V t 1 1 ^ ^ t€ 3
Suy ra cac ham so deu dong bien, nen de dang suy ra x = y = z .
Khi do phuong trinh thii nhat ciia h^ thanh - 1 hay
7x^ + 5 - 3 + N / X^ - 1 - (X2 - 4 = 0. Su dyng bieu thuc lien hgfp ta dugc:
x ^ - 4 x - 2 / ^-v „ - _ ^ + - p = ^ - ( x - 2 ) ( x + 2) = 0 Vx^+S+S V x - 1 + 1 Do do ( x - 2 ) x + 2 - x - 2 = 0 (2) \r. ^ ^ 3 , x + 2 x + 2 1 Vi x > - nen , < , , — 2 7x^+5 + 3 5 V x ^ + 1 < 1 . 1R9 x + 2 1 ^ 4x + 3 „ Do do , + . X - 2 < < 0. 7x^+5 + 3 V x - 1 + 1 5 ,
Phuong trinh (2) tuong duong voi x = 2. Vay nghi^m can tim ciia hf (x; y; z) = (2; 2; 2).
2x3 + 3x2--18 = y3 + y Vi dv 5. Giai h^ phuong trinh: + 3y2 -18 = z3 + z 2z3 + 3 z 2 --18 = x3 + x
Giai
-1- 'vv
Xet ham so f (t) = 2t3 + 3t^ -18 va g(t) = t^ +1 voi t € M
R6 rang ham so g (t) dong bien tren R . £ - . - ^ 0 4 x , . y . /
'f(x) = g(y)
f ( y ) e
Khong giam tinh tong quat, gia su x = max(x;y;z) thi . ^ ' ^ ^ ' x > z Viet he da cho ve dang
nen
| g W ^ g ( y ) | g W ^ f ( x ) , suy ra •
g W ^ g ( z )
Phan tich thanh h'ch ta duqyc:
hay x3 + x ^ 2 x 3+ 3 x 2- 1 8 z3 + z<2z3+ 3 7^-18 hay x < 2 < z (x-2) ( x 2 +5x + 9J<0 ( z - 2) ( z 2 + 5 z + 9 Ma X > z nen x = z, suy ra X = y = z.
Phuong trinh thu nhat ciia h? thanh: x^ + 3x2 - x -18 = 0 j^^y ^ ^ 2. da cho CO nghi^m (x; y; z) = (2;2;2).