: o)+ o G+ zoi l o c= zo t+ zo u oG
x2-(nt+4)r +3m+3:0 :
As I(o \\ C
@6 dAng bhn Ddc san Totitt hoc ,-ri Tu6li mi Stí.1, trang 14)
Luu y ring, hd đ cho ld hq c6 dpng ho6n vi vdng quanh. Vai tro cua x, y, z kh6ng binh
iling, nOn khdng thO gi6 su ring x > y > z
duqị Xhu v6y, l<ri giai đ ndu ld khdng 6n!
MQt ldi gi6i dirng cho bdi to6n ndy dugc trinh bdy nhu sau:
He dA cho fucrng ducrng vdi
lxt _ 2x*",&J _ 3: !\ý -2y * Jii -3 = z \ý -2y * Jii -3 = z t" l_, ._, J=_3=x. lz -LZT X6t him sO f (t):t3 -2t+\81-3 (r > 1). 1 Ta c6 .f '(t):3t2 -2+ -- > 0(Vr > 1), 2'lt - 1 suy ra hdm s6 /(r) d6ng bilin tr6n [1;+cc).
Kh6ng m6t tinh t6ng qu5t, gie su rittg
x: max{x;y;z}.
Trtdng hW 1. x> y> z = f(x)>-f(i>--f(z)
=y>7)Y)x:.Y=2.
Trudng hW 2. x) z ) ,," = .f(x)>f(z)>fb,)
> Y>x)z ) x:Y=2.
VAy x, y,z ldnghiOm cua PT
1 :7t -V a,fi _1-3, hay t3 _3t +,lA-3=0 (1)
Xdt hem sO g(l) :t3 -3t+\r:l _3, ta thdY
s'(1)= 3(rt - r1* -!(vI> l).
2Jt-1
Suy ra hirm so -q(r) dong bi6n tr6n [l;+co). Lpi thAl'g(2) : 0. nen PT (1) co nghiem duy
nh6t r: 2. Tu do he da cho co nghiem duy
nhdt (.t:-r';: 1 = 12: 2: 2).
NG9C HtdN
Ldr crru sA cruUix cnua I
Trong pC ttri tuy6n sinh vio lop 10 ciia tinh
HAi Duorg n[m hgc 2006-2007 c6 bdi to6n
Cho phwtng trinh
x2-(nt+4)r +3m+3:0.-: -:
Tim m de phmrng trinh cri hai nghiem x1, x2
thoa mfrn xi + xir > 0 .
Nhi6u hgc sinh dA gi6i nhu sau :
Ta co
L=(m+4)2 -4(3m+3)
=m2 -4m+4=(m-2)220;
n€n phuong trinh lu6n c6 hai nghiQm xl, x2.
Vi xl - fr>O n6n (x1 +d(t -kx2+x3)>0.
' (
r? -x1x2*1.;)*1,,;'lu x( -x1r2 *ri =[ri -x1x2-t 'lu x( -x1r2 *ri =[ri -x1x2-t
4*, )* 4,=( *, -Lr,) *lr; > o =( *, -Lr,) *lr; > o
. 2-) 4'
suy ra x1*4)-Q, tric li m+4> 0 e m> -4.
VQy v6i m> -4 thi phucrng trinh
*'*(*+4)x +3m+3=0
c6 hai nghiQm xyx2thoilmdn xf + x)20 .
c6 hai nghiQm xyx2thoilmdn xf + x)20 .
DINH VAN DONG
(GV THCS Thanh An, Thanh Hd, HAiDmng)