X goila idean nguyen t6 ngu• va chi lieu veil u, u E tich
1 II — Gia sit co mot da flute
y xit „. ynn
la hang tit cao nhgt cim nci ; Ming li luan nhtt del veil fi (x i , xn ) ta ductc
c 3 e2 3 3 c
vie (b 1 , b 2 , bn ) > (c 1 , c2 , ..., en)
Qua trinh d6 kheng the dai ra vO tan vi the Win tit (a l , an ) , (b i , bn ) , (e 1 , ., cn ) ma to dude la nhung phan tit cim tap him M nun han trong be de 4. Vay qua trinh Mtn phai them dill, We la sau mat se Mu han Jou& ta se phai dupe
0
= Tx (xi x„) A CI? 12 4-13
Ta suy ra tit dd rang
f(x i , x2 , " xn) = a a e22.
1 —1 / 1
Cnn 6.11 2 C22 3 ... Unn .
Vay da thee h(xi , x2 , ..„ xn) can tim la da fink
11.(2C i , .X2 , 2., x,) = et/ a1 a3 x,241, xb b 2 xbz —63
2
1 I I —1 1 Gia sit co mot da flute Gia sit co mot da flute
le(x i , xn )
hit; , otn) = f(xi ,
an) = an).
tip dung he qua dm b'O' de 5 ta co
h(x i xn ) = h'(x i , xn ). n sao cho
Phep chdng minh coa dinh Ii 4 !Mang nhitng cho ta bigt stt ton tai Na duy nhgt ciut da thtic h(x , xn) , nd don cho ta
mat phuong phtip thuan tin de thitnh lap h(x i ,.:., xn). Vide tim da thlic h(x i , xn) sao cho f(x t , xn ) = h51 ,..,an) got la
bidu thi da thug diti ring xn) qua cac da thile d61 ring co ban C1 , ..., dn .
Vi du. Trong vanh da tittle Z [x i , x2 , x3] trig ha s6 nguyen hay bidu thi da thac ddi ring
x3 + x2 2 X3 + 2x 1 x2 + 2x x + 2x 3 1 3
2 qua the da tittle d61 ring co ban 61, 62 , 43 .
Truoc hgt ta nhan xet rang
2x 1 x2 + ?xi x3 + 2x2 x3= 262
\ray chi can xet da tittle dai ring
f(x i , x2 , x 3) = + .32 +x3.
Hang tit cao nhgt cua nd la xi x3 t x° x3°- 2 Theo dinh li 4 ta lap da thug
/3 (xi , x2 , x3) = f(x 1 , x2 , x3) — Oro -0 a(3) = + + x33 _ (x, + x2 + x3)3
—(3x2i x2 + 3x2i x3 + 3x 1 x2,2 + 3x 1 x2 + 3x2 x3 + 3x2 xj + fix, x2 x3).
Van theo dinh li 4, ta lai sap xgp nd theo lot tit dign va tim hang tit cao nhgt elm 11 (x1, x2 , x3) . Nhung trong trUting hap Cu the' nay ta nhan xet rang
xj x2 + xi r3 + xl r22 + + x x3 + x2 .3 = = (x1 + x2 + x3) (x, x2 +xl x, + x2 x3) - 3x1 x2 x3
Do do fi (x i , x2, x3) = —361 62 + 3a3 .
Duel cling
x3 ) =61 —36162+363
2x 1 x2 + 2x 1 x3 ± X3 2x2 = va
= 62 + 3153 + 252 •
Trong thgc tign ngdOi ta con (Ida ra mat s6 nhan xet de viac bigu din dirge nhanh clidng hon. Ta hay xet try& Mt mat da Vide dgi ximg clang cap xn) EA[x 1 xnl cd hang td cao MI& la
a a
a X Il X22
Vay bac dm /(x, xn) la + a2 + + an . VI cac da thilc dal ring ca ban al , 62, ..., 6 la Bang cap cd bac theo thd ta la 1, 2, ..., n, nen da thdc tieh
a - 012 - 43
Ming clang cap va cd bac la
al - a2 + 2(a2 - a3) + + nan = al + a2 + + Do dd
11(X1 , X2 , ...; xn) = fix]. 7 x2 , d - a bai l 42 012 -43
Ming (tang cap va ed bac la a1 + a2 + + an ngu nd khac O.
Sap xap fl (xi ,..., xn) theo 161 td din va gia six /34 xinn la
hang tit cao nhat cua nd thg nil
6 1 + b2 + + bn = al + a2 + an
va (a 1 , a2 ,.., an ) > , 6 2 , .
Theo dinh li 4 ta cd mat day hdu han phan td cna lc" (6) (a 1 ,..., an )• > ton) > (c 1 , c 2i) > .
thent man tinh chat
a1 et. an
(7) b i bn
va trong truing hop 6 day
(8) al + + an = 6 1 + + b„ =
Do cd them tinh chat (8) nen s6 phtin tit aim day (6) giam di nhieu. Tap hop can phetn tit aim day (6) la met be phttn cot% tap hop him han
M = , tin ) , (tmi , tmn)} trong dci
> va.
iii + ti2 + + tin = cti + a2 + + an
vet i = 1, 2, rn. Theo dinh 11 4 f(x i , xn ) via dyed clang
E
f(X i 3Cn ) = upr
i -1
vet ti e A vet ti = 0 nen (tti , , tin) khOng cd mat trong day (6) hay quay lai da thee
f(xi , x2 , x3) = 4 + Tap hop M d day la
= {(3, 0, 0), (2, 1, 0), (1, 1, 1)1 Do dd
fix, , x2 , x3) = t1 61- 0 602 - 0 + r2 621 -1 621 - 0 6-03 + r3 ' 621 = 6-; 22 61 62 + C3 ;
ti la lig tit eim hang tit cao nhat tha /Oil , x2 , x3), do dO gi 1. Mutin co r2 , r3, ngubi to thay x 1 , x2 , x3 theo thd to
bang 1, 1, 0 ta Male
2 = 8 + 2g2
do do r2 = -3. Sau do thay x 1 , x2 , x3 theo the to bang 1, 1,
-1 ta co
1 = 1 + 3.- 13
hay a3 =
3
Vay x2 , - x3) = - 36162 + 363
Pinning 'tip vita thing goi lit phoong phap vgri h6 td Mt dinh.
Chit 9 : Khiing phiti bao gib pinning phap nay clang co higu lye Mu A la mat vanh hay Mtn nila la mat min nguyen, nhung
hitu han.
Trong truang Mp da that xn) khang phai la clang
cap thi ta nhan xet rang the hang tit ding mot cap cos. no lap thanh mat da tilde d6i xting clang cap do do ta co f(x i , xn) la tang dm nhung da Hide d6i ring ding cap, chang him trong vi da tren
4 + YZ + x3 ÷ 2,1„2 2.1x3 2x2x3 la tang aim da thde dal xdng (tang cap bac 3
x3 + X3 X x3 1 2 3
va da th e dgi ming ding cap. be 2 2•T 1z2 2t1i3 + 2%2'3
ling dung. Tim the s6 nguygn a, p, y sao cho a p y = 6,
a3 p3 + y3 = 36, •
a + fi + y = 6. Theo vi du tren ta co
P3 + Y3 = + P + y) 3 — 3(a + + (a,8 ay + PO+ 3a/3Y 'Pa suy ra co + ay + fly = 11.
Mat khac xet da thdc f(x) E Z [x]
f(x) = afix fifix r). GM sit a E Z, ta c6
f(a) = (a — a)(a — (3)(a — y).
Vi f(a) = 0 khi va chi khi mat trong the thus 86
a —a, a —p,a —y bang 0, cho nen the nghiam cua f(x) la
a, p, y. Khai trign f(x) ta Mtge
fix) = x3 — + fi + 11)? + (a/3 + ay + fiY)x afiY = x3 — 6x2 + th -6
vi a + /3 + y = 6, 4 + ay + fly = 11, apy = 6.
Da thtic f(x) co tang cite hO sd bang 0, do d6 f(x) co mat nghiem la 1. Theo he qua dm dinh li 4 trong (§1, 4), f(x) chia het cho x — 1. Chia f(x) cho x — 1 ta duce
f(x) = (x — 1)(x 2 — 5x + 6).
Da thtic x2 — 5x + 6 cho ta hai nghiem la 2 va 3. Vay Cite s6 nguyen a, p y can tim la 1, 2, 3.
BAI TAP
1. Trong vanh Z ix1 , x2 , x3] biau din da thiic xi +4 3
qua the da thdc deli ming co ban.
2. Prong vanh Z/2Z [x l , x2 , x3] biau din da thdc
lxi + 1xZ + 1x3 qua the da thdc dgi rung CO ban 3. GUI h@ phuong trinh
x +y +z = —3 x3 + y3 + z3 = —27 xd yd z4 =
113
cHUONG V