解多元高次不定方程组

1、一元幷次方程代数解研究好石一元幷次方程代数解研究好石解三元二次不定方程组已知方程组成孙 比中a、b,K为未知数，求abK的一组解.3町叽391 厂 b；+(b；2b)abh 严 3K(k22-2k1k)a2+(2kJ2k-k,k/+k2kJ+3bkJ-bk!k2)a+b2k2+(klik2-k!k-2kb 一3 讣扎+b；+3b；=3K：解法一：令式中（可-2昵=伏 则冇3k扎-3b厂b；=3K、解得b = (kf-2k2)a/k!K=(3k:k-3k-k/)/3将解得的b、K代入式.整理后就可以得到一条关丁p的一元二次方程，解之可加的值 解法二*由式得K=3*厂3居一爲+(财一2為)Q-纨/

2、3,代入式,整理得(k22-2klki)-(k,2-2k2y/3a+ (2k t kjk3bk3bk k2)2(3k lk23kk (k /2k:)/3-2kl(kl22k2)b/3 al(爲一& 3)十(嘉促厂嘉嘉一 2财)十 2(3klk2-3ki-klt)k,/3b-3ktk1k+k/3kJ2-(3klk2-3kJ-klt)i/3 =0令式中a的系数(k22-2klk,)-(kf2-2k2Y/3a+ (2k ktk jk22 + kJt t-3bk (bk 2(3klki3k fki.k l1-2k)/3-i-2kl(kl22ki,)b/3=0则式化为(k-kl2/3)b2+(kl2k-

3、klk-2k2(3klk-3k-klt)kJ/3b-3klk2kk3k/-(3kJk2-3kJ-kJty/3=Q式是一条关于b的一元二次方程，解捡的值則a,K确定注：不定方程组是右无穷多组解的，其它通解就右待各位同好自行参悟吧!解四元三次不定方程组(一)已知方程组成立,其中Sb,sd为未知数，求心d的一组解.3ak+5kik29dk)a 2bk23bkI26d2+6kf12dk:+(4k:a+3bkr)c = 0 cki+(3k/4k2b)c* + (8adk23kI,5akk23abk128ak26bdkJ-2bkJk2)c-ak1+(4bk2-i-6dkJ2klk2)ax + (bklk2

4、9dk1-i-10dkJk,klk22-3bkJt)a+b3kjt+(4dk-4kf)b2(2k,2k-6dkb+4di-12d2kl2dk/+k,4-4k/=0解：由得d=(4kJ+3akl)/4,代入式,整理得(4kic-3akl2/2i-4klk2)a(2bk-3kl2+3klc)b=0(7)令式中的系数分别为0,即4k2c-3akl2/2i-4k,k2=02bft2-3b； + 3和=0解得a=2kxklk)/3kl2b = (3k,2-3k,c)/2k2将解得的Q,b,d代入式，整理后就町得到一条关丘的-元三次方程，解之即得c的值.注：不定方程组是右无穷多组解的，其它通解就右待各位同

5、好自行参悟吧!一元川次方程代数解研究好石解四元三次不定方程组（二）已知方程组成立，其中a.b.c.d为未知数.求a、b心d的一组解.4d-4kJ-3ak22bkl-2kl2=0(4)(貯一3他免+3财)+从浪2+“爲一 2财)c-9A/+5怡人+為亀a+(财+2人)尸+ (3ck26dkJ+2k!ki3k22kli)b i-c2kl5klk2c+6d2+(6kl2 12ks)d+kl44kl2kJ+4klk22+6k2 = 0(3klk2k-k/)ai-(klkf-2kl2k4ki2)b-(klzk2+5k2k)c-(2kls-6klk6k22)d-2klki2-k/ka2I +b2k2ki+

6、(4cklk3ck22dklk2-8k1k2ki3k/)b2c2kIk(8dki-4dk,klk8k2)c-9d2k2+(10k2k3+2k,2kd+3k12k2k-klk/-k2ka+ (kf-2klk3)b+-ck!k(2k,4ki)d-4k4kl!k-2k1k22b!(6)+-4k3c2+(6dk2+2kl2k2+2k2k3)c-6d!kl(4k:k3-6k-4kli)d-2kI,k3kJ2k/+2k!kf+2k；k3b -c,k2+(2dkt+2klk3+3kf)c2(2kkJi-10dk!k-3k2,-kl,k2)c+4屮+(6财一 Z2阳才+(8嘉石+22可一 8他篦+2他)d+2

7、為：-4局嘉3+财一 4材=0解：由式彼/=(3必,+2地+祕厂2财)/4,代入式得(k；-3k,k33k/)a2+bk,k2+(4k-2kc-9k2(3ak2+2bkJ+4kr2k,2)/4Sk2k3+k2k2ia(k+2k3)b2 3ck2-6k1(3ak2+2bk+4k3-2kl2)/4+2kik3-3k/-2klib+c2k-5k!k,c+6(3ak2bkl4k-2kl2)2/42+(6klt-12k3)(3ak2bkl+4k-2kli)/4+kl4-4klik+4k!k/+6k/=0(ft ；3 ft 占 3+3 b ；)(T + bk lk2-(4ks2klt)c32k2(3ak2

8、+,2bkl4kJ2kl2)/ 22+Skj- k,2k,a+ (貯+2居)X+3ckj3kl(3ak2+2bkt4kj2kl2)/2+2 怡上厂3A；2 昇b +c 亀5怡上#+3( 3a打+26他+4 嘉2/j：)/ 2(3kl26kJ)(3ak2-2bk. + 4kf2kl2) / 2+k. ： 4b ； b 4b h ；+6b ； = 0(b ：3 h 见+3 b 2、a + bk ,k2-(4kJ2k/)c3ik2a/213k2kIb/2-32k2ki-32k2kl2/2 + 5 k 2k kk :a + (钉 +-l3ck23klk/i/23 k /b6k kj3 k tk /+2

9、 k k j3 k/2 k / jb+c k j5 k ,k+3(3ak2-2bkl + 4ki2kl2)2/2 -3ak2(3k126kJ) / 2 + bkJ(3kl26kf) + 2k k 6k k .2 (3 k . 6k + h ：4b；bj+4b；h； + 6b： = 0(k,3-3klki+3k/)a2-3ikfa2/22(klki-32klki)b + (4ki-2kl2)cl0kl2k2-13kikJa+ (财+2 处)；一3 可F+(3M 厂处岛-3/-Q 2-3h；2b；)b 十 c 鼻厂 Sbjt* +332a2k22b2k(4kr2ki2y22-3akJbkl+2-3

10、ak2(4k- 2可)+22 bk,(4k-2k/23 -bk!(3k12-6kJ)2k(3kl2-6k)-k!X3k!t-6k3)k!4-4kfk4kIk/+6k；=0(kli-3klki3k/)a2-3ik22a2/22+(klk-32klk,)b + (4k(-2kl2)clOk2k-13kikla+ (kl2+2kJ)bz-3k/b2(3ck-4klki-3k-kl2-3k22-2kls)b+c2k-5klk2c+3/防/2+30 快：/2+3(2 他一财 r/2+3a 爲必 j/2+3% 爲(2 免一财)/2+3 以/(2 免一貯)+bkl(3k/6kJ)+2kJ(3ki26kJ)k

11、t!(3kl!6kJ) kl44kl kJ+4klk2!+6k/=0(A/- 3k,ks3k/)a2+33a2k/23-33k；a2/22+ (k!k2-32k1k2)b(4kJ-2kl2)cl0kl?kt-13k,kia-3zak2bkl/2+3!aki(2krkl2)/2+ (kl2ki)b3b:klI/2-3kl2b(3ck4klk-3k,-k,2-3k/-2kl)b3bkl(2ki-k/)-bkl(3kl2-6ki)+cklSklkic+3(2ki-kf)2/2+2kJ(3kl2-6k)-kl2(3k/6k 3) +b：4b：b+4bh；+6b； = 0(k,t-3klk-3k/2,)

12、a2kIkJ)+ (4k-2kl2)c-2k2kllktik2/2)a+ (2kk/ 22)b (3ck24kkJ3kIk/3k22k/)bc2kSklk-2,3k2,3kki+3k*/ 2+2,3kl2kJ22,3k/3k-6kl2kJ-kl44kliks+4klkf6kji=0(kJi-3k,k-3k/2i)a2klk(4k-2kl2)c-2k2ks+llkl2k2/2a(2k-kl2/22)b2i-(3k2c4klkJ3klkl23k:tt2kl,)b-i-ckl5klk2ckl1/ 22,3kl2-2kl2kJ-i-4kk 6k (2 0 &D,=kl3-3klks-3k/2 D,=k

13、,k2, Dt=4k-2k, D,= -2kJt,+1 lk,2k2/2, Ds=2k-k/2 D=3k2, D,= -4讣m3b；-2h；, Ds=-Sk,k2t 0=-财/223财+2跳亀+你岛，+6财, 则式化为jD：a+(D 上 +/)+jDJa +D/ + (D&+Z)7)b+c经 j+D#+Z)g=O(8)由式得D + (D+c+,)/2 丁-(D/+D0Dj2/2*+Dy+(6：+%+c亀+亠6=0Dl,/ia(D2b+Dc+D1)/2Dl,/2Y-(D!2bi+DJic2+D/+2D2D3bc+2DiD2D)D4c)/2iDlD/2a+(D+Q3C+Q,)/-D/b2/2iD-

14、D/c2/22D-D/2iD-D:Dibc/2D-0.0720-D.D/ZDj+D +(DD 7)b+c kD gC- D ,=0一元川次方程代数解研究好石D；/2a + (D2b + DjC+0)/2DT +Dsb2-D/b72!D,(DDJ)b-DJ)Jbc/2Dl-D2Dtb/2Dl-c2klDy/2:Dlt Dc -DQg/2D j D9-D42/22D,=0D/zo + (D+Djc+DJ/2D/22+(D-D/2zD/)+(6+D 厂 DQj/2DJc-DQ,/2D +伙厂 Q；/2Jc+(D 厂 DQ,/2DJc+D 厂 ZV/2 严 0&ED-D/22Dlt EDD-DJDs/

15、2Dn E产一EkrD/22DtED-D/20瓦=厂；/2力八则式化为D/方十(Q+0c+Z)/2fT+EF+(E#+Ej)b+EF+(Q 厂 QQj/2DJc+E=0(10)由(10)式得 D/2a4-(D+DJc+DJ/2D/22+E/26+(E2c+E,)/2/2-(2c4-EJ)722/+/+ (D 厂 DQ,/2DJc-E严 0Dll/2a(D+Dic+D4)/2Dll/i2E,/2b(EJcEJ)/2El,/12-E/ci/22E-E2Eic/2E-E/21E,+Ec-(DD(D4/2D/)c+=0/+(+d#+)/2 F+E %+(E +E)/2E “F(ID (E4-E/22E

16、l)c2-V(Ds-DfD4/2D-Efi/2El)c+E6-EJI/22El=0令QI)式中(E-E；/2E)c+(D厂厶/2EJc+E厂E；/2迟=0,这是一条关于c的一 元二次方程，设G足方程的一个根，则)式化为 严a+(+)+DJ/2DT+E/勺+(E+Ej)/2E/Tu0D；/2a+(D+Q+D J/2DT=一 E,t/2b+(E+EJ/2ET (12)02)式两边同时开方得D% + (D/+Q+DJ/2D“=iE/9 + (E+EJ/2E 严Dl,/2a-D/2D,/2Dicl/2Dl,/2-D4/2D,/2-i(E2cE/2El,/2=iEll/!bDa/2D；-iE 严 b =

17、 i(E*+E)/2E 丫一0心/21)严一0/20严_。纭b=i(E+EJ/2E 严-D/2D 严-0/2D 严一 严a/(0/2Q 严一适严)将解得的b,c,d代入式中，整理后就可以得到一条关于“的一元三次方程，解之即嵌的值注：不定方程组是右无穷多组解的，其它通解就右待各位同好自行参悟吧!一元ri次方程代数解研究好石解五元三次不定方程组（一）已知方程组成立，其中a,b,cde为未知数，求abcde的一组解.(2kl24kl)a3bk22ckl+4e+5klkJ=0(kl4+6k4kfk)+4kik/)a2+ (5k2k-k!k2)b-V(2klk3k22kJ)c5dkIk, + (6kJ2

18、12kJ)e+3k/-2k!k2kJ+klka+ (3可“ ：一3為爲0+(毗“厂7爲财一 4可一 2可+必岛)6+(财+2居)c 一 (3爲臥厂7怡上$)c+嘉屮 -4k2k3ck 2- 2k 3k / 6k k d 6e?+15k sk 2 9bk26ckJ)e+昇一 5他為+5貯貯+5嘉可+5氏篦=0 (财一 4材+2貯可一 4嘉為篦)d+(3為篦打一化财-怡比；)6+“；+2嘉可+2财打一2财打比+ (2klk2kl3k2lklik:)d-(12k2+2kl48klkj+8klk22)e3klk2kj+k2fk)a+ (2“/-貯打0(4- (3k2i-8klkzk)c(klk2+8k

19、Jt)d-(l Ob 扎+2h ；k)e3b；：4b ；b； +11 k tk k fb +4c2k,2k-(2k,k22-i-4kf2)c2+(2k7kf + 2kl2k2)d + (4k,kJ-6k22-4kli)e-2k,2k2kf-i-6k7kJ2+klk/c (2k ,k t- 3k/)d -(4k)k ,28 k / k j k k / 1 Oek ,kd + (6k / 12k j)2+(6k /+2k kt4k tk Ji()e +2kl,k-4kl2k!2ki+k,kf-6klkk/ka-(3klk!k-k/)bi+ (4k/-2kl2ki+klkf)c-(SkJkJ+k,2

20、kd+(6k/-6k,kt+2kli)e+2klk6kl2kikkb2 (6) + c2kikJ-v(4klki-3k/)d2eklk4klikt-12klk/+k/k-2k,2kfc+2d!klk2+ (8k- 4k；)e+2k；A,+2k,kjk3k/d -9e2k2+(16k,2kt- 14k,k/-8k/- 4h；+3h；h扎-h；h； -2k/k3+4klk2k/b-(k/-2klk3)c3+-dklk2+(2kl2+4k3)e+6klkk-3k/c2+1 4dk(8kJ24kfk+5kjk,-6k2)d6ekJ(6kl2k,- “札&Je2 打匕+农/可一 9k.k2Jk3 -3k

21、*4kJt-8k/kjcdik2+(7k2k32k.k2-2ekl)d+ (4ki,12k;k36k)e+3ks k2kJ4k.k:, 11 打可町&Jd+4e + J 5eklki+(10k,kj +1 Ok2 kj-2k j 1 Okj k- 1 Ok； k5kk2k+Sk/k/kjk/0解：令式中(2貯-4打)a-32=0, -2% + 4e + 5Eb=0，解得b=(2k,2-4kJ)a/3k:e=(2ck-5klk2)/4将解得的b “代人式得(嘉+6怡；4财他+ 4匕可)矿+(5kJt3+kllk2)(2k-4kJ)a/3k2 + (2k!k-3k/-2kJi)c-5dklk2+(

22、6k12-12k3)(2ck-5k!k2)/4 +3k2i2ktk2kj+kj3k2o+ (3kf+kti-3kIks)(2k,2-4k3ya2/3!k22+(8k!2k-7kik22-4kf-2k!ck,k2)(2k!2-4k)a/3k2+ (耐 A-2k)c(3kt2k2 Jkkjjc+kfd A-(4kJ2kl2)(2kl4ki)al3k: -3ck2+2kli3k226k:k3d 6(2ck-5klk2Y/4i+15ktk2-9k2(2klx-4k3)a/3k2-6ck,(2ck-Sk,k2)/44昇一 5為亀+5昇财+5化町+5嘉亀=0(kl4+6kf-4kJ2kJ + 4kk22)

23、a2(5k2k3+kJ2k2)(2k12-4k3)a/3k, + (3k/+klt-3k1ki)(2kI2-4k3ya2/3!k22 -i-(2klk33k222kl,)c5dklk2-(3kI!/23k)2ck15klk2(3kl2/23k) + 3 k 2l2k ,k 2k 3+k / k za + (8k.k 7klk224k322kl4+ck,k2)(2kI24k)a/3k2 -(4k2k)(2kl24kJ)ad/3k2 -3(2kl2-4ka(2ck-5klk2)/4+kld2(3ck2k1i-3k22-6k,kJ)d+ (k2ky-(3k,2k-7ktkJ)c+3(22c2kJ2-

24、20k12k2cSkl2k/)/2i (15klk2-6ckl)(2ck-5klk/4A-k,s-5k,3ki+Sk,2kSk,k5k/ki=0貯+6材一 4昇打+4嘉貯+(5A上,+嘉亿)(2貯一 4打)/3免+ (3防+貯一3觥力)(2财一 4打M3快 +(5=2貯一3昇一6处钻，尸2免一貯/2, 0=3亀亀/2+7怡岛，D9=k-Sk,3ks-Sd9kl2k/23+Sk,kSk；kn 则式化为ZM + ()/:+/)d+D )a+d +(3&并+)、)d+Z)0+Z)7c+Z=O由式得D/2 +(0c+ZM+D)/2)f2-(DaD4+0)72/+Ed2 + (3 嘉 c+DJd+D0+

25、D+D 严 0Da(Dzc-VD,dD4)l2D!mY-(DyDd2D422DpcdA-2D2D2Dp4d)l22DIkld2+ (3 k 2c+D s)d+D 6c2+D jC + D s*= 0-9-一元幷次方程代数解研究好石D/2a + (D/ l DDt)l2D!，2Y-Dyi22Dr D；d2!22D-D/22D-DQQ2D, D(D/H2D (3kc-vDd+D4eS D/:+D,= OD/,7tf+(D/+D/f+DJ/2D/zz+*A DfdfDzGhf+DM-DQAQD厂DPM/2D+D6c2- D/c2/2:Dl-D7c-DiD4c/2Dl +DS- D/2?D;=0D,l

26、!2a + (D/: I Dx/4D4)/2Dllt22A-(k,- Dt2/2:Dl)d2+l(3k-DfiJ2Dl)c-D,-DlDJ2Dld(9) + (Dt-D22/2rDl)c2(D-D2D4/2Dl)c+Dt-D/2!Dl=Q设E尸跳一Ee=3k-DDJ2D Q,DQ/% +(+/+, )/2/T+Efd+(E名+耳)/2匸0Dtl,2a + (+)+,)/2/吁=-E,l,2d+(%+)/2时丁(式两边同时开方得/+(+)/+, )/2/=4/匀+(+3)/2竹D/S+D 易/2Dr+D/2D 严+)丿2=运/%+讽+耳)/2/“D3d/2Dl,t2-iEll/2d=i(E3cE

27、s/2Ef2-Dtl,2a-Dcl/2Dlia-DJ2Dlt,2 d=i(EE2ED,l2a-DJ2D,ll2-DJ2D/aKDJ2D/,2-iE,t,2) 将解得的b,c,d,e分别代人，整理后就可以得到一条关于。的一元三次方程，解之即得a的值.注：不定方程组是右无穷多组解的，其它通解就右待各位同好自行参悟吧!一9一解四元三次不定方程组（三）已知方程组成立.其中abc.d为未知数.求absd的 组解.-c匕+ 4d+Qb c打+0/+(D,d+DM+QJc+4d+DM2+0M+09=O解：由式得d=(-Dta-DJb +c&厂 D J/4将d代人式得D/z+lQsb+QeC+Q,Q#+ch/

28、QJ/4 +0a+DJ)2Jr-Dllf+Du(r)Ial2b-ck1r)J)/4 + D1b +c嘉+D”(0a+M 厂 Q3)/4+Q“c+ 6(-DlaDb+ckiDi)2/42+Dlf(DiaDb+cklDt)/4-i-Dlli=0D4a2+(DsbDlic-DlD/i/4-D2D7b/4+cklD7/4-D3D7/4D,)a+D9b2+ (Di(fi-DlDlla/4-D2Dnb/4+cklDl/4-DDll/4 +0jb+%+ (-D,Dla/4-D!Dlb/4cklDu/4-DiDn/4+DH)c+3(Dl2a+D22b2+c2kD2+2DlD!ab-2Dlk,ac+2DfDJa

29、-2D!klbc2Dfih-2klDic)/2i -DPM-DQM+DQJ4 - DDJ4+D 日D4a2-DlD1a2/4(D-Dfl7/4-DlDll/4)b + (D6+ktD7/4-DlDn/4)c-DtD7/4Dlla +D9b2-D2Dl,b2/4(Dl9+klDn/4-D2Dls/4)c-D9DJ4Dl2b+c2k2c2krDIJ/4+(-D9DlJ4Dl4)c +3Dl2a2l2i+3D.b2/2i3c2k/2s3D/l23+3DlDab/22-3Dlklacl223DlDJa/22-3D2k,bcl22 +3D2D)b/22-3klDJc/22-DlDl,a/4-D2Dl,b

30、/4+Dlsckl/4-DtDl5/4+Dl6=0-11-一元川次方程代数解研究好石(Ds-DfiJ/4-DtD/43DtD2/22)b+(D6+kfl7/4-DlDlJ4-3Dlk,/22)c+3D Q J2-D Q J4 + D 厂 D Q J4a+D-DQT4+3D；bf+ UDg+bQM4-DQM4-3DJiJ22)c+3DQJ22-DQJ4 Dn-DQJ4b+c2h2+3cl2+clQJ4 +(D、bJ4-3kQJ2JDQJ4+DJc+3D：/2DQJ4+DM=Q(D厂 DQJ4+3D；/2W+(D-DJJJ4-D,D/43D,l22)b + (Dk!D7/4-DlDlJ4-3D!k

31、l/22)c +3D!DJ22DJD7/4-Da-D!DlJ4a+ (D9-DQ“/4+3D；/2)F +(DlktDll/4-Dftl/4-3D2k,/22)c3D!DJ/22-DfDll/4+Dl2-D!Dls/4b+(虬+3h：/2+hpj4 记+(DAJ4-3bfDJ22Dpj4+DJc+3D；/2-DDJ4+DM=0&Et=D1-DlD7/4-V3Dl2/2,t E2D-Dfl,14-0,0430flJ2E 严 D&+kQJ4 - DQJ43D 局/2 E 产 SDQW，-QQ/4+D厂E产D厂D/)/4+3D畀2，EDl0+kiDit/4-D2D13/4-3D2k1/22,E产3D

32、QJ2JDQJ4+D旷DQJ4, =打+3貯/2+嘉0屛，E9=D旗丿SkQjF-DQM+Dg目产325皿4+D“,则式化为Esa+(EJb+EjC +EJa +E/*+(E+E?)b+8+农+“=0(8)由式得E1,/2a+(E2b+E3c+E4)/2Hl,/2i-(E2b-hEcEy/22Hl+E+(E(cHJ)b+Esc2+EH10=0E；/2a(EbEfirE/2Ell/2Y-Eb2/2zE,-Ec2/22Ef-E；/22E,-Ef)Eic/2Ei-E!Efi/2E, -EfifC/2E/+，7+(E6+E7)+E2+E9c+E/tt=0El,/Ja(E2bE3c+E4)/2El,/2

33、Y-E5bI-E2Jb72IEl+(Eip+E7)b-E3bc/2E-E!EJ)/2El Effi2-Ec2/2iEl-E.c-EiEs/2EE!C-E!/22El|E/ + (E+E 人+EJ/2E 严+(E 厂 E2/2E)/+(E 厂 EQ/2EJc+ElEE,/2E(9)+ (Et-E/22E,)c2(E.-EiE4/2El)cEl0-E/22El=Q设尺=耳-；/2迟,FE-Efit/2En FrE-Efit/2En F尸E厂E；/2E、F,=E9-EfE4/2E F,=E,0-E/22En则式化为Elt/2a + (E2b+EJc+E4)/2E,t/22+Flb2+(FJc+Ff)

34、b+F4c2+Fic+F6=0的一元川次方程代数解研究好石由(10)式得E 严a+(E+Ee+EJ/2E/T+F/作+(f#+FJ/2F丁-(Ff+F/2”+F/+F$c+F6=OE/a + 3#+E/:+Z)/2& 丁+F络+ (F# + FJ/2F 丁-巧公/2 巧一 Ff#/2F 厂 F；/22巧 +心+几=0E.l/2a + (E,b+EJc+E4)/2E!t/22+Ft,/2b(FxF.)/2Fl/22(11)+ (F厂尸:/公咖+屮厂吋血+F厂町/2学严0令(11)式中(F厂Fj/2另)J+(F厂FH/2F)c+F厂町/2中严0,这是条关于啲元二次方程，设G是方程的一个根，则(1D

35、式化为E/ + (E/ + E 孙+)/23丁+巧叫 + (尸切 + 兀)/2巧丁=0E/!a + (Eb+E 心 +EJ/2E严I，=- F/作 + (F易+FJ/2FT(12)02)式网边同时开方得E/%+(E#+E 扔+EJ/2E 严=iF/作+(F易+FJ/2F 严&%+E#/2E严+疋丹/2严+&/2&=庐/勺+,(尸旳+几)/2尸严E2b/2El,/2-iFl,/Ib=i(F2c,-Fi)/2Fl,/2-Etcl/2E,/2-E4/2E,/2-E,/2ab=lg+F/2F：_E 仪/2E：EJ2E：n_E：畑八 EJ2E：7F：B将解得的b,c,d分别代入，幣理后就可以得到条关丁。

36、的元三次方程，解之即得a的值.一元n次方程代数解研究好石-#-一元n次方程代数解研究好石解五元三次不定方程组（二）已知方程组成立.其中abcde为未知数，求abcde的一组解.1上+/）厂0E4d+E0,则有&一地+0=0,解得b=Dta/D2d=(D fi+D/kt将以代入式,得EJa2 + (E2(D,a/D2) +EJc+E(DJcD4)/k!+EifiE6)a+E7(DM/D2)+ (E3c+E9(D jC+D/kf+E+EjjjiDja/D2) +E32c2+(E”()e+0)/h+E/+&$)c+爲(D+DJ，/町+(/+7)(“+0)/他+6/+3产0Elai+(-E2Dla/D

37、2EfirDtE/:/k:D4E4/kl+EseJrEa-E1Dla/D2一(上:+)04/觥+七/%+&声+E)M/0+E$+(Q“c/打+D“/%+E/+&Jc+爲(f+2Dx：+ZV)/财+(Q 迟/e+Q占话為+6/+E/+E“=0Elai-E1Diai/D2+ED1Et/kt)c+EEDtEi/k-E,/Da-(目+3&/打)C+E/+D 厶/处+ 皿/0 + (/占”/打)+(/+门/%+&5),/处一厂 DQ 运/爲0, Gj-E-Et/DG4=Ei+DtEt/k-ErDl/D-DtE)l/DJi-EltD!/Di,Gi=El2+Dll/k,+k!D1i/kl29G=ElDtEu

38、/k,tGJ=DIJ/k,+Ets+2kflpyk,Dn/kltGt=E,D4Elt/knGkiD/k,2+D4EIT/kEl9t 则式化対G,a+(G2c+G+ +(G(teJrG7)c+6e+Glle+G9=0 (8)由式，得G/ a + (GjC+G04)/2Gf! 22 (GjC + Gte+ G1 /+ (G6e+G7)c+6f*4-Ge+GOIG/ a + iGc+G2GyG2cf/2GlGi2e,/2GlG41/2G/2G2cGie/2Gl2G2cG4/2?G, -2GjeGq/ 2Gj + G0 +(G“+G7)c+6e+G“ + G9=0G,/2a + (G2cG3e-G4)

39、/2Gl,/2iGsc2-G22c72!GJ + (G+G7)c-G2cG3e/2G-G2cG4/2GI +6e2-Gf2e3/2!Gl-GKeGteG4/2Gl+Gl) GJ/2G, 0G/ a + (G/:GjtG4)/2G； T+(G$Gj/2G；)c2 + (G+G； G2G/2Gt)eG2G4/2G)c+ (6G/ 2G Je+ (G8 GfG4/ 2G)e + G9 G ： / 2Gt0G 严a+(G*+G*+GJ/2G 严+ (Gf-G272!G,),/tc(G6+G-G2Gi/2Gl)e-G2G4/2Gl/2(Gf-G2!/22Gl),/r -l(GG-G2G/2Gl)e-G2

40、G./2Gl,/2J(G-G2t/Gl)(6-Gl72,Gl)e?+(G9GG4/2Gl)eA-G9G/2Crl=0G/ + (G 疋+GGJ/2G 丁+ (G5-G2,/2?G/)/ 7c + (G6+G7G2Gj/2G/)-G?G/2GJ/2(G5G2,/2G/)/,(,-(G6-i G- GzGj/2G,)*/2GS G/2G；)+2(Gs+G7 G2G3/2Gt)eG2G4/2Gt/22(GS G7/2*G；)-G：G：/2G/2 (Gs/ 2Gl)+(6Gjr/2G”)e+(G 厂 GG,/2G Je + G?G：/2G ,0G/ a + (G疋+ G“+GJ/2G, T+ (GGr/2G J叫 + (G 厂 G2G3/2G J G22G J/2( Gs-Gr/2?Gy 十4-(6G/2 G/)e (G6+G7G2Gf/2Gt) e/2GSG22/2rG,)+(GG- G:G3/2Gl)eG2G,/22GI( G 厂 GJ/2G J+(G 厂 G2./2G Je+ G9 G; G/2*G/(Gv Gi /2GJG,/2 G,0-15-一元”次方程代数解研究好石G严a + (G* + G/n GJ/2