Mt s phng php gii phng trnh nghim nguyn

Trong qu trnh ging dy v lm ton, ti h thng c mt s phng php gii phng trnh nghim nguyn, hi vng s gip cc em hc sinh bit la chn phng php thch hp khi gii bi ton loi ny.Phng php 1 :a v dng tchBin i phng trnh v dng : v tri l tch ca cc a thc cha n, v phi l tch ca cc s nguyn.Th d 1 :Tm nghim nguyn ca phng trnh :y3- x3= 91 (1)Li gii :(1) tng ng vi (y - x)(x2+ xy + y2) = 91 (*)V x2+ xy + y2> 0 vi mi x, y nn t (*) => y - x > 0.Mt khc, 91 = 1 x 91 = 7 x 13 v y - x ; x2+ xy + y2u nguyn dng nn ta c bn kh nng sau :y - x = 91 v x2+ xy + y2= 1 ; (I)y - x = 1 v x2+ xy + y2= 91 ; (II)y - x = 3 v x2+ xy + y2= 7 ; (III)y - x = 7 v x2+ xy + y2= 13 ; (IV)n y, bi ton coi nh c gii quyt.Phng php 2 :Sp th t cc nNu cc n x, y, z, ... c vai tr bnh ng, ta c th gi s x y z ... tm cc nghim tha mn iu kin ny. T , dng php hon v => cc nghim ca phng trnh cho.Th d 2 :Tm nghim nguyn dng ca phng trnh :x + y + z = xyz (2).Li gii :Do vai tr bnh ng ca x, y, z trong phng trnh, trc ht ta xt x y z.V x, y, z nguyn dng nn xyz 0, do x y z => xyz = x + y + z 3z => xy 3 => xy thuc {1 ; 2 ; 3}.Nu xy = 1 => x = y = 1, thay vo (2) ta c : 2 + z = z, v l.Nu xy = 2, do x y nn x = 1 v y = 2, thay vo (2), => z = 3.Nu xy = 3, do x y nn x = 1 v y = 3, thay vo (2), => z = 2.Vy nghim nguyn dng ca phng trnh (2) l cc hon v ca (1 ; 2 ; 3).Th d 3 :Tm nghim nguyn dng ca phng trnh :1/x + 1/y + 1/z = 2 (3)Li gii :Do vai tr bnh ng ca x, y, z, trc ht ta xt x y z. Ta c :2 = 1/x + 1/y + 1/z 3.1/x => x 3/2 => x = 1.Thay x = 1 vo (3) ta c :1/y + 1/z + 1 = 2 => 1 = 1/y + 1/z 2/y => y 2=> y = 1 => 1/z = 0 (v l)hoc y = 2 => 1/z = 2 => z = 2.Vy nghim nguyn dng ca phng trnh (3) l cc hon v ca (1 ; 2 ; 2).Phng php 3 :S dng tnh cht chia htPhng php ny s dng tnh cht chia ht chng minh phng trnh v nghim hoc tm nghim ca phng trnh.Th d 4 :Tm nghim nguyn ca phng trnh :x2- 2y2= 5 (4)Li gii :T phng trnh (4) ta => x phi l s l. Thay x = 2k + 1 (k thuc Z) vo (4), ta c :4k2+4k + 1 - 2y2= 5tng ng 2(k2+ k - 1) = y2=> y2l s chn => y l s chn.t y = 2t (t thuc Z), ta c :2(k2+ k - 1) = 4t2tng ng k(k + 1) = 2t2+ 1 (**)Nhn xt :k(k + 1) l s chn, 2t2+ 1 l s l => phng trnh (**) v nghim.Vy phng trnh (4) khng c nghim nguyn.Th d 5 :Chng minh rng khng tn ti cc s nguyn x, y, z tha mn :x3+ y3+ z3= x + y + z + 2000 (5)Li gii :Ta c x3- x = (x - 1).x.(x + 1) l tch ca 3 s nguyn lin tip (vi x l s nguyn). Do : x3- x chia ht cho 3.Tng t y3- y v z3- z cng chia ht cho 3. T ta c : x3+ y3+ z3- x - y - z chia ht cho 3.V 2000 khng chia ht cho 3 nn x3+ y3+ z3- x - y - z 2000 vi mi s nguyn x, y, z tc l phng trnh (5) khng c nghim nguyn.Th d 6 :Tm nghim nguyn ca phng trnh :xy + x - 2y = 3 (6)Li gii :Ta c (6) tng ng y(x - 2) = - x + 3. V x = 2 khng tha mn phng trnh nn (6) tng ng vi:y = (-x + 3)/(x - 2) tng ng y = -1 + 1/(x - 2).Ta thy : y l s nguyn tng ng vi x - 2 l c ca 1 hay x - 2 = 1 hoc x - 2 = -1 tng ng vi x = 1 hoc x = 3. T ta c nghim (x ; y) l (1 ; -2) v (3 ; 0).Ch :C th dng phng php 1 gii bi ton ny, nh a phng trnh (6) v dng : x(y + 1) - 2(y + 1) = 1 tng ng (x - 2)(y + 1) = 1.Phng php 4 :S dng bt ng thcDng bt ng thc nh gi mt n no v t s nh gi ny => cc gi tr nguyn ca n ny.Th d 7 :Tm nghim nguyn ca phng trnh :x2- xy + y2= 3 (7)Li gii :(7) tng ng vi (x - y/2)2= 3 - 3y2/4V (x - y/2)2 0 => 3 - 4y2/4 0=> -2 y 2 .Ln lt thay y = -2 ; 2 ; -1 ; 1 ; 0 vo phng trnh tnh x. Ta c cc nghim nguyn ca phng trnh l :(x ; y) thuc {(-1 ; -2) ; (1 ; 2) ; (-2 ; -1) ; (2 ; 1) ; (-1 ; 1) ; (1 ; -1)}.Chc chn cn nhiu phng php gii phng trnh nghim nguyn v cn nhiu th d hp dn khc. Mong cc bn tip tc trao i v vn ny. Cc bn cng th gii mt s phng trnh nghim nguyn sau y :Bi 1 :Gii cc phng trnh nghim nguyn :a) x2- 4 xy = 23 ;b) 3x - 3y + 2 = 0 ;c) 19x2+ 28y2=729 ;d) 3x2+ 10xy + 8y2= 96.Bi 2 :Tm x, y nguyn dng tha mn :a) 4xy - 3(x + y) = 59 ;b) 5(xy + yz + zx) = 4xyz ;c) xy/z + yz/x + zx/y = 3 ;d) 1/x + 1/y + 1/z = 1/1995.

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