2.2. Phˆan th´u . ch˜u . uty ’ 55 (DS.  x 2 − √ 5 − 1 2 x +1  x 2 + √ 5+1 2 x +1  ) Chı ’ dˆa ˜ n. D˘a . t x 2 l`am th`u . asˆo ´ chung rˆo ` id`ung ph´ep d ˆo ’ ibiˆe ´ n y = x + 1 x 7) x 2n − 1(DS. (x 2 − 1) n−1  k=1 (x 2 − 2x cos kπ n + 1)) 8) x 2n+1 − 1(DS. (x − 1) n  k=1  x 2 −2x cos 2kπ 2n +1 +1  ) 2.2 Phˆan th´u . ch˜u . uty ’ Mˆo . t h`am sˆo ´ x´ac di . nh du . ´o . ida . ng thu . o . ng cu ’ a hai d ath´u . cd a . isˆo ´ ta . i nh˜u . ng diˆe ’ mm`amˆa ˜ usˆo ´ khˆong triˆe . t tiˆeu go . i l`a phˆan th´u . ch˜u . uty ’ . R(x)= P (x) Q(x) ,Q(x) =0. Nˆe ´ u degP<degQ th`ı R(x)go . il`aphˆan th´u . ch˜u . uty ’ thu . . csu . . .Nˆe ´ u degP  degQ th`ı R(x)d u . o . . cgo . il`aphˆan th´u . ch˜u . uty ’ khˆong thu . . csu . . . Nˆe ´ u degP  degQ th`ı b˘a ` ng c´ach thu . . chiˆe . nph´ep chia P(x)cho Q(x) ta thu du . o . . c P (x) Q(x) = W (x)+ P 1 (x) Q(x) (2.12) trong d ´o W (x)l`adath´u . c, c`on P 1 (x) Q(x) l`a phˆan th´u . ch˜u . uty ’ thu . . csu . . . Vˆe ` sau ta chı ’ x´et c´ac phˆan th´u . ch˜u . uty ’ l`a thu . o . ng cu ’ ahaidath´u . c da . isˆo ´ v´o . ihˆe . sˆo ´ thu . . c (phˆan th´u . cnhu . vˆa . ydu . o . . cgo . i l`a phˆan th´u . ch˜u . u ty ’ v´o . ihˆe . sˆo ´ thu . . c). Phˆan th´u . c thu . . cd o . n gia ’ n nhˆa ´ t (c`on go . i l`a phˆan th´u . cco . ba ’ n) l`a nh˜u . ng phˆan th´u . cdu . o . . cbiˆe ’ udiˆe ˜ ntˆo ´ i gia ’ nbo . ’ imˆo . t trong hai da . ng sau dˆa y I. A (x −α) m ; II. Bx + C (x 2 + px + q) m ; A, B, C, p, q ∈ R. 56 Chu . o . ng 2. D - ath´u . c v`a h`am h˜u . uty ’ T`u . d i . nh l´y Gauss v`a c´ac hˆe . qua ’ cu ’ a n´o ta c´o D - i . nh l´y. Mo . i phˆan th´u . ch˜u . uty ’ thu . . csu . . P (x) Q(x) hˆe . sˆo ´ thu . . cv´o . imˆa ˜ u sˆo ´ c´o da . ng Q(x)=(x −α) r (x −β) s ···(x 2 + p 1 x + q 1 ) m × × (x 2 + p 2 x + q 2 )  ···(x 2 + p s x + q s ) n (2.13) d ˆe ` u c´o thˆe ’ biˆe ’ udiˆe ˜ ndu . ´o . ida . ng tˆo ’ ng h˜u . uha . n c´ac phˆan th´u . cco . ba ’ n da . ng I v`a II P (x) Q(x) = A (x − α) r + B (x − α) r−1 + ···+ C x − α + + D (x − β) s + E (x −β) s−1 + ···+ F x − β + + Gx + H (x 2 + p 1 x + q 1 ) m + Ix + H (x 2 + p 1 x + q 1 ) m−1 + ···+ Lx + M x 2 + p 1 x + q 1 + + Nx+ P (x 2 + p s x + q s ) n + Qx + R (x 2 + p s x + q s ) n−1 + ···+ Sx + T x 2 + p s x + q s , (2.14) trong d ´o A,B, l`a nh˜u . ng h˘a ` ng sˆo ´ thu . . c. Nhu . vˆa . y c´ac phˆan th´u . cco . ba ’ no . ’ vˆe ´ pha ’ icu ’ a (2.14) s˘a ´ pxˆe ´ p theo t`u . ng nh´om tu . o . ng ´u . ng v´o . i c´ac th`u . asˆo ´ o . ’ vˆe ´ pha ’ icu ’ a (2.13), trong d´o sˆo ´ sˆo ´ ha . ng cu ’ amˆo ˜ i nh´om b˘a ` ng sˆo ´ m˜ucu ’ alu˜yth`u . acu ’ ath`u . asˆo ´ tu . o . ng ´u . ng. Cˆa ` nlu . u´yr˘a ` ng khi khai triˆe ’ n phˆan th´u . ccu . thˆe ’ theo cˆong th´u . c (2.14) mˆo . tsˆo ´ hˆe . sˆo ´ c´o thˆe ’ b˘a ` ng 0 v`a do d´osˆo ´ sˆo ´ ha . ng trong mˆo ˜ i nh´om c´o thˆe ’ b´e ho . nsˆo ´ m˜ucu ’ ath`u . asˆo ´ tu . o . ng ´u . ng. Trong thu . . c h`anh, d ˆe ’ t´ınh c´ac hˆe . sˆo ´ A,B, ta s˜e su . ’ du . ng c´ac phu . o . ng ph´ap sau. 2.2. Phˆan th´u . ch˜u . uty ’ 57 I. Gia ’ su . ’ d ath´u . c Q(x)chı ’ c´o c´ac nghiˆe . m thu . . cd o . n, t´u . cl`a Q(x)= n  j=1 (x − a j ),a i = a j ∀i = j. Khi d´o P (x) Q(x) = n  j=1 A j x −a j · (2.15) D ˆe ’ x´ac di . nh A k ta nhˆan hai vˆe ´ cu ’ a (2.15) v´o . i x − a k v`a thu du . o . . c P (x) n  j=1 j=k (x − a j ) = A k +  A 1 x −a 1 + ···+ A k−1 x −a k−1 + A k+1 x − a k+1 + ···+ A n x − a n  (x −a k ). (2.16) Thay x = a k v`ao (2.16) ta c´o A k = P (a k ) n  j=1 j=k (a k − a j ) · (2.17) Nhu . vˆa . yd ˆe ’ t´ınh hˆe . sˆo ´ A k cu ’ a phˆan th´u . c A k x −a k ta x´oa th`u . asˆo ´ (x − a k ) kho ’ imˆa ˜ usˆo ´ cu ’ a P (x) Q(x) v`a tiˆe ´ p theo l`a thay x = a k v`ao biˆe ’ u th ´u . c c`on la . i. V`ıvˆa . yphu . o . ng ph´ap n`ay d u . o . . cgo . il`aphu . o . ng ph´ap x´oa. II. Nˆe ´ u Q(x) c´o nghiˆe . mbˆo . i th`ı cˆong th´u . c (2.17) khˆong c`on su . ’ du . ng du . o . . c. Gia ’ su . ’ Q(x)=g m , trong d´o h o ˘a . c g = x − α ho˘a . c g l`a t´ıch c´ac th `u . asˆo ´ l`a tam th´u . cbˆa . c hai v´o . i hai biˆe . tsˆo ´ ˆam. Trong tru . `o . ng ho . . p n`ay ta cˆa ` n khai triˆe ’ n P (x) theo c´ac lu˜y th`u . acu ’ a g: P (x)=a 0 + a 1 g + a 2 g 2 + 58 Chu . o . ng 2. D - ath´u . c v`a h`am h˜u . uty ’ trong d´o a 0 ,a 1 , l`a h˘a ` ng sˆo ´ nˆe ´ u g = x − α v`a l`a dath´u . cbˆa . c khˆong vu . o . . t qu´a 1 trong tru . `o . ng ho . . pth´u . hai (trong tru . `o . ng ho . . p n`ay ta cˆa ` n thu . . chiˆe . n theo quy t˘a ´ cph´ep chia c´o du . ). III. Dˆo ´ iv´o . i tru . `o . ng ho . . ptˆo ’ ng qu´at, ta nhˆan hai vˆe ´ cu ’ a (2.14) v´o . i dath´u . c Q(z) v`a s˘a ´ pxˆe ´ p c´ac sˆo ´ ha . ng o . ’ vˆe ´ pha ’ id˘a ’ ng th´u . cthudu . o . . c th`anh dath´u . cv`athudu . o . . c dˆo ` ng nhˆa ´ tth´u . c gi˜u . ahaidath´u . c: mˆo . tda th ´u . cl`aP (x), c`on d ath´u . c kia l`a d ath´u . cv´o . ihˆe . sˆo ´ A,B, chu . ad u . o . . c x´ac d i . nh. Cˆan b˘a ` ng c´ac hˆe . sˆo ´ cu ’ a c´ac lu˜y th`u . ac`ung bˆa . c ta thu du . o . . c hˆe . phu . o . ng tr`ınh tuyˆe ´ n t´ınh v´o . iˆa ’ nl`aA,B, Gia ’ ihˆe . d ´o, ta t`ım du . o . . c c´ac hˆe . sˆo ´ A,B, Phu . o . ng ph´ap n`ay go . i l`a phu . o . ng ph´ap hˆe . sˆo ´ bˆa ´ tdi . nh. Ta c´o thˆe ’ x´ac di . nh hˆe . sˆo ´ b˘a ` ng c´ach kh´ac l`a cho biˆe ´ n x trong dˆo ` ng nhˆa ´ tth´u . cnh˜u . ng tri . sˆo ´ t`uy ´y (ch˘a ’ ng ha . n c´ac gi´a tri . d ´o l`a nghiˆe . m thu . . c cu ’ amˆa ˜ usˆo ´ ). C ´ AC V ´ IDU . V´ı d u . 1. Khai triˆe ’ n c´ac phˆan th´u . ch˜u . uty ’ sau th`anh tˆo ’ ng c´ac phˆan th ´u . cco . ba ’ n 1) 2x 3 +4x 2 + x +2 (x − 1) 2 (x 2 + x +1) , 2) x 2 − 2x (x − 1) 2 (x 2 +1) 2 · Gia ’ i. 1) V`ı tam th´u . cbˆa . c hai x 2 + x+1 khˆong c´o nghiˆe . m thu . . cnˆen R 1 (x)= 2x 3 +4x 2 + x +2 (x − 1) 2 (x 2 + x +1) = B 1 (x −1) + B 2 (x −1) 2 + Mx+ N x 2 + x +1 · Quy dˆo ` ng mˆa ˜ usˆo ´ ta c´o 2x 3 +4x 2 + x +2 (x − 1) 2 (x 2 + x +1) = B 1 (x 3 − 1) + B 2 (x 2 + x +1)+(Mx + N)(x 2 −2x +1) (x − 1) 2 (x 2 + x +1) · 2.2. Phˆan th´u . ch˜u . uty ’ 59 Cˆan b˘a ` ng hˆe . sˆo ´ cu ’ a x 0 , x 1 , x 2 v`a x 3 trong c´ac tu . ’ sˆo ´ ta thu d u . o . . chˆe . phu . o . ng tr`ınh x 3    B 1 + B 2 + N =2, x 2    B 2 + M −2N =1, x 1    B 2 + N −2M =4, x 0    B 1 + M =2. Gia ’ ihˆe . phu . o . ng tr`ınh ta c´o B 1 =2,B 2 =3,M =0,N =1. T`u . d´o R 1 (x)= 2 x − 1 + 3 (x − 1) 2 + 1 x 2 + x +1 · 2) Ta c´o R 2 = x 2 − 2x (x − 1) 2 (x 2 +1) 2 = A 1 x −1 + A 2 (x −1) 2 + M 1 x + N 1 x 2 +1 + M 2 x + N 2 (x 2 +1) 2 · Quy d ˆo ` ng mˆa ˜ usˆo ´ v`a cˆan b˘a ` ng c´ac tu . ’ sˆo ´ ta c´o x 2 − 2x = A 1 (x −1)(x 2 +1) 2 + A 2 (x 2 +1) 2 +(M 1 x + N 1 )(x −1) 2 (x 2 +1) +(M 2 x + N 2 )(x − 1) 2 . So s´anh c´ac hˆe . sˆo ´ cu ’ a c´ac lu˜y th`u . ac`ung bˆa . co . ’ hai vˆe ´ ta thu du . o . . c x 5    A 1 + M 1 =0, x 4    − A 1 + A 2 − 2M 1 + N 1 =0, x 3    2A 1 +2M 1 −2N 1 + M 2 =0, x 2    − 2A 1 +2A 2 − 2M 1 +2N 1 +2N 1 − 2M 2 + N 2 =1, x 1    A 1 + M 1 − 2N 1 + M 2 − 2N 2 = −2, x 0    − A 1 + A 2 + N 1 + N 2 =0. T`u . d ´o suy ra A 1 = 1 2 ,A 2 = − 1 4 ,M 1 = − 1 2 , N 1 = − 1 4 ,M 2 = − 1 2 ,N 2 =1 60 Chu . o . ng 2. D - ath´u . c v`a h`am h˜u . uty ’ v`a do vˆa . y x 2 − 2x (x − 1) 2 (x 2 +1) 2 = 1 2 x −1 + − 1 4 (x −1) 2 + − 1 2 x − 1 4 x 2 +1 + − 1 2 x +1 (x 2 +1) 2 · V´ı d u . 2. C˜ung ho ’ inhu . trˆen 1) R 1 (x)= x 4 x 4 +5x 2 +1 ;2)R 2 (x)= 1 x 4 +1 · Gia ’ i. 1) R 1 (x) l`a phˆan th´u . ch˜u . uty ’ khˆong thu . . csu . . nˆen d ˆa ` u tiˆen cˆa ` n thu . . chiˆe . n ph´ep chia: x 4 x 4 +5x 2 +4 =1− 5x 2 +4 x 4 +5x 2 +4 =1+R 3 (x). Ch´u´yr˘a ` ng x 4 +5x 2 +4=(x 2 + 1)(x 2 + 4), do d´o R 3 = − 5x 2 +4 (x 2 + 1)(x 2 +4) = M 1 x + N 1 x 2 +1 + M 2 x + N 2 x 2 +4 · Quy d ˆo ` ng mˆa ˜ usˆo ´ v`a so s´anh hai tu . ’ sˆo ´ ta thu d u . o . . c −5x 2 − 4=(M 1 x + N 1 )(x 2 +4)+(M 2 x + N 2 )(x 2 +1) v`a tiˆe ´ p theo l`a cˆan b˘a ` ng c´ac hˆe . sˆo ´ cu ’ a c´ac lu˜y th`u . ac`ung bˆa . ccu ’ a x ta thu du . o . . chˆe . phu . o . ng tr`ınh x 3    M 1 + M 2 =0, x 2    N 1 + N 2 = −5,    ⇒ M 1 = M 2 =0,N 1 = 1 3 ,N 2 = − 16 3 · x 1    4M 1 + N −2=0, x 0    4N 1 + N −2=−4 Vˆa . y R 1 (x)=1+ 1 3 · 1 x 2 +1 − 16 3 · 1 x 2 +4 · 2.2. Phˆan th´u . ch˜u . uty ’ 61 2) V`ı x 4 +1=(x 2 +1) 2 −2x 2 =(x 2 + √ 2x + 1)(x 2 − √ 2x +1) nˆen R 2 = 1 x 4 +1 = M 1 x + N 1 x 2 + √ 2x +1 + M 2 x + N 2 x 2 − √ 2x +1 · T`u . dˆo ` ng nhˆa ´ tth´u . c 1 ≡ (M 1 x + N 1 )(x 2 − √ 2x +1)+(M +2x + N 2 )(x 2 + √ 2x +1), tiˆe ´ n h`anh tu . o . ng tu . . nhu . trˆen ta c´o M 1 = −M 2 = 1 2 √ 2 ,N 1 = N 2 = 1 2 · Do d ´o 1 x 4 +1 = 1 2 √ 2 x + √ 2 x 2 + √ 2x +1 − 1 2 √ 2 x − √ 2 x 2 − √ 2x +1 · V´ı d u . 3. T`ım khai triˆe ’ n phˆan th´u . c 1) R 1 (x)= x +1 (x − 1)(x − 2)x ;2)R 2 (x)= x 2 +2x +6 (x − 1)(x −2)(x −4) · Gia ’ i. 1) V`ı mˆa ˜ usˆo ´ chı ’ c´o nghiˆe . md o . n0, 1, 2nˆen x +1 x(x −1)(x −2) = A 1 x + A 2 x − 1 + A 2 x − 2 · ´ Ap du . ng cˆong th´u . c (2.17) ta du . o . . c A 1 = x +1   x=0 (x −1)(x −2)   x=0 = 1 2 ; A 2 = x +1 x(x −2)    x=1 = −2,A 3 = x +1 x(x − 1)    x=2 = 3 2 · Vˆa . y R 1 (x)= 1 2x + −2 x −1 + 3 2(x − 2) · 62 Chu . o . ng 2. D - ath´u . c v`a h`am h˜u . uty ’ 2) Tu . o . ng tu . . ta c´o R 2 (x)= x 2 +2x +6 (x −1)(x −2)(x −4) = A 1 x − 1 + B x − 2 + C x − 3 V`ımˆa ˜ usˆo ´ cu ’ a R 2 (x)chı ’ c´o nghiˆe . mdo . nnˆen A = x 2 +2x +6 (x − 2)(x − 4)    x=1 =3, B = x 2 +2x +6 (x − 1)(x − 4)    x=2 = −7, C = x 2 +2x +6 (x − 1)(x − 2)    x=4 =5. Do d ´o R 2 (x)= 3 x −1 − 7 x − 2 + 5 x − 4 · Nhˆa . nx´et. Trong mˆo . tsˆo ´ tru . `o . ng ho . . pd ˘a . cbiˆe . t, viˆe . c khai triˆe ’ n phˆan th ´u . ch˜u . uty ’ c´o thˆe ’ thu du . o . . cdo . n gia ’ nho . n v`a nhanh ho . n. Ch˘a ’ ng ha . n, dˆe ’ khai triˆe ’ n phˆan th´u . c 1 x 2 (1 + x 2 ) 2 th`anh tˆo ’ ng c´ac phˆan th´u . cco . ba ’ n ta c´o thˆe ’ thu . . chiˆe . nnhu . sau: 1 x 2 (x 2 +1) 2 = (1 + x 2 ) − x 2 x 2 (x 2 +1) 2 = 1 x 2 (x 2 +1) − 1 (x 2 +1) 2 = (1 + x 2 ) − x 2 x 2 (x 2 +1) − 1 (x 2 +1) 2 = 1 x 2 − 1 x 2 +1 − 1 (x 2 +1) 2 ·  V´ı d u . 4. Khai triˆe ’ n c´ac phˆan th´u . ch˜u . uty ’ sau: 1) x 4 +5x 3 +5x 2 − 3x +1 (x +2) 5 ;2) x 5 +3x 4 + x 3 − 2x 2 +2x +3 (x 2 + x +1) 3 · Gia ’ i. 1) D ˘a . t g =(x + 2). Khi d´ob˘a ` ng c´ach khai triˆe ’ ntu . ’ sˆo ´ theo c´ac lu˜y th`u . acu ’ a x +2 b˘a ` ng c´ach ´ap du . ng cˆong th´u . c nhi . th ´u . c Newton 2.2. Phˆan th´u . ch˜u . uty ’ 63 ta thu du . o . . c x 4 +5x 3 +5x 2 − 3x +1 (x +2) 5 = = [(x +2)− 2] 4 + 5[(x +2)−2] 3 + 5[(x +2)− 2] 2 − 3[(x +2)− 2)] + 1 (x +2) 5 = 3+5g − g 2 − 3g 3 + g 4 g 5 = 3 g 5 + 5 g 4 − 1 g 3 − 3 g 2 + 1 g = 3 (x +2) 5 + 5 (x +2) 4 − 1 (x +2) 3 − 3 (x +2) 3 + 1 x +2 · 2) D˘a . t g = x 2 + x +1. D´o l`a tam th´u . cbˆa . c hai khˆong c´o nghiˆe . m thu . . c. ´ Ap du . ng thuˆa . t to´an chia c´o du . ta c´o P (x)=x 5 +3x 4 + x 3 −2x 2 +2x +3 =(x 2 + x + 1)(x 3 +2x 2 −2x −2) + 6x +5 t´u . cl`a P = g ·q 1 + r 1 ,q 1 = x 3 +2x 2 − 2x − 2,r 1 =6x +5. Ta la . i chia q 1 cho g v`a thu du . o . . c q 1 = gq 2 + r 2 , degq 2 < deg(g) q 2 = x +1,r 2 = −4x −3. Nhu . vˆa . y P = gq 1 + r 1 = r 1 + g(r 2 + gq 2 ) = r 1 + r 2 g + q 2 g 2 . T`u . d ´o suy ra P g 3 = r 1 g 3 + r 2 g 3 + q 2 · 1 g = 6x +5 (x 2 + x +1) 3 − 4x +3 (x 2 + x +1) 2 + x +1 x 2 + x +1 ·  64 Chu . o . ng 2. D - ath´u . c v`a h`am h˜u . uty ’ B ` AI T ˆ A . P Trong c´ac b`ai to´an sau d ˆay, h˜ay khai triˆe ’ n phˆan th´u . ch˜u . uty ’ d˜a cho th`anh tˆo ’ ng h˜u . uha . n c´ac phˆan th´u . cco . ba ’ n thu . . c. 1. 2x − 3 x(x 2 − 1)(x 2 − 4) (DS. − 3 4x + 1 6(x − 1) + 5 6(x +1) + 1 24(x − 2) − 7 24(x +2) ) 2. x +1 x 3 − 1 (DS. 2 3(x − 1) − 2x +1 3(x 2 + x +1) ) 3. 1 x 3 (x − 1) 4 (DS. 10 x + 4 x 2 + 1 x 3 − 10 x −1 + 6 (x −1) 2 − 3 (x − 1) 3 + 1 (x − 1) 4 ) 4. 1 (x 4 − 1) 2 (DS. − 3 16(x − 1) + 1 16(x − 1) 2 + 3 16(x +1) + 1 16(x +1) 2 + 1 4(x 2 +1) + 1 4(x 2 +1) 2 ) 5. 2x − 1 (x +1) 3 (x 2 + x +1) (D S. 2 x +1 − 1 (x +1) 2 − 3 (x +1) 3 − 2x − 1 x 2 + x +1 ) 6. 1 x(x 2 +1) 3 (DS. 1 x + x (x 2 +1) 3 − x (x 2 +1) 2 − x x 2 +1 ) 7. x 2 +3x +1 x 4 (x 2 +1) (DS. 1 x 4 + 3 x 3 − 3 x + 3x x 2 +1 ) 8. x 5 +3x 3 − x 2 +4x −2 (x 2 +1) 3 (DS. 2x − 1 (x 2 +1) 3 + x − 1 (x 2 +1) 2 + x x 2 +1 ) 9. x 5 +2x 3 − 6x 2 − 3x − 9 (x 2 + x +2) 3 (DS. 1 (x 2 + x +2) 3 + x −1 (x 2 + x +2) 2 + x − 2 x 2 + x +2 ) 10. 2x − 1 x(x +1) 2 (x 2 + x +1) 2 [...]... −7 15  1 5  1 14) 12 11 2 T´ t´ c´c ma trˆn ınh ıch a a     5 2 1 1 3 −2     1) 5 2 3  3 −4 −5 6 5 2 2 1 3     3 4 9 5 6 4     2) 2 1 6  8 9 7  5 3 5 −4 −5 3    1 2 −2  1 3 1 1 −2 4     3)   2 3 2 1 −2 5  3 1 4 1 3 −2  2  4  4)  −2 1    1 3 1 2 1       2  1 3 1 2 1   1 3 2   (DS  5 10 9 ) −5 0 −7   11 9 13   (DS −22 −27 17 )... 17 ) 29 32 26   1 2 0  4 6 6   (DS  )  12 3 20 1 5 2   1 7   (DS  ) 3 9 Chu.o.ng 3 Ma trˆn D nh th´.c a -i u 82   1 1 1 3 3 1 1 2  0 0   5) )   (DS 1 3 −5 1 1 1  0 0 1 −2     3 2 1 1     6) 2 3 2 1 (DS 6 4 2) 9 6 3 3 ´ 3 T´ c´c t´ AB v` BA nˆu ınh a ıch a e   1 3 0 −2 1 1  5 1 3 1   ıch 1) A =  (DS T´ AB , B =  3 0 −2 2 0 1 4 4 1 2 ` ı... ˜ ˜ = B n + C1 B n 1 B = B n + nB n 1 B = n3n 1 0 3n 0 = n 0 3 0 n3n 1 0 1 0 0 0 n3n 1 3n 0 3n n3n 1 + = = 0 0 0 3n 0 3n 2) Tu.o.ng tu nhu trˆn ta c´ e o A= B 1 1 3 0 4 1 ˜ = B + B + = 0 0 0 3 0 3 m 3 0 = 0 3 1 1 ˜ Bm = 0 0 m = 3m 0 , 0 3m = 1 1 0 0 m ∀m (3. 3) 1 (3. 4) Chu.o.ng 3 Ma trˆn D nh th´.c a -i u 80 ˜ ˜ ’ ´ e o ea o u Tiˆp theo do B B = BB nˆn ta c´ thˆ ´p dung cˆng th´.c e 1 2 ˜ ˜ ˜ ˜ An... n 1 B + Cn B n−2 B 2 + · · · + B n (3. 5) ˜ a o Ta t´ Cn B n−k B k Theo (3. 3) v` (3. 4) ta c´ ınh k k Cn 3n−k 0 0 3n−k n−k k k Cn 3n−k Cn 3n−k 1 1 3n−k k 3 = = Cn 0 0 0 0 0 0 (3. 6) T` (3. 6), (3. 3) v` (3. 5) ta thu du.o.c u a n k k Cn 3n−k Cn 3n−k 3n 0 + A = 0 0 0 3n k =1   n n k k Cn 3n−k 0 + Cn 3n−k 3n +  = k =1 k =1 n 0 3 n V` 3n + ı n n k Cn 3n−k = (3 + 1) n = 4n v` 0 + a k =1 k Cn 3n−k = k =1 3n... ty a u u ’ 7 3 3x + 2 1 6x + 2 + − 2 (DS − + − 2 ) 2 x x + 1 (x + 1) x + x + 1 (x + x + 1) 2 x2 11 2 (x + 1) (x2 + x + 1) 2 1 1 x (DS 2 ) + 2 − 2 x + 1 x + x + 1 (x + x + 1) 2 1 12 5 4 + x3 − x2 + x − 1 x −x 1 2x + 1 1 1 − − ) (DS 2+x +1 2 − x + 1) 3( x − 1) 6 x 2(x 65 Chu.o.ng 3 - Ma trˆn Dinh th´.c a u 3 .1 Ma trˆn 67 a 3 .1. 1 - Dinh ngh˜ ma trˆn 67 ıa a 3 .1. 2 ´ C´c ph´p...   am b1 am b2 am bn am Chu.o.ng 3 Ma trˆn D nh th´.c a -i u 74 ´ V´ du 3 T´ AB v` BA nˆu ı ınh a e   1 3 2 1   1) A = , B = 3 0 1 2 3   1 0 1 4 1   , B =  1 3  2) A = 2 0 1 1 1 ´ ’ Giai 1) Theo quy t˘c nhˆn c´c ma trˆn ta c´ a a a a o   1 3 2 1   3 1+ 2 3 +1 3 12 AB = = 3  = 0 1 2 0 1+ 1 3+ 2 3 9 3 ` T´ BA khˆng tˆn tai v` ma trˆn B khˆng tu.o.ng th´ v´.i ma ıch o o ı a o... = 4n − 3n , do vˆy a An = 4n 4n − 3n 0 3n ` ˆ BAI TAP ´ 1 T´ A + B, AB v` BA nˆu ınh a e 1 3  1  2) A = 2 3 1) A = 2 , 4 B=  1 0  1 1 , 1 2 4 −4 ; 0 i   −2 1 2   B= 0 4 5  2 3 7 n k=0 k Cn 3n−k − 3 .1 Ma trˆn a 81 (DS 1) A + B = BA = 5 −2 4 −4 + 2i , AB = , 3 4+i 12 12 + 4i −8 −8 ; 3 4i  1  2) A + B =  2 5  6  BA =  23 17    0 2 1 3 3    5 6, AB = −2 3 16 , −4... trˆn B Do d´ o a   −2 0 1 4 1   AB =  1 3 2 0 1 1 1 = 1 · (−2) + 4 · 1 + ( 1) ( 1) 1 · 0 + 4 · 3 + ( 1) · 1 2 · (−2) + 0 · 1 + (1) · ( 1) 2·0+0 3 +1 1 = 3 11 −5 1 Tu.o.ng tu., ma trˆn B tu.o.ng th´ch v´.i a ı o  −2 −8  BA =  7 4 1 −4 V´ du 4 1) Cho ma trˆn A = ı a ho´n v´.i A (AX = XA) a o ma trˆn A v` a a  2  2 2 0 1 T` moi ma trˆn X giao ım a 0 0 3 .1 Ma trˆn a 75 2) T` moi... khˆng tu.o.ng th´ch v´.i ma trˆn B; BA = o o a o ı o a 10 15 −5 ) 11 10 10   2 0 1 −4   2) A =  (DS T´ AB khˆng ıch o , B = 5 1 0 3 3 1  0 1 ` ı tˆn tai v` A khˆng tu.o.ng th´ch v´.i B; BA = 11 1 ) o o ı o  3) A = 1 6 1 2 3 4  , B =  1 2 1 −2 3 3  5 3 8 2    2 1 0 1 28 27 8 ` , t´ BA khˆng tˆn tai) ıch o o 15 14 13 4) A = cos α − sin α cos β − sin β , B= cos α cos α sin... = a a 1 2 5 6 1+ 5 2+6 6 8 + = = 3 4 7 8 3+ 7 4+8 10 12 2) λA = 3 · 1 2 1 4 0 1 = 1 · 3 2 · 3 1 · 3 4 3 0 3 1 3 = 3 .1 Ma trˆn a 73 3 6 3 12 0 3 V´ du 2 Trong tru.`.ng ho.p n`o th` ı o a ı: ’ nhˆn bˆn phai mˆt ma trˆn h`ng v´.i mˆt ma trˆn cˆt ? ’ o a o 1) c´ thˆ a e o e o a a o i mˆt ma trˆn h`ng ? ’ ’ o a a 2) c´ thˆ nhˆn bˆn phai mˆt ma trˆn cˆt v´ o e a e o a o o ’ i 1) Ma trˆn . 1) 4 (DS. 10 x + 4 x 2 + 1 x 3 − 10 x 1 + 6 (x 1) 2 − 3 (x − 1) 3 + 1 (x − 1) 4 ) 4. 1 (x 4 − 1) 2 (DS. − 3 16 (x − 1) + 1 16(x − 1) 2 + 3 16 (x +1) + 1 16(x +1) 2 + 1 4(x 2 +1) + 1 4(x 2 +1) 2 ) 5. 2x − 1 (x +1) 3 (x 2 +. trˆa . n ta c´o  12 34  +  56 78  =  1+ 5 2+6 3+ 7 4+8  =  68 10 12  . 2) λA =3  12 1 40 1  =  1 · 32 · 3 1 · 3 4 · 30 · 31 3  = 3 .1. Ma trˆa . n 73  36 3 12 0 3  . V´ı d u . 2 x +1) (D S. 2 x +1 − 1 (x +1) 2 − 3 (x +1) 3 − 2x − 1 x 2 + x +1 ) 6. 1 x(x 2 +1) 3 (DS. 1 x + x (x 2 +1) 3 − x (x 2 +1) 2 − x x 2 +1 ) 7. x 2 +3x +1 x 4 (x 2 +1) (DS. 1 x 4 + 3 x 3 − 3 x + 3x x 2 +1 ) 8. x 5 +3x 3 −