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Bài tập Toán cao cấp 1 Trương Đại học Thương mại có lời giải chi tiết
I: Ma tr nh th c MA TR NH TH C T v ma tr n Hai ma tr n b ng Hai ma tr p A = (aij) Ma tr cg , B = (bij) ng n nt A=B ng c ng aij = bij ( i, j) ng, tr hai ma tr n Cho hai ma tr A = (aij) , B = (bij) T ng c A + B = (aij + bij) nv im ts a ma tr n A v i m t s ij) ij) n n c m x p: A = (aij) c m x n u: A.B = C = (cij) (bij) t ma tr n n A.B ch th c hi c s c t c a ma tr a ma tr n B A.B B.A N u A.B = B.A = In n ngh oc c l i nh th c iv Cho A iv nh th c c p 2: L det(A) = nh th c c p cao (n = a11.a22 a12.a21 = const 3) I: Ma tr nh th c nh th c c p c Scrame: Vi cc i ho c k nh th nt y d u c ng (+) nt ta l y d u tr (-) p 3: A = et (A) = = a11.a22.a33 + a12.a23.a31 + a13.a21.a32 a13.a22.a31 a12.a21.a33 a11.a23.a32 c c t) = - Const n nt c Bi u th c n nt c S ph c n iv nt c a11 + - nh th th c nh th a12 a + - a13 nh th c thu c R thu c C tri nh th c c p cao (c p n): ng d Ch nh th c c cc c c t : us ti tri n mb c trung gian c c t c nh th c xu t hi n nhi u s c ch N u ma tr khai tri n nh th c b nt I: Ma tr nh th c i gauss th 1: N i d u i gauss th 2: N ik i gauss th 3: L nh th kl n nh th i III H ng c a ma tr n ng c a ma tr n b nh th c nh th c c p k c 1: tc c 2: c a A Gi s nh th c c p k nh th c c p k + c a A ch k nh th c Dk X y kh : nh th c c a A, x y k = min{m, n} (A) = k = min{m, n} Thu T tc nh th c c p k + c a A ch nh th c Dk u b ng r(A) = k Thu T nt im Khi pl nh th c c p k + c c v i Dk+1 thay cho Dk ng h p (1) ho k+1 ch nh th c Dk ti p t c nh n x y ng c a ma tr n b i sau g i ch a ma tr n: t ts ts b tk r ic IV Ma tr n ngh c o t c a ma tr n ngh Ma tr ngh Ma tr ngh N u hai ma tr o - nh nh t - - = A ngh - - - I: Ma tr nh th c - N n ngh o nh th c c a ma tr c 1: ngh n ngh ng: nh th c c a ma tr n A N u det(A) = n ngh N u det(A) n ngh c 2: L p ma tr n chuy n v o A-1 chuy c : A* = (Aij )nn v i A = Aij n A-1 = o A-1 c a A c 3: L p ma tr n ph h p c c 4: oc i s c a ph n t c t j ma tr n A A* I: Ma tr nh th c P n Th c hi n a b d c e L i gi i a = = b tA= V i n = 1: A = V i n = 2: A = = = = c = = d = nh th c e tA= V i n = 1: A = V i n = 2: A = = = = = Cho A = a ;B= B b B.(2A c hi i sao? L i gi i a 2A = = = = = 2A = ).B = = b B= c t c a B b ng nh th c (2A + )= a (2A + B.(2A + c hi 1, c tc ) b ng ng s a (2A + ) x2, x3, x 4: L i gi i X : L i gi i a - b nh th c - c nh th c sau: a d b e c L i gi i nh th c ) a =0 b = ( 1)2 + + ( 1)4 + = = c ( d ) ) = ) =( ) 10 nh th c ( ) =( e = = = nh th c sau: a b L i gi i 11 t bi n x y (1, 2) bi t f (x, y) = ln(x2 + 2y2) L i gi i x (x, y) = (ln (x2 + 2y2 x (x, y) = y (x, y) = (ln (x2 + 2y2 y (x, y) = (1, 2) = x y x y (1,2) = i Cho f(x, y) = x (0, 0) L i gi i x (0, 0) = tt= x = suy t (0,0) = = (Quy t x y (1, 2) bi t f (x, y) = ( L i gi i x (x, y) = (( )x x (x, y) = y.( x (1, 2) = 10 =y = y (x, y) = ( +y y (x, y) = 25.( ) 267 t bi n Cho f(x, y) = x y (x, y) L i gi i x (x, y) = ( u th y x= x i x ng v (x, y) = = i ch Cho f(x, y) = x x y (0, 0) L i gi i x (x, y) = ( )x= x x x (0, 0) = (1, 1) = (0, 0) Ta s d = n t i gi i h y : = ih i h n ph (0, 0) = = ng =0 (x) L i gi i (x) = = x= = Cho z = = = + 4xy 10 xy L i gi i 268 t bi n x = (2 =8 + 4xy x= + 4y = x y= xy (8 (2 ( x x x ( x + 4y y y= = (8 y =0+4=4 y (C) ix=1b L i gi i V i gia c = is t = = = V i x = 0: f (x) = L i gi i )= = = = = = ) f(x) kh vi t i x = Cho f(x) = L i gi i - +1 +x kh vi 2sinx + kh vi 269 t bi n 1) f(x) = 3sin(x 1) kh vi T ix= f ( 1) = = 1+ 2cos1 = f (1+) = =3 f (1+) f ( 1) n t i x = vi t i x = T i x = 0: f( )= = f( = = = = f( +2 = = 1+ f( =1+ =1 =1 v t ix=0 kh ; (0; 1); (1; + vi t i x = 0; x = f(x) = c i x = L i gi i = ( a = a = vi t =3 =f vi c ) = a = f(x) t i x = 0: f(x) = L i gi i = = =3 270 t bi n Do + 3x x f = D th =6 f vi t i x = i x = 1: f(x) = L i gi i = f = a f(x) kh vi t i x = f = i h n: L i gi i = = = = cc theo m: f(x) = L i gi i = x > 0: f(x) = x < 0: f(x) = a nh v i cv i nh x < 271 t bi n x = 0: f(0) = a + sin(2.0) = a = = = N u a= ct ix=0 ct i Cho f(x) = x.(x + 1).(x+2) (x + n) L i gi i = = n! Cho f(x) = uh nt i = L i gi i = = n = 0: + u h n n t i n > 1: Cho Kh a f t i x0 = L i gi i = ct i =0 = i = 272 t bi n g ng c a L i gi i : f(x) = = t = f (1 + f(1) = 0,02 :Y= (x > 5) L i gi i ln y = ln = 2.ln(x - 2) + ln(x + 1) = + 3.ln(x - 5) - + - ) p hai f(1, 1) bi t: f(x, y) = L i gi i .(1+xy) = p hai : + 273 t bi n f(x, y) = [ + 2.(1 + xy)dxdy + f(1, 1) = e(d ] + 4dxdy + d ) p1 A= L i gi i Ch Ch g nv dx = =x : x0 = 1, y0 = x0 = 1,03 = 0,03 dy = dx + f(1,03; 1,98) f(1; dy 2).( 0,02) = f(1,03; 1,98) + (0,03) + ( 0,02) = 2,98 L i gi i Ta c t t i 15,8 = 16 = 16, + = 0,2 f( = c: B 0,2 = =2 = 0,00625 1,9938 p f = f(u) = , u = sin(xy) L i gi i 274 t bi n F = f(x, y) = = 2u ycos(xy) = 2sin(xy) ycos(xy) = 2u xcos(xy) = 2sin(xy) xcos(xy) bi t f = f(u, v) = v + ln(uv), u = ,v= L i gi i v + ) p c a f(x, y) = +( + ).sin(2x) + 3xy + y L i gi i + 3xy + y + 3xy + y p1c a , bi t f(x, y) = + 10 L i gi i y) = = p c a f(x, y) = y.sinxy L i gi i cosxy 275 t bi n p c a f(x,y) = (1; 2) (1; 2) L i gi i = p c a: a z(x, y) = + b z(x, y) = ln(tan L i gi i a ( ) b = = = dx + dx 276 t bi n p ng d ng Cho n xu t ng n h n Q = f (L) = 24 L t c L = 64 t qu L i gi i MPPL = 24 ( T i m c L = 64 = 24 MPPL = 24 = 0,5 N px T i m c L = 125 s n ph m MPPL N px M bi n xu s n ph ng c nh tranh v n xu t Q = 12 d s n ph m (L) = 40USD nh m c s ng cho l i nhu n t L i gi i i nhu n c n xu = 20.12 40L = 240 40L BBT: Q + Theo BBT, ta suy c 64 t GTLN t i L = 64 l i s n xu t m t chi 277 t bi n xu c c iv im cs n u n u m c s n xu t L i gi i n xu t Q s n ph m: c - Khi y, n Q t ng 7, n sinh b quay quanh Ox L i gi i Di n b ng t ng hai di sinh b i n ng i quay quanh Ox N N Trong c ng h V c: S= =4 = = = H 278 NH TH C T ma tr n nh th c III H ng c a ma tr n IV Ma tr n ngh o P b n p b sung 23 T 43 II S c l p III H thu c 44 47 c a vector 51 P b n p b sung 57 63 279 t bi n C T 86 I Gi i h II Bi n lu n s nghi 88 90 108 P b n p b sung T m 138 n d ng t c, chu n t c nh d u 141 nh d u 145 P n 150 p b sung 165 187 p ng d ng , GI I H C T I S th 189 m t bi n s II Gi i h III S cc 197 207 m t bi n s P b n 212 280 t bi n p b sung 225 T BI N T m 250 c cao III 244 251 ng d ng c P b n p b sung p ng d ng 255 267 277 281