1990 IMO 1990/1/15 1:00-4:00 1 N N N 1, N 1 N 2, N 2 N 3 N 3 2 x x + 52 = 3 x x , A, B, C 3,, A B, C 2,,,, 7, A, B, C
|
|
|
- ゆいと そめや
- 9 years ago
- Views:
Transcription
1 0 9 ( ) 10 (2000 ) SAT ACT 1
2 1990 IMO 1990/1/15 1:00-4:00 1 N N N 1, N 1 N 2, N 2 N 3 N 3 2 x x + 52 = 3 x x , A, B, C 3,, A B, C 2,,,, 7, A, B, C 3, 1 2, ABCD AB, BC, CD, DA 5 : 2, P, Q, R, S AQ BR W, BR CS X, CS DP Y, DP AQ Z W XY Z 6 2, 3, 5 90,,,, n ( 2 ) n 8 f { n 3, n 1000 f(n) = f(f(n + 7)), n < 1000, f(90) 2
3 9 n, a n = 50 + n 2 n, a n a n+1 d n n ( ), d n 10 x, y, z x + y + z = 1, 1 x + 4 y + 9 z 11 ( ), 3 ( ) 12, , {a n } a 1 = 1, n 1 a n+1 = a n + 1, a 100 [a 100 ] a n 3
4 1990 IMO 1990/2/11 1 (xyz ) E A 1, A 2, A 3, A 4, A 5 E, (1), (2) (1) A 1 A 2 A 3 A 4 A 5 = E (2) i j A i A j = φ, A 1, A 2, A 3, A 4, A 5 2 n 3, a 0, a 1,, a n 1 a 0 < a 1 < < a n 2n 3 a i + a j = a k + a l = a m i, j, k, l, m 3 X, (1), (2) (1) x X 4x X (2) x X [ x ] X, [a] a, X 4 n n a 1, a 2,, a n K < a 1 + a a n < G a 1 + a 2 a 2 + a 3 a n + a 1 K G 5 A, B n 2n P (n) P (n), 2n k, k A, k B Q(n) ( ) Q(8) ( ) Q(n) 4
5 /1/15 1:00-4:00 1 A = (81 9) A 2 2 x x 5 = 0 (199 ) ABC G GA = 2 3, GB = 2 2, GC = 2 ABC 4 1 x y + 1 (x + 1)y = (x, y) 5 8 (x, y, z) (x, y, z 0 6) P (e, π, 5) P (6 ) [ n ] [ n ] 6 n, f(0) = 0, f(1) = 1, f(n) = f( ) + n f(n) 0 n 1991 f(n) [x], x 7 n = n 5 n 8 n = 2 i 3 j 5 k (i, j, k ), 10 4 < n r n 63 (n, r), n! nc r = r!(n r)! 0C 0 =
6 11 A, B 15 : ( ) AA 5, AB, BA, BB 3 AABBAAAABAABBBB, AA 5, AB 3, BA 2, BB 4, ( 12 ) 6
7 /2/11 1 ABC BC, CA, AB t : (1 t) P, Q, R AP, BQ, CR K, K ABC L L (t ) 2 N N N p, q p(1) = 2, p(2) = 3, p(3) = 4, p(4) = 1, n 5 p(n) = n q(1) = 3, q(2) = 4, q(3) = 2, q(4) = 1, n 5 q(n) = n (1) N N f, n N f(f(n)) = p(n) + 2 f (2) N N f, n N f(f(n)) = q(n) A 16 A, A 4, 4 10, 14, 1 140,, 0 1 :, A n (n 2) [ A n ] ( ), A 3, [x] x 7
8 /1/15 1:00-4:00 1 {a n }, a 0 = 1, a 1 = 2, a n+2 = a n + (a n+1 ) 2 a x 2 + x + 1 = 0 ω ω 2k (ω + 1) 2k = k 3 y 2 = x x 8019 E 2 (3, 9), (4, 53), E x 4 A 1), 2) 1) 2, 3, 5, 7, 11, 13 2) 2 2, 3 2, 5 2, 7 2, 11 2, , 1 A A n n ,,,,, 7 6 ABC, BC, CA, AB 3 : (n 3) D, E, F (, n > 6) AD, BE, CF 4, 49, n 7 x, y, x 4 + y 4 x + y < m 1/3 n < 10 3 m, n, n n 8
9 9 A, B A = { (x, y) x, y 1 x 20, 1 y 20 } B = { (x, y) x, y 2 x 19, 2 y 19 } A, 219, 180 B, (1, 1), (1, 20), (20, 1), (20, 20) 2,, : 2, 2, 2 ( 1 ) 237, ( 1 ) 10 n n, ABC S T, S 441, T 440, AC + CB A C S B A T C B 12 A = { 1, 2,, 10 } A A f, 1) x A, f 30 (x) = x 2) k, 1 k 29, f k (a) a a A, x A, f 1 (x) = f(x), f 2 (x) = f(f 1 (x)),, f k+1 (x) = f(f k (x)), 9
10 /2/11 1 x y, xy 1, n, x + y x n + y n 2 1 ABC AB, AC D, E, BE, CD P BCED P BC 2 D, E AB, AC, P DE 3 n 2, n 1 k=1 n n k 1 2 k 1 < 4 4 A (m, n) 1) m n 2) 0 1 3) f { 1,, m } { 1,, n }, (i,f(i)) 0 1 i m, S { 1,, m } T { 1,, n } (1) i S, j T (i, j) 0 (2) (S ) + (T ) > n, f 1 i 1 i 2 m f(i 1 ) f(i 2 ) 5 n 2, a 1, a 2, a 3, a 4 : i) i = 1, 2, 3, 4, n a i ii) k = 1,, n 1 (ka 1 ) n + (ka 2 ) n + (ka 3 ) n + (ka 4 ) n = 2n 10
11 , (a 1 ) n, (a 2 ) n, (a 3 ) n, (a 4 ) n, n 2,, (a 1 ) n + (a j ) n = n j, 2 j 4, a (a) n, a n 11
12 /1/15 1:00-4:00 1 n , 120 n 2 12, 1, 2 6 2, 1, 3 1 4,, 4 3 (0, 0), (276, 153), (a, b) 2, a, b (a, b), 5 1 ABCD P, Q, AP + BP + P Q + CQ + DQ 6 A = {1, 2, 3, 4, 5, 6, 7} A f, (1),(2), (1) j A, j = f(k) k A, k j (2) A A g, j A, f(j) = g(g(j)) 7 3, A,B 2 9 A 6 3, 3 9 8, A A = { (x, y, z) 2xy z 2, x + y 1, x 0, y 0 }, B B = { (u, v, w) A (x, y, z), 0 ux+vy +wz 1 }, B, A 9 S = {1, 11, 31, 51, 71}, {a n } (1),(2),(3) (1) a 1 S, 12
13 (2) a n+1 1 a n + 1 S, (3) 10 n a n = 1993 {a n }, a ,,, 11, , 3,
14 /2/11 1 a, b, c,, x, y, z, 2, kyonkyon 8 2, 2 W 1 W 2, 1, 1, W 1 W 2 2 n, n d(n), D(n) T (n) : D(n) = d(1) + d(2) + + d(n) T (n) = n, 3D(n) = 2T (n) n 3 x, y,, 2 2 3, x, y, x, y,,,, ( ) ( ) 4 S 5 l 1,,l 5, 3 l 1,,l , 5 5 (n a 1,, a n ) C : n a 1,,a n, n n max x a j C n max x a j 0 x 2 j=1 0 x 1 j=1, max f(x) α x β f(x), b j α x β b 1,, b n b 1 b n n j=1 14
15 /1/15 1:00-4:00 1 (x, y), x, y y = 3 7 x a = 2 + 3, 2 a 3 ABCD-EF GH AF H BDE θ (0 θ 90 ) cos θ D C A B H G E F 4 P P (a, b) a + b 4 0, 1, 2, 3, P,,, 1 P 0 10 (0, 10) P 0 5 ABC AB, AC D, E, BE, CD P ADE, BP D, CEP 5, 8, 3, ABC , 15
16 7 5 5,, 8 A = {0, 1, 2, 3, 4, 5, 6, 7}, (1), (2) A A f (1) i, j A, i j f(i) f(j) (2) i, j A, i + j = 7 f(i) + f(j) = 7 9 a > b a, b x n = a 2 n 2 + 2bn x, {x} x (0 {x} < 1), lim { x n } n, lim a n n a n n 10 A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, (1), (2) A S (1) S 5 (2) S, 1, x, y 1 f(x, y) { f(x, y) + f(y, x) = 0 f(x, x + y) + f(y, x + y) = km, 1km 11, 11 y = m (m = 5, 4,, 4, 5), x = n (n = 5, 4,, 4, 5) 5 A k (k = 1, 2,, 5),, (x k, y k ) ( 5, 13), (2, 45), (44, 3), (4, 1), ( 27, 2) 1, S(x, y) (x, y) 16
17 /2/11 1 n, n a n b n = n+a n, b n (n = 1, 2, ), {c n } c n n , A 0 A 1 A 2 P 0, P 1,, P 6, : i = 0, 1,, 5, i 3 k, P i P i+1 A k, (1) P 0 = P 6 (2) : i = 0, 1,, 5, P i P i+1 A 0 A 1 A 2 P 0 4 ABC BC M MAC = 15 B 5 N 1 N, N 1 N 2, A, 1 2 A B A B A, 1 1 2,, 2,, 17
18 a = 3 ( ) a 1995/1/15 1:00-4:00 3, x 3 x 2,,, 1cm,,, 1cm 1cm 1cm 1cm 3 ABC O, OA, BC M, N B = 4 OMN, C = 6 OMN OMN 4 x 2 3x + 3 = 0 x = α α n, k α 1995 = kα n n k 5 (1), (2), p, q x : (1) x q, 1 p 1 (2) x p, 1 2 p q x p 10 x p 18
19 6 4 8, 7 5 A, B, C, D, E, x 2 y 2 + y 2 = 26x (x, y) 9 m m a 1, a 2,, a m, a i 1 4, : a i = a j a i+1 = a j+1 i = j a 1, a 2,,a m m 10, S S = {(x, y, z) x 2 4y 2 + z 2 12xy = 20} 2x + 3y + z = 3, S 11 4, 12 f(x, y, z) x, y, z, x 4, f(x, y, z) { f(x, z 2, y) + f(x, y 2, z) = 0, f(z 3, y, x) + f(x 3, y, z) = 0 19
20 1995 i=1 1995/2/11 1 n 2, r n g n r n 1 ri = 1 (n g), n 2 x x, x x 2 x f(x) a {f(x)} 2 a = f(x 2 ) a f(x) x, x 3 5 ABCDE, AC, AD BE S, R, CA, CE BD T, P CE, AD Q ASR, BT S, CP T, DQP, ERQ 1, (1) 5 P QRST (2) 5 ABCDE 4 {a 1, a 2, } a 2n = a n, a 2n+1 = ( 1) n, P (1) P 0, P P 0 x 1 P 1 (2) P i P, a i , P i+1, i = 1, 2, P 5 k n 1 k n, a 1, a 2,, a k a 1 + a a k = n, a a a 2 k = n, a k 1 + a k a k k = n 20
21 (x + a 1 )(x + a 2 ) (x + a k ) = x k + n C 1 x k 1 + n C 2 x k n C k i (i 1) (i j + 1), i C j 2, j (j 1)
22 /1/15 1:00-4:00 1 xyz- 4 (0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1) n a n = 10 2n 10 n +1 2 a n 4 a x 3 x 1 = 0, a 2 5 A = { 1, 2, 3, 4, 5, 6 } f: A A, f f f f f 6 N f: N N (1), (2), (3) (1) f(xy) = f(x) + f(y) 1 x, y (2) f(x) = 1 x (3) f(30) = 4 f(14400) 7 (a, b) LCM(a, b) + GCD(a, b) + a + b = ab a b LCM(a, b), GCD(a, b) a b, 8 0 f, f(0) = 0 [ x f(x) = f( ) + 10] [ ] 10 log 10 x 10 [ ] x x 1996 f(x) x 22
23 , x [x] x 9 xyz- 2 ( 2, 0, 1) (2, 0, 1) 4 T, xy 2 B T B, z = 1 2, xyz- A, A 2 P, Q, P Q A 10 n, S = { 1, 2,, n } S ( ) A, B, C, D A B C D = S A B C = φ (A, B, C, D) , 1, 2,, 1996, OFF k, P k, k ON/OFF P k, k = 1, 2,, 1996, ON 12 n 2, 4n (i, j) (i = 1, 2,, n j = 1, 2, 3, 4) L n L n (1, 1), (1, 1) a n : (2, 1) : (1, 1) L n : x y a 2 = 1, a 3 = 2 a 12 23
24 /2/ ABC P QR, ABC P QR, 0, T,, T, ABC, T A, B, C, ABC A, B, C θ ABC T, T, ABC, θ 2 GCD(m, n) = 1 m,n GCD(5 m + 7 m, 5 n + 7 n ) GCD(m, n) m n 3 x x > 1 a n = [x n+1 ] x[x n ] (n = 1,2,3, ) {a n }, n a p+n = a n p [x] x 4 Γ θ Γ θ 5 q < q < 2 n 2 2 n = 2 k + a k 1 2 k a a 0 ( a i = 0 1) p n p n = q k + a k 1 q k a 1 q + a 0, k ( ): p 2k < p l < p 2k+1 l 24
25 /1/15 1:00-4: !, 0? 2 30,, 3 xyz- xy- 13, yz- 6, zx- 18, 4 A = {1, 2, 3, 4, 5 } f: A A ( ) f(f(f(x))) = x x A 5 ABC BC = 6, CA = 5, AB = 4 AB, AC D, E, ADE BC D, E AB, AC, DE 6 a 3 a 1 = 0, a + 2, n, n 0, n 1,, k + 1 n k (0 k 9) n 8 f(x) 5, 5 f(x) + 1 = 0 x = 1 3, f(x) 1 = 0 x = 1 3 f(x) 9 f(x) x,, (1) (4) (1) 0 f(x) 1996 (x ) (2) f(x ) = f(x) (x ) (3) f(xy) f(x)f(y) mod 1997 (x, y ) 25
26 (4) f(2) = 999 f(x),, f(x) 1000 mod 1997 x a b mod n, a, b n 10 1 n, 2,,, 1 n = 5 2, 4, 3, 5 1, n = 7 2, 4, 6, 5, 1, 7 3 n = ABCD, AB = CD = a, AC = BD = b, AD = BC = c, 12 n ( ) 1 i, j 19 (i, j), 1, 2,, n, ABCD, A C, B D,, (10, 10) (, 1,, n, ) 26
27 :00 17: , 10, 2 2 a, b, c, (b + c a) 2 (c + a b)2 (a + b c)2 (b + c) a2 (c + a) b2 (a + b) 2 + c P 1,, P n, 2 ( ) , G 9, : G 5, 5 2 G 4 A, B, C, D 4,,, 3 AX + BX + CX + DX A, B, C, D X = X 0, AX 0 B = CX 0 D 5 2 n A B n n 27
28 /1/15 1:00-4:00 1 A, B, 10 A, B A, B A, B, 8 B n n AB CD ABCD, AB = BC = DA = 1, CD = AD E : E, A DC DE 4, 10 1, 2,, 5 xy- 4 A : (3, 0), B : (3, 2), C : (0, 2), D : (0, 0) ABCD uv- (u, v), ABCD (x, y) 0 ux + vy 1 (u, v) S S 6 8, 2 4, 2 4, 8 7 a n = n 1 (1 n 100) a 1,, a 100, 1 8 n,, , 1998 n 28
29 9 ABC B, C AC, AB D, E, ABC : BDE : CED = 2 : 3 : 4 A x 3 y 2 z 10 x, y, z x 6 + y 6 + z 6 n 11 a 1, a 2,, a n m = a i, i=1 1 n i a i, n,, i a i ( ) 1) 2) k, k 3), 2),, m, a 1,, a n (, ) 12 a n = n 3 5n 2 + 6n, b n = n (n = 1, 2, 3, ), a n b n d n a n = 0 d n = b n d 1, d 2, d 3, d d n = d n 29
30 :00 17: p 3 p, 1, 2, 2 3,, p 1 p 1 p,, , A, B, C, AB, BC, CA, 3 P 1, P 2,, P n P 1 P 2 P n P 1,,, 3 1 P 1, P 2,, P n, P 1, P i, P i 180 P i 1 P i P i+1, P i, P i (180 P i 1 P i P i+1 ) ( (180 P i 1 P i P i+1 ) < 0) P 0 = P n, P n+1 = P 1 720, 4 (1, 2,, n) A = (a 1, a 2,, a n ) n, n, 1, 2,, n, 1 k n, k a k, A = (a 1, a 2,, a n ), 1, 2, n A, A,, f(a) 0 m, f(a) = m n 30
31 A c n,m, P n (t) = c n,m t m m=0 Q n (t) = 1(1 + t)(1 + t + t 2 ) (1 + t + t t n 1 ) P n (t) = Q n (t) ,,, ( ):, 0, 1, 12, 11,, 31
32 , 50, 100 (0 ), (X, Y ) 3x + 5y = 7, X + Y x, y n 1999 n, n 3 n 3 3 n ( ) n = 1990, = 7, 880, 599, 000 = = ABCD-EF GH, AG, 5 1, 1, 2, 3, 2, 4, 5 1, 6 0 n, k p n (k) p n(n k) p n (n k) (0 k n) 6 3 AB = 4, BC = 6, AC = 5 ABC BC P, P 2 AB, AC M, N M, N P P 0 BP 0 32
33 A M N B P C ! 10 n n, 1999! 10 n 8 ABC, A = 60, B = 20 1, AB = 1, BC AC E F y 60 C x A y x B abc + abd + acd + bcd 1 9 n = a b c abcd d > 1 (a, b, c, d), a n, i = 1, α = cos 2π n + i sin 2π n m 33
34 1 m n, 1 n 1 k=0 α mk x α k 12 n ( 3) (1), (2), (3), (1) A, B A B, B A (2) A B, B A A, B (3) C,, C C 34
35 :00 17: ,, f(x) = x n, f(x) 3 n, 3 n+1 x 3 2n + 1, 2n 4 f(x) = (x )(x )(x ) (x 2 + n 2 ) ABCDEF, max{ AD, BE, CF }, min{ AD, BE, CF } 35
iii 1 1 1 1................................ 1 2.......................... 3 3.............................. 5 4................................ 7 5................................ 9 6............................
熊本県数学問題正解
00 y O x Typed by L A TEX ε ( ) (00 ) 5 4 4 ( ) http://www.ocn.ne.jp/ oboetene/plan/. ( ) (009 ) ( ).. http://www.ocn.ne.jp/ oboetene/plan/eng.html 8 i i..................................... ( )0... (
IMO 1 n, 21n n (x + 2x 1) + (x 2x 1) = A, x, (a) A = 2, (b) A = 1, (c) A = 2?, 3 a, b, c cos x a cos 2 x + b cos x + c = 0 cos 2x a
1 40 (1959 1999 ) (IMO) 41 (2000 ) WEB 1 1959 1 IMO 1 n, 21n + 4 13n + 3 2 (x + 2x 1) + (x 2x 1) = A, x, (a) A = 2, (b) A = 1, (c) A = 2?, 3 a, b, c cos x a cos 2 x + b cos x + c = 0 cos 2x a = 4, b =
( )
18 10 01 ( ) 1 2018 4 1.1 2018............................... 4 1.2 2018......................... 5 2 2017 7 2.1 2017............................... 7 2.2 2017......................... 8 3 2016 9 3.1 2016...............................
2 (1) a = ( 2, 2), b = (1, 2), c = (4, 4) c = l a + k b l, k (2) a = (3, 5) (1) (4, 4) = l( 2, 2) + k(1, 2), (4, 4) = ( 2l + k, 2l 2k) 2l + k = 4, 2l
ABCDEF a = AB, b = a b (1) AC (3) CD (2) AD (4) CE AF B C a A D b F E (1) AC = AB + BC = AB + AO = AB + ( AB + AF) = a + ( a + b) = 2 a + b (2) AD = 2 AO = 2( AB + AF) = 2( a + b) (3) CD = AF = b (4) CE
1 1 3 ABCD ABD AC BD E E BD 1 : 2 (1) AB = AD =, AB AD = (2) AE = AB + (3) A F AD AE 2 = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD 1 1
ABCD ABD AC BD E E BD : () AB = AD =, AB AD = () AE = AB + () A F AD AE = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD AB + AD AB + 7 9 AD AB + AD AB + 9 7 4 9 AD () AB sin π = AB = ABD AD
ORIGINAL TEXT I II A B 1 4 13 21 27 44 54 64 84 98 113 126 138 146 165 175 181 188 198 213 225 234 244 261 268 273 2 281 I II A B 292 3 I II A B c 1 1 (1) x 2 + 4xy + 4y 2 x 2y 2 (2) 8x 2 + 16xy + 6y 2
0 (18) /12/13 (19) n Z (n Z ) 5 30 (5 30 ) (mod 5) (20) ( ) (12, 8) = 4
0 http://homepage3.nifty.com/yakuikei (18) 1 99 3 2014/12/13 (19) 1 100 3 n Z (n Z ) 5 30 (5 30 ) 37 22 (mod 5) (20) 201 300 3 (37 22 5 ) (12, 8) = 4 (21) 16! 2 (12 8 4) (22) (3 n )! 3 (23) 100! 0 1 (1)
x, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y)
x, y x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 1 1977 x 3 y xy 3 x 2 y + xy 2 x 3 + y 3 = 15 xy (x y) (x + y) xy (x y) (x y) ( x 2 + xy + y 2) = 15 (x y) ( x 2 y + xy 2 x 2 2xy y 2) = 15 (x y) (x + y) (xy
29
9 .,,, 3 () C k k C k C + C + C + + C 8 + C 9 + C k C + C + C + C 3 + C 4 + C 5 + + 45 + + + 5 + + 9 + 4 + 4 + 5 4 C k k k ( + ) 4 C k k ( k) 3 n( ) n n n ( ) n ( ) n 3 ( ) 3 3 3 n 4 ( ) 4 4 4 ( ) n n
2 2 MATHEMATICS.PDF 200-2-0 3 2 (p n ), ( ) 7 3 4 6 5 20 6 GL 2 (Z) SL 2 (Z) 27 7 29 8 SL 2 (Z) 35 9 2 40 0 2 46 48 2 2 5 3 2 2 58 4 2 6 5 2 65 6 2 67 7 2 69 2 , a 0 + a + a 2 +... b b 2 b 3 () + b n a
0.6 A = ( 0 ),. () A. () x n+ = x n+ + x n (n ) {x n }, x, x., (x, x ) = (0, ) e, (x, x ) = (, 0) e, {x n }, T, e, e T A. (3) A n {x n }, (x, x ) = (,
[ ], IC 0. A, B, C (, 0, 0), (0,, 0), (,, ) () CA CB ACBD D () ACB θ cos θ (3) ABC (4) ABC ( 9) ( s090304) 0. 3, O(0, 0, 0), A(,, 3), B( 3,, ),. () AOB () AOB ( 8) ( s8066) 0.3 O xyz, P x Q, OP = P Q =
A(6, 13) B(1, 1) 65 y C 2 A(2, 1) B( 3, 2) C 66 x + 2y 1 = 0 2 A(1, 1) B(3, 0) P 67 3 A(3, 3) B(1, 2) C(4, 0) (1) ABC G (2) 3 A B C P 6
1 1 1.1 64 A6, 1) B1, 1) 65 C A, 1) B, ) C 66 + 1 = 0 A1, 1) B, 0) P 67 A, ) B1, ) C4, 0) 1) ABC G ) A B C P 64 A 1, 1) B, ) AB AB = 1) + 1) A 1, 1) 1 B, ) 1 65 66 65 C0, k) 66 1 p, p) 1 1 A B AB A 67
さくらの個別指導 ( さくら教育研究所 ) A 2 P Q 3 R S T R S T P Q ( ) ( ) m n m n m n n n
1 1.1 1.1.1 A 2 P Q 3 R S T R S T P 80 50 60 Q 90 40 70 80 50 60 90 40 70 8 5 6 1 1 2 9 4 7 2 1 2 3 1 2 m n m n m n n n n 1.1 8 5 6 9 4 7 2 6 0 8 2 3 2 2 2 1 2 1 1.1 2 4 7 1 1 3 7 5 2 3 5 0 3 4 1 6 9 1
i
i 3 4 4 7 5 6 3 ( ).. () 3 () (3) (4) /. 3. 4/3 7. /e 8. a > a, a = /, > a >. () a >, a =, > a > () a > b, a = b, a < b. c c n a n + b n + c n 3c n..... () /3 () + (3) / (4) /4 (5) m > n, a b >, m > n,
1/68 A. 電気所 ( 発電所, 変電所, 配電塔 ) における変圧器の空き容量一覧 平成 31 年 3 月 6 日現在 < 留意事項 > (1) 空容量は目安であり 系統接続の前には 接続検討のお申込みによる詳細検討が必要となります その結果 空容量が変更となる場合があります (2) 特に記載
1/68 A. 電気所 ( 発電所, 変電所, 配電塔 ) における変圧器の空き容量一覧 平成 31 年 3 月 6 日現在 < 留意事項 > (1) 空容量は目安であり 系統接続の前には 接続検討のお申込みによる詳細検討が必要となります その結果 空容量が変更となる場合があります (2) 特に記載のない限り 熱容量を考慮した空き容量を記載しております その他の要因 ( 電圧や系統安定度など ) で連系制約が発生する場合があります
高校生の就職への数学II
II O Tped b L A TEX ε . II. 3. 4. 5. http://www.ocn.ne.jp/ oboetene/plan/ 7 9 i .......................................................................................... 3..3...............................
入試の軌跡
4 y O x 4 Typed by L A TEX ε ) ) ) 6 4 ) 4 75 ) http://kumamoto.s.xrea.com/plan/.. PDF) Ctrl +L) Ctrl +) Ctrl + Ctrl + ) ) Alt + ) Alt + ) ESC. http://kumamoto.s.xrea.com/nyusi/kumadai kiseki ri i.pdf
(1) 3 A B E e AE = e AB OE = OA + e AB = (1 35 e ) e OE z 1 1 e E xy e = 0 e = 5 OE = ( 2 0 0) E ( 2 0 0) (2) 3 E P Q k EQ = k EP E y 0
(1) 3 A B E e AE = e AB OE = OA + e AB = (1 35 e 0 1 15 ) e OE z 1 1 e E xy 5 1 1 5 e = 0 e = 5 OE = ( 2 0 0) E ( 2 0 0) (2) 3 E P Q k EQ = k EP E y 0 Q y P y k 2 M N M( 1 0 0) N(1 0 0) 4 P Q M N C EP
x () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x
[ ] IC. f(x) = e x () f(x) f (x) () lim f(x) lim f(x) x + x (3) lim f(x) lim f(x) x + x (4) y = f(x) ( ) ( s46). < a < () a () lim a log xdx a log xdx ( ) n (3) lim log k log n n n k=.3 z = log(x + y ),
Solutions to Quiz 1 (April 20, 2007) 1. P, Q, R (P Q) R Q (P R) P Q R (P Q) R Q (P R) X T T T T T T T T T T F T F F F T T F T F T T T T T F F F T T F
Quiz 1 Due at 10:00 a.m. on April 20, 2007 Division: ID#: Name: 1. P, Q, R (P Q) R Q (P R) P Q R (P Q) R Q (P R) X T T T T T T F T T F T T T F F T F T T T F T F T F F T T F F F T 2. 1.1 (1) (7) p.44 (1)-(4)
i I II I II II IC IIC I II ii 5 8 5 3 7 8 iii I 3........................... 5......................... 7........................... 4........................ 8.3......................... 33.4...................
1 29 ( ) I II III A B (120 ) 2 5 I II III A B (120 ) 1, 6 8 I II A B (120 ) 1, 6, 7 I II A B (100 ) 1 OAB A B OA = 2 OA OB = 3 OB A B 2 :
9 ( ) 9 5 I II III A B (0 ) 5 I II III A B (0 ), 6 8 I II A B (0 ), 6, 7 I II A B (00 ) OAB A B OA = OA OB = OB A B : P OP AB Q OA = a OB = b () OP a b () OP OQ () a = 5 b = OP AB OAB PAB a f(x) = (log
x = a 1 f (a r, a + r) f(a) r a f f(a) 2 2. (a, b) 2 f (a, b) r f(a, b) r (a, b) f f(a, b)
2011 I 2 II III 17, 18, 19 7 7 1 2 2 2 1 2 1 1 1.1.............................. 2 1.2 : 1.................... 4 1.2.1 2............................... 5 1.3 : 2.................... 5 1.3.1 2.....................................
6 2 2 x y x y t P P = P t P = I P P P ( ) ( ) ,, ( ) ( ) cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ y x θ x θ P
6 x x 6.1 t P P = P t P = I P P P 1 0 1 0,, 0 1 0 1 cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ x θ x θ P x P x, P ) = t P x)p ) = t x t P P ) = t x = x, ) 6.1) x = Figure 6.1 Px = x, P=, θ = θ P
Part () () Γ Part ,
Contents a 6 6 6 6 6 6 6 7 7. 8.. 8.. 8.3. 8 Part. 9. 9.. 9.. 3. 3.. 3.. 3 4. 5 4.. 5 4.. 9 4.3. 3 Part. 6 5. () 6 5.. () 7 5.. 9 5.3. Γ 3 6. 3 6.. 3 6.. 3 6.3. 33 Part 3. 34 7. 34 7.. 34 7.. 34 8. 35
1. 2 P 2 (x, y) 2 x y (0, 0) R 2 = {(x, y) x, y R} x, y R P = (x, y) O = (0, 0) OP ( ) OP x x, y y ( ) x v = y ( ) x 2 1 v = P = (x, y) y ( x y ) 2 (x
. P (, (0, 0 R {(,, R}, R P (, O (0, 0 OP OP, v v P (, ( (, (, { R, R} v (, (, (,, z 3 w z R 3,, z R z n R n.,..., n R n n w, t w ( z z Ke Words:. A P 3 0 B P 0 a. A P b B P 3. A π/90 B a + b c π/ 3. +
(, Goo Ishikawa, Go-o Ishikawa) ( ) 1
(, Goo Ishikawa, Go-o Ishikawa) ( ) 1 ( ) ( ) ( ) G7( ) ( ) ( ) () ( ) BD = 1 DC CE EA AF FB 0 0 BD DC CE EA AF FB =1 ( ) 2 (geometry) ( ) ( ) 3 (?) (Topology) ( ) DNA ( ) 4 ( ) ( ) 5 ( ) H. 1 : 1+ 5 2
(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0
1 1 1.1 1.) T D = T = D = kn 1. 1.4) F W = F = W/ = kn/ = 15 kn 1. 1.9) R = W 1 + W = 6 + 5 = 11 N. 1.9) W b W 1 a = a = W /W 1 )b = 5/6) = 5 cm 1.4 AB AC P 1, P x, y x, y y x 1.4.) P sin 6 + P 1 sin 45
空き容量一覧表（154kV以上）
1/3 A. 電気所 ( 発電所, 変電所, 配電塔 ) における変圧器の空き容量 覧 < 留意事項 > (1) 空容量は 安であり 系統接続の前には 接続検討のお申込みによる詳細検討が必要となります その結果 空容量が変更となる場合があります (2) 熱容量を考慮した空き容量を記載しております その他の要因 ( や系統安定度など ) で連系制約が発 する場合があります (3) 表 は 既に空容量がないため
2/8 一次二次当該 42 AX 変圧器 なし 43 AY 変圧器 なし 44 BA 変圧器 なし 45 BB 変圧器 なし 46 BC 変圧器 なし
1/8 A. 電気所 ( 発電所, 変電所, 配電塔 ) における変圧器の空き容量一覧 < 留意事項 > (1) 空容量は目安であり 系統接続の前には 接続検討のお申込みによる詳細検討が必要となります その結果 空容量が変更となる場合があります (2) 特に記載のない限り 熱容量を考慮した空き容量を記載しております その他の要因 ( や系統安定度など ) で連系制約が発生する場合があります (3)
OABC OA OC 4, OB, AOB BOC COA 60 OA a OB b OC c () AB AC () ABC D OD ABC OD OA + p AB + q AC p q () OABC 4 f(x) + x ( ), () y f(x) P l 4 () y f(x) l P
4 ( ) ( ) ( ) ( ) 4 5 5 II III A B (0 ) 4, 6, 7 II III A B (0 ) ( ),, 6, 8, 9 II III A B (0 ) ( [ ] ) 5, 0, II A B (90 ) log x x () (a) y x + x (b) y sin (x + ) () (a) (b) (c) (d) 0 e π 0 x x x + dx e
０４年度LS民法Ⅰ教材改訂版.PDF
?? A AB A B C AB A B A B A B A A B A 98 A B A B A B A B B A A B AB AB A B A BB A B A B A B A B A B A AB A B B A B AB A A C AB A C A A B A B B A B A B B A B A B B A B A B A B A B A B A B A B
1 26 ( ) ( ) 1 4 I II III A B C (120 ) ( ) 1, 5 7 I II III A B C (120 ) 1 (1) 0 x π 0 y π 3 sin x sin y = 3, 3 cos x + cos y = 1 (2) a b c a +
6 ( ) 6 5 ( ) 4 I II III A B C ( ) ( ), 5 7 I II III A B C ( ) () x π y π sin x sin y =, cos x + cos y = () b c + b + c = + b + = b c c () 4 5 6 n ( ) ( ) ( ) n ( ) n m n + m = 555 n OAB P k m n k PO +
x V x x V x, x V x = x + = x +(x+x )=(x +x)+x = +x = x x = x x = x =x =(+)x =x +x = x +x x = x ( )x = x =x =(+( ))x =x +( )x = x +( )x ( )x = x x x R
V (I) () (4) (II) () (4) V K vector space V vector K scalor K C K R (I) x, y V x + y V () (x + y)+z = x +(y + z) (2) x + y = y + x (3) V x V x + = x (4) x V x + x = x V x x (II) x V, α K αx V () (α + β)x
21 2 26 i 1 1 1.1............................ 1 1.2............................ 3 2 9 2.1................... 9 2.2.......... 9 2.3................... 11 2.4....................... 12 3 15 3.1..........
1 12 ( )150 ( ( ) ) x M x 0 1 M 2 5x 2 + 4x + 3 x 2 1 M x M 2 1 M x (x + 1) 2 (1) x 2 + x + 1 M (2) 1 3 M (3) x 4 +
( )5 ( ( ) ) 4 6 7 9 M M 5 + 4 + M + M M + ( + ) () + + M () M () 4 + + M a b y = a + b a > () a b () y V a () V a b V n f() = n k= k k () < f() = log( ) t dt log () n+ (i) dt t (n + ) (ii) < t dt n+ n
17 ( ) II III A B C(100 ) 1, 2, 6, 7 II A B (100 ) 2, 5, 6 II A B (80 ) 8 10 I II III A B C(80 ) 1 a 1 = 1 2 a n+1 = a n + 2n + 1 (n = 1,
17 ( ) 17 5 1 4 II III A B C(1 ) 1,, 6, 7 II A B (1 ), 5, 6 II A B (8 ) 8 1 I II III A B C(8 ) 1 a 1 1 a n+1 a n + n + 1 (n 1,,, ) {a n+1 n } (1) a 4 () a n OA OB AOB 6 OAB AB : 1 P OB Q OP AQ R (1) PQ
() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)
0. A A = 4 IC () det A () A () x + y + z = x y z X Y Z = A x y z ( 5) ( s5590) 0. a + b + c b c () a a + b + c c a b a + b + c 0 a b c () a 0 c b b c 0 a c b a 0 0. A A = 7 5 4 5 0 ( 5) ( s5590) () A ()
漸化式のすべてのパターンを解説しましたー高校数学の達人・河見賢司のサイト
https://www.hmg-gen.com/tuusin.html https://www.hmg-gen.com/tuusin1.html 1 2 OK 3 4 {a n } (1) a 1 = 1, a n+1 a n = 2 (2) a 1 = 3, a n+1 a n = 2n a n a n+1 a n = ( ) a n+1 a n = ( ) a n+1 a n {a n } 1,
S K(S) = T K(T ) T S K n (1.1) n {}}{ n K n (1.1) 0 K 0 0 K Q p K Z/pZ L K (1) L K L K (2) K L L K [L : K] 1.1.
![S K(S) = T K(T ) T S K n (1.1) n {}}{ n K n (1.1) 0 K 0 0 K Q p K Z/pZ L K (1) L K L K (2) K L L K [L : K] 1.1.](/thumbs/91/105754396.jpg "S K(S) = T K(T ) T S K n (1.1) n {}}{ n K n (1.1) 0 K 0 0 K Q p K Z/pZ L K (1) L K L K (2) K L L K [L : K] 1.1.")
() 1.1.. 1. 1.1. (1) L K (i) 0 K 1 K (ii) x, y K x + y K, x y K (iii) x, y K xy K (iv) x K \ {0} x 1 K K L L K ( 0 L 1 L ) L K L/K (2) K M L M K L 1.1. C C 1.2. R K = {a + b 3 i a, b Q} Q( 2, 3) = Q( 2
a (a + ), a + a > (a + ), a + 4 a < a 4 a,,, y y = + a y = + a, y = a y = ( + a) ( x) + ( a) x, x y,y a y y y ( + a : a ) ( a : a > ) y = (a + ) y = a
[] a x f(x) = ( + a)( x) + ( a)x f(x) = ( a + ) x + a + () x f(x) a a + a > a + () x f(x) a (a + ) a x 4 f (x) = ( + a) ( x) + ( a) x = ( a + a) x + a + = ( a + ) x + a +, () a + a f(x) f(x) = f() = a
.1 A cos 2π 3 sin 2π 3 sin 2π 3 cos 2π 3 T ra 2 deta T ra 2 deta T ra 2 deta a + d 2 ad bc a 2 + d 2 + ad + bc A 3 a b a 2 + bc ba + d c d ca + d bc +
.1 n.1 1 A T ra A A a b c d A 2 a b a b c d c d a 2 + bc ab + bd ac + cd bc + d 2 a 2 + bc ba + d ca + d bc + d 2 A a + d b c T ra A T ra A 2 A 2 A A 2 A 2 A n A A n cos 2π sin 2π n n A k sin 2π cos 2π
(1) (2) (1) (2) 2 3 {a n } a 2 + a 4 + a a n S n S n = n = S n
. 99 () 0 0 0 () 0 00 0 350 300 () 5 0 () 3 {a n } a + a 4 + a 6 + + a 40 30 53 47 77 95 30 83 4 n S n S n = n = S n 303 9 k d 9 45 k =, d = 99 a d n a n d n a n = a + (n )d a n a n S n S n = n(a + a n
取扱説明書 -詳細版- 液晶プロジェクター CP-AW3019WNJ
B A C D E F K I M L J H G N O Q P Y CB/PB CR/PR COMPONENT VIDEO OUT RS-232C LAN RS-232C LAN LAN BE EF 03 06 00 2A D3 01 00 00 60 00 00 BE EF 03 06 00 BA D2 01 00 00 60 01 00 BE EF 03 06 00 19 D3 02 00
A
A 2563 15 4 21 1 3 1.1................................................ 3 1.2............................................. 3 2 3 2.1......................................... 3 2.2............................................
4 4 4 a b c d a b A c d A a da ad bce O E O n A n O ad bc a d n A n O 5 {a n } S n a k n a n + k S n a a n+ S n n S n n log x x {xy } x, y x + y 7 fx
4 4 5 4 I II III A B C, 5 7 I II A B,, 8, 9 I II A B O A,, Bb, b, Cc, c, c b c b b c c c OA BC P BC OP BC P AP BC n f n x xn e x! e n! n f n x f n x f n x f k x k 4 e > f n x dx k k! fx sin x cos x tan
15 mod 12 = 3, 3 mod 12 = 3, 9 mod 12 = N N 0 x, y x y N x y (mod N) x y N mod N mod N N, x, y N > 0 (1) x x (mod N) (2) x y (mod N) y x
A( ) 1 1.1 12 3 15 3 9 3 12 x (x ) x 12 0 12 1.1.1 x x = 12q + r, 0 r < 12 q r 1 N > 0 x = Nq + r, 0 r < N q r 1 q x/n r r x mod N 1 15 mod 12 = 3, 3 mod 12 = 3, 9 mod 12 = 3 1.1.2 N N 0 x, y x y N x y
I A A441 : April 15, 2013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida )
I013 00-1 : April 15, 013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida) http://www.math.nagoya-u.ac.jp/~kawahira/courses/13s-tenbou.html pdf * 4 15 4 5 13 e πi = 1 5 0 5 7 3 4 6 3 6 10 6 17
(1) θ a = 5(cm) θ c = 4(cm) b = 3(cm) (2) ABC A A BC AD 10cm BC B D C 99 (1) A B 10m O AOB 37 sin 37 = cos 37 = tan 37
4. 98 () θ a = 5(cm) θ c = 4(cm) b = (cm) () D 0cm 0 60 D 99 () 0m O O 7 sin 7 = 0.60 cos 7 = 0.799 tan 7 = 0.754 () xkm km R km 00 () θ cos θ = sin θ = () θ sin θ = 4 tan θ = () 0 < x < 90 tan x = 4 sin
DVIOUT-HYOU
() P. () AB () AB ³ ³, BA, BA ³ ³ P. A B B A IA (B B)A B (BA) B A ³, A ³ ³ B ³ ³ x z ³ A AA w ³ AA ³ x z ³ x + z +w ³ w x + z +w ½ x + ½ z +w x + z +w x,,z,w ³ A ³ AA I x,, z, w ³ A ³ ³ + + A ³ A A P.
1 n A a 11 a 1n A =.. a m1 a mn Ax = λx (1) x n λ (eigenvalue problem) x = 0 ( x 0 ) λ A ( ) λ Ax = λx x Ax = λx y T A = λy T x Ax = λx cx ( 1) 1.1 Th
1 n A a 11 a 1n A = a m1 a mn Ax = λx (1) x n λ (eigenvalue problem) x = ( x ) λ A ( ) λ Ax = λx x Ax = λx y T A = λy T x Ax = λx cx ( 1) 11 Th9-1 Ax = λx λe n A = λ a 11 a 12 a 1n a 21 λ a 22 a n1 a n2
() n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (5) (6 ) n C + nc + 3 nc n nc n (7 ) n C + nc + 3 nc n nc n (
3 n nc k+ k + 3 () n C r n C n r nc r C r + C r ( r n ) () n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (4) n C n n C + n C + n C + + n C n (5) k k n C k n C k (6) n C + nc
, x R, f (x),, df dx : R R,, f : R R, f(x) ( ).,, f (a) d f dx (a), f (a) d3 f dx 3 (a),, f (n) (a) dn f dx n (a), f d f dx, f d3 f dx 3,, f (n) dn f
,,,,.,,,. R f : R R R a R, f(a + ) f(a) lim 0 (), df dx (a) f (a), f(x) x a, f (a), f(x) x a ( ). y f(a + ) y f(x) f(a+) f(a) f(a + ) f(a) f(a) x a 0 a a + x 0 a a + x y y f(x) 0 : 0, f(a+) f(a)., f(x)
I II
I II I I 8 I I 5 I 5 9 I 6 6 I 7 7 I 8 87 I 9 96 I 7 I 8 I 9 I 7 I 95 I 5 I 6 II 7 6 II 8 II 9 59 II 67 II 76 II II 9 II 8 II 5 8 II 6 58 II 7 6 II 8 8 I.., < b, b, c, k, m. k + m + c + c b + k + m log
[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s
![[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s](/thumbs/94/118579915.jpg "[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s")
[ ]. lim e 3 IC ) s49). y = e + ) ) y = / + ).3 d 4 ) e sin d 3) sin d ) s49) s493).4 z = y z z y s494).5 + y = 4 =.6 s495) dy = 3e ) d dy d = y s496).7 lim ) lim e s49).8 y = e sin ) y = sin e 3) y =
1 8, : 8.1 1, 2 z = ax + by + c ax by + z c = a b +1 x y z c = 0, (0, 0, c), n = ( a, b, 1). f = n i=1 a ii x 2 i + i<j 2a ij x i x j = ( x, A x), f =
1 8, : 8.1 1, z = ax + by + c ax by + z c = a b +1 x y z c = 0, (0, 0, c), n = ( a, b, 1). f = a ii x i + i
4.6: 3 sin 5 sin θ θ t θ 2t θ 4t : sin ωt ω sin θ θ ωt sin ωt 1 ω ω [rad/sec] 1 [sec] ω[rad] [rad/sec] 5.3 ω [rad/sec] 5.7: 2t 4t sin 2t sin 4t
![4.6: 3 sin 5 sin θ θ t θ 2t θ 4t : sin ωt ω sin θ θ ωt sin ωt 1 ω ω [rad/sec] 1 [sec] ω[rad] [rad/sec] 5.3 ω [rad/sec] 5.7: 2t 4t sin 2t sin 4t](/thumbs/70/62705042.jpg "4.6: 3 sin 5 sin θ θ t θ 2t θ 4t : sin ωt ω sin θ θ ωt sin ωt 1 ω ω [rad/sec] 1 [sec] ω[rad] [rad/sec] 5.3 ω [rad/sec] 5.7: 2t 4t sin 2t sin 4t")
1 1.1 sin 2π [rad] 3 ft 3 sin 2t π 4 3.1 2 1.1: sin θ 2.2 sin θ ft t t [sec] t sin 2t π 4 [rad] sin 3.1 3 sin θ θ t θ 2t π 4 3.2 3.1 3.4 3.4: 2.2: sin θ θ θ [rad] 2.3 0 [rad] 4 sin θ sin 2t π 4 sin 1 1
2016 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 1 16 2 1 () X O 3 (O1) X O, O (O2) O O (O3) O O O X (X, O) O X X (O1), (O2), (O3) (O2) (O3) n (O2) U 1,..., U n O U k O k=1 (O3) U λ O( λ Λ) λ Λ U λ O 0 X 0 (O2) n =
y π π O π x 9 s94.5 y dy dx. y = x + 3 y = x logx + 9 s9.6 z z x, z y. z = xy + y 3 z = sinx y 9 s x dx π x cos xdx 9 s93.8 a, fx = e x ax,. a =
[ ] 9 IC. dx = 3x 4y dt dy dt = x y u xt = expλt u yt λ u u t = u u u + u = xt yt 6 3. u = x, y, z = x + y + z u u 9 s9 grad u ux, y, z = c c : grad u = u x i + u y j + u k i, j, k z x, y, z grad u v =
I, II 1, A = A 4 : 6 = max{ A, } A A 10 10%
1 2006.4.17. A 3-312 tel: 092-726-4774, e-mail: [email protected], http://www.math.kyushu-u.ac.jp/ hara/lectures/lectures-j.html Office hours: B A I ɛ-δ ɛ-δ 1. 2. A 1. 1. 2. 3. 4. 5. 2. ɛ-δ 1. ɛ-n
(2018 2Q C) [ ] R 2 2 P = (a, b), Q = (c, d) Q P QP = ( ) a c b d (a c, b d) P = (a, b) O P ( ) a p = b P = (a, b) p = ( ) a b R 2 {( ) } R 2 x = x, y
![(2018 2Q C) [ ] R 2 2 P = (a, b), Q = (c, d) Q P QP = ( ) a c b d (a c, b d) P = (a, b) O P ( ) a p = b P = (a, b) p = ( ) a b R 2 {( ) } R 2 x = x, y](/thumbs/101/151465237.jpg "(2018 2Q C) [ ] R 2 2 P = (a, b), Q = (c, d) Q P QP = ( ) a c b d (a c, b d) P = (a, b) O P ( ) a p = b P = (a, b) p = ( ) a b R 2 {( ) } R 2 x = x, y")
(2018 2Q C) [ ] R 2 2 P = (a, b), Q = (c, d) Q P QP = a c b d (a c, b d) P = (a, b) O P a p = b P = (a, b) p = a b R 2 { } R 2 x = x, y R y 2 a p =, c q = b d p + a + c q = b + d q p P q a p = c R c b
HITACHI 液晶プロジェクター CP-AX3505J/CP-AW3005J 取扱説明書 -詳細版- 【技術情報編】
B A C E D 1 3 5 7 9 11 13 15 17 19 2 4 6 8 10 12 14 16 18 H G I F J M N L K Y CB/PB CR/PR COMPONENT VIDEO OUT RS-232C LAN RS-232C LAN LAN BE EF 03 06 00 2A D3 01 00 00 60 00 00 BE EF 03 06 00 BA D2 01
IA 2013 : :10722 : 2 : :2 :761 :1 (23-27) : : ( / ) (1 /, ) / e.g. (Taylar ) e x = 1 + x + x xn n! +... sin x = x x3 6 + x5 x2n+1 + (
IA 2013 : :10722 : 2 : :2 :761 :1 23-27) : : 1 1.1 / ) 1 /, ) / e.g. Taylar ) e x = 1 + x + x2 2 +... + xn n! +... sin x = x x3 6 + x5 x2n+1 + 1)n 5! 2n + 1)! 2 2.1 = 1 e.g. 0 = 0.00..., π = 3.14..., 1
( 12 ( ( ( ( Levi-Civita grad div rot ( ( = 4 : 6 3 1 1.1 f(x n f (n (x, d n f(x (1.1 dxn f (2 (x f (x 1.1 f(x = e x f (n (x = e x d dx (fg = f g + fg (1.2 d dx d 2 dx (fg = f g + 2f g + fg 2... d n n
Chap9.dvi
.,. f(),, f(),,.,. () lim 2 +3 2 9 (2) lim 3 3 2 9 (4) lim ( ) 2 3 +3 (5) lim 2 9 (6) lim + (7) lim (8) lim (9) lim (0) lim 2 3 + 3 9 2 2 +3 () lim sin 2 sin 2 (2) lim +3 () lim 2 2 9 = 5 5 = 3 (2) lim
(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y
[ ] 7 0.1 2 2 + y = t sin t IC ( 9) ( s090101) 0.2 y = d2 y 2, y = x 3 y + y 2 = 0 (2) y + 2y 3y = e 2x 0.3 1 ( y ) = f x C u = y x ( 15) ( s150102) [ ] y/x du x = Cexp f(u) u (2) x y = xey/x ( 16) ( s160101)
1 1.1 ( ). z = a + bi, a, b R 0 a, b 0 a 2 + b 2 0 z = a + bi = ( ) a 2 + b 2 a a 2 + b + b 2 a 2 + b i 2 r = a 2 + b 2 θ cos θ = a a 2 + b 2, sin θ =
1 1.1 ( ). z = + bi,, b R 0, b 0 2 + b 2 0 z = + bi = ( ) 2 + b 2 2 + b + b 2 2 + b i 2 r = 2 + b 2 θ cos θ = 2 + b 2, sin θ = b 2 + b 2 2π z = r(cos θ + i sin θ) 1.2 (, ). 1. < 2. > 3. ±,, 1.3 ( ). A
.3. (x, x = (, u = = 4 (, x x = 4 x, x 0 x = 0 x = 4 x.4. ( z + z = 8 z, z 0 (z, z = (0, 8, (,, (8, 0 3 (0, 8, (,, (8, 0 z = z 4 z (g f(x = g(
06 5.. ( y = x x y 5 y 5 = (x y = x + ( y = x + y = x y.. ( Y = C + I = 50 + 0.5Y + 50 r r = 00 0.5Y ( L = M Y r = 00 r = 0.5Y 50 (3 00 0.5Y = 0.5Y 50 Y = 50, r = 5 .3. (x, x = (, u = = 4 (, x x = 4 x,
2000年度『数学展望 I』講義録
2000 I I IV I II 2000 I I IV I-IV. i ii 3.10 (http://www.math.nagoya-u.ac.jp/ kanai/) 2000 A....1 B....4 C....10 D....13 E....17 Brouwer A....21 B....26 C....33 D....39 E. Sperner...45 F....48 A....53
211 [email protected] 1 R *1 n n R n *2 R n = {(x 1,..., x n ) x 1,..., x n R}. R R 2 R 3 R n R n R n D D R n *3 ) (x 1,..., x n ) f(x 1,..., x n ) f D *4 n 2 n = 1 ( ) 1 f D R n f : D R 1.1. (x,
function2.pdf
2... 1 2009, http://c-faculty.chuo-u.ac.jp/ nishioka/ 2 11 38 : 5) i) [], : 84 85 86 87 88 89 1000 ) 13 22 33 56 92 147 140 120 100 80 60 40 20 1 2 3 4 5 7.1 7 7.1 1. *1 e = 2.7182 ) fx) e x, x R : 7.1)
高等学校学習指導要領解説 数学編
5 10 15 20 25 30 35 5 1 1 10 1 1 2 4 16 15 18 18 18 19 19 20 19 19 20 1 20 2 22 25 3 23 4 24 5 26 28 28 30 28 28 1 28 2 30 3 31 35 4 33 5 34 36 36 36 40 36 1 36 2 39 3 41 4 42 45 45 45 46 5 1 46 2 48 3
.5 z = a + b + c n.6 = a sin t y = b cos t dy d a e e b e + e c e e e + e 3 s36 3 a + y = a, b > b 3 s363.7 y = + 3 y = + 3 s364.8 cos a 3 s365.9 y =,
[ ] IC. r, θ r, θ π, y y = 3 3 = r cos θ r sin θ D D = {, y ; y }, y D r, θ ep y yddy D D 9 s96. d y dt + 3dy + y = cos t dt t = y = e π + e π +. t = π y =.9 s6.3 d y d + dy d + y = y =, dy d = 3 a, b
20 9 19 1 3 11 1 3 111 3 112 1 4 12 6 121 6 122 7 13 7 131 8 132 10 133 10 134 12 14 13 141 13 142 13 143 15 144 16 145 17 15 19 151 1 19 152 20 2 21 21 21 211 21 212 1 23 213 1 23 214 25 215 31 22 33
6kg 1.1m 1.m.1m.1 l λ ϵ λ l + λ l l l dl dl + dλ ϵ dλ dl dl + dλ dl dl 3 1. JIS 1 6kg 1% 66kg 1 13 σ a1 σ m σ a1 σ m σ m σ a1 f f σ a1 σ a1 σ m f 4
35-8585 7 8 1 I I 1 1.1 6kg 1m P σ σ P 1 l l λ λ l 1.m 1 6kg 1.1m 1.m.1m.1 l λ ϵ λ l + λ l l l dl dl + dλ ϵ dλ dl dl + dλ dl dl 3 1. JIS 1 6kg 1% 66kg 1 13 σ a1 σ m σ a1 σ m σ m σ a1 f f σ a1 σ a1 σ m
i
009 I 1 8 5 i 0 1 0.1..................................... 1 0.................................................. 1 0.3................................. 0.4........................................... 3
2011de.dvi
211 ( 4 2 1. 3 1.1............................... 3 1.2 1- -......................... 13 1.3 2-1 -................... 19 1.4 3- -......................... 29 2. 37 2.1................................ 37
1 1 n 0, 1, 2,, n n 2 a, b a n b n a, b n a b (mod n) 1 1. n = (mod 10) 2. n = (mod 9) n II Z n := {0, 1, 2,, n 1} 1.
1 1 n 0, 1, 2,, n 1 1.1 n 2 a, b a n b n a, b n a b (mod n) 1 1. n = 10 1567 237 (mod 10) 2. n = 9 1567 1826578 (mod 9) n II Z n := {0, 1, 2,, n 1} 1.2 a b a = bq + r (0 r < b) q, r q a b r 2 1. a = 456,
学習の手順
NAVI 2 MAP 3 ABCD EFGH D F ABCD EFGH CD EH A ABC A BC AD ABC DBA BC//DE x 4 a //b // c x BC//DE EC AD//EF//BC x y AD DB AE EC DE//BC 5 D E AB AC BC 12cm DE 10 AP=PB=BR AQ=CQ BS CS 11 ABCD 1 C AB M BD P
1 (1) ( i ) 60 (ii) 75 (iii) 315 (2) π ( i ) (ii) π (iii) 7 12 π ( (3) r, AOB = θ 0 < θ < π ) OAB A 2 OB P ( AB ) < ( AP ) (4) 0 < θ < π 2 sin θ
1 (1) ( i ) 60 (ii) 75 (iii) 15 () ( i ) (ii) 4 (iii) 7 1 ( () r, AOB = θ 0 < θ < ) OAB A OB P ( AB ) < ( AP ) (4) 0 < θ < sin θ < θ < tan θ 0 x, 0 y (1) sin x = sin y (x, y) () cos x cos y (x, y) 1 c
(2016 2Q H) [ ] R 2 2 P = (a, b), Q = (c, d) Q P QP = ( ) a c b d (a c, b d) P = (a, b) O P ( ) a p = b P = (a, b) p = ( ) a b R 2 {( ) } R 2 x = x, y
![(2016 2Q H) [ ] R 2 2 P = (a, b), Q = (c, d) Q P QP = ( ) a c b d (a c, b d) P = (a, b) O P ( ) a p = b P = (a, b) p = ( ) a b R 2 {( ) } R 2 x = x, y](/thumbs/98/136512264.jpg "(2016 2Q H) [ ] R 2 2 P = (a, b), Q = (c, d) Q P QP = ( ) a c b d (a c, b d) P = (a, b) O P ( ) a p = b P = (a, b) p = ( ) a b R 2 {( ) } R 2 x = x, y")
(2016 2Q H) [ ] R 2 2 P = (a, b), Q = (c, d) Q P QP = a c b d (a c, b d) P = (a, b) O P a p = b P = (a, b) p = a b R 2 { } R 2 x = x, y R y 2 a p =, c q = b d p + a + c q = b + d q p P q a p = c R c b
13 0 1 1 4 11 4 12 5 13 6 2 10 21 10 22 14 3 20 31 20 32 25 33 28 4 31 41 32 42 34 43 38 5 41 51 41 52 43 53 54 6 57 61 57 62 60 70 0 Gauss a, b, c x, y f(x, y) = ax 2 + bxy + cy 2 = x y a b/2 b/2 c x