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

Similar documents
( )

熊本県数学問題正解

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 ) = (,

18 ( ) ( ) [ ] [ ) II III A B (120 ) 1, 2, 3, 5, 6 II III A B (120 ) ( ) 1, 2, 3, 7, 8 II III A B (120 ) ( [ ]) 1, 2, 3, 5, 7 II III A B (

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 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 +

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

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 :

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


Part y mx + n mt + n m 1 mt n + n t m 2 t + mn 0 t m 0 n 18 y n n a 7 3 ; x α α 1 7α +t t 3 4α + 3t t x α x α y mx + n

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

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,

29

1 θ i (1) A B θ ( ) A = B = sin 3θ = sin θ (A B sin 2 θ) ( ) 1 2 π 3 < = θ < = 2 π 3 Ax Bx3 = 1 2 θ = π sin θ (2) a b c θ sin 5θ = sin θ f(sin 2 θ) 2

高等学校学習指導要領解説 数学編

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 +

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

入試の軌跡

[ ] Table

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

さくらの個別指導 ( さくら教育研究所 ) A AB A B A B A AB AB AB B

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

さくらの個別指導 ( さくら教育研究所 ) A 2 2 Q ABC 2 1 BC AB, AC AB, BC AC 1 B BC AB = QR PQ = 1 2 AC AB = PR 3 PQ = 2 BC AC = QR PR = 1

O E ( ) A a A A(a) O ( ) (1) O O () 467

) 9 81

4STEP 数学 B( 新課程 ) を解いてみた 平面上のベクトル 6 ベクトルと図形 59 A 2 B 2 = AB 2 - AA æ 1 2 ö = AB1 + AC1 - ç AA1 + AB1 3 3 è 3 3 ø 1

(4) P θ P 3 P O O = θ OP = a n P n OP n = a n {a n } a = θ, a n = a n (n ) {a n } θ a n = ( ) n θ P n O = a a + a 3 + ( ) n a n a a + a 3 + ( ) n a n

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

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

untitled

数学Ⅲ立体アプローチ.pdf

04年度LS民法Ⅰ教材改訂版.PDF

(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

高校生の就職への数学II

70 : 20 : A B (20 ) (30 ) 50 1

function2.pdf

76 3 B m n AB P m n AP : PB = m : n A P B P AB m : n m < n n AB Q Q m A B AQ : QB = m : n (m n) m > n m n Q AB m : n A B Q P AB Q AB 3. 3 A(1) B(3) C(

5. F(, 0) = = 4 = 4 O = 4 =. ( = = 4 ) = 4 ( 4 ), 0 = 4 4 O 4 = 4. () = 8 () = 4

18 ( ) I II III A B C(100 ) 1, 2, 3, 5 I II A B (100 ) 1, 2, 3 I II A B (80 ) 6 8 I II III A B C(80 ) 1 n (1 + x) n (1) n C 1 + n C

PSCHG000.PS

B. 41 II: 2 ;; 4 B [ ] S 1 S 2 S 1 S O S 1 S P 2 3 P P : 2.13:

数論入門

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

> > <., vs. > x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D > 0 x (2) D = 0 x (3

名古屋工業大の数学 2000 年 ~2015 年 大学入試数学動画解説サイト

.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 =,

空き容量一覧表(154kV以上)

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

アンリツ株式会社様

入試の軌跡

さくらの個別指導 ( さくら教育研究所 ) A a 1 a 2 a 3 a n {a n } a 1 a n n n 1 n n 0 a n = 1 n 1 n n O n {a n } n a n α {a n } α {a

さくらの個別指導 ( さくら教育研究所 ) 1 φ = φ 1 : φ [ ] a [ ] 1 a : b a b b(a + b) b a 2 a 2 = b(a + b). b 2 ( a b ) 2 = a b a/b X 2 X 1 = 0 a/b > 0 2 a

DVIOUT-HYOU

【】 1次関数の意味

A B 5 C mm, 89 mm 7/89 = 3.4. π 3 6 π 6 6 = 6 π > 6, π > 3 : π > 3

n ( (


繖 7 縺6ァ80キ3 ッ0キ3 ェ ュ ョ07 縺00 06 ュ0503 ュ ッ 70キ ァ805 ョ0705 ョ ッ0キ3 x 罍陦ァ ァ 0 04 縺 ァ タ0903 タ05 ァ. 7


4 4 θ X θ P θ 4. 0, 405 P 0 X 405 X P 4. () 60 () 45 () 40 (4) 765 (5) 40 B 60 0 P = 90, = ( ) = X

untitled

学習の手順

BD = a, EA = b, BH = a, BF = b 3 EF B, EOA, BOD EF B EOA BF : AO = BE : AE, b : = BE : b, AF = BF = b BE = bb. () EF = b AF = b b. (2) EF B BOD EF : B

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)


.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 +

122 6 A 0 (p 0 q 0 ). ( p 0 = p cos ; q sin + p 0 (6.1) q 0 = p sin + q cos + q 0,, 2 Ox, O 1 x 1., q ;q ( p 0 = p cos + q sin + p 0 (6.2) q 0 = p sin

zz + 3i(z z) + 5 = 0 + i z + i = z 2i z z z y zz + 3i (z z) + 5 = 0 (z 3i) (z + 3i) = 9 5 = 4 z 3i = 2 (3i) zz i (z z) + 1 = a 2 {

直交座標系の回転

f (x) x y f(x+dx) f(x) Df 関数 接線 x Dx x 1 x x y f f x (1) x x 0 f (x + x) f (x) f (2) f (x + x) f (x) + f = f (x) + f x (3) x f

(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

2001 Mg-Zn-Y LPSO(Long Period Stacking Order) Mg,,,. LPSO ( ), Mg, Zn,Y. Mg Zn, Y fcc( ) L1 2. LPSO Mg,., Mg L1 2, Zn,Y,, Y.,, Zn, Y Mg. Zn,Y., 926, 1

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】

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

取扱説明書[L-02E]

(, Goo Ishikawa, Go-o Ishikawa) ( ) 1

取扱説明書 -詳細版- 液晶プロジェクター CP-AW3019WNJ

( )

() 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)

(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

表紙(社会系)/153024H

13ィェィ 0002ィェィ 00ィヲ1 702ィョ ィーィ ィイ071 7ィ 06ィヲ02, ISSN

a n a n ( ) (1) a m a n = a m+n (2) (a m ) n = a mn (3) (ab) n = a n b n (4) a m a n = a m n ( m > n ) m n 4 ( ) 552

II 2 3.,, A(B + C) = AB + AC, (A + B)C = AC + BC. 4. m m A, m m B,, m m B, AB = BA, A,, I. 5. m m A, m n B, AB = B, A I E, 4 4 I, J, K

あさひ indd

linearal1.dvi

行列代数2010A

mobius1

HITACHI 液晶プロジェクター CP-EX301NJ/CP-EW301NJ 取扱説明書 -詳細版- 【技術情報編】 日本語

2016年度 京都大・文系数学

untitled

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

(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


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

( )

2018年度 神戸大・理系数学

x x x 2, A 4 2 Ax.4 A A A A λ λ 4 λ 2 A λe λ λ2 5λ + 6 0,...λ 2, λ 2 3 E 0 E 0 p p Ap λp λ 2 p 4 2 p p 2 p { 4p 2 2p p + 2 p, p 2 λ {

( a 3 = 3 = 3 a a > 0(a a a a < 0(a a a

Transcription:

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 = CO + OE = BA + AF = AB + AF = a + b B C a A b O D F E [1] ABCD O AB 1 : 2 E, BC F AB, AD (1) AO (2) DO (3) AF (4) EF [2] x, y a, b (1) x + 2( x b) = 4( b 2 a) a (2) 1 2 ( b + x) + 3( x 1 2 b) = 0 (3) 2 x + 3 y = a + b, 3 x + 2 y = b 1

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 2k = 4 l = 6, k = 8 (2) 1 a 1 1 a = (3, 5) = (3, 5) 32 + 52 34 [3] a = ( 1, 4), b = (3, 2), c = (0, 5) (1) 3 a + b (2) a 2 b + 3 c (3) 6( c 2 a) 5( 3 b + a) [4] (1) a = (1, 1), b = ( 1, 3), c = ( 3, 5) c = l a + k b l, k (2) ABCD A(1, 2), B(3, 5), C( 4, 0) D [5] a = (3, 1), b = (1, 2), c = a + t b t (1) c = 15 t (2) c t [6] A, B, C AB = a, AC = b ABC a, b 2

3 (1) a = (x 1, y 1 ), b = (x 2, y 2 ), a b θ a b cos θ = x 1 x 2 + y 1 y 2 (2) a = (1, 1), b = (2, x) x (1) a b (2) a b 120 (3) a b (1) O(0, 0), A(x 1, y 1 ), B(x 2, y 2 ) AB 2 = OA 2 + OB 2 2OA OB cos θ (x 2 x 1) 2 + (y 2 y 1) 2 = (x 2 1 + y1) 2 + (x 2 2 + y2) 2 2 a b cos θ a b cos θ = x 1x 2 + y 1y 2 (( )) a b = a b cos θ = x 1 x 2 + y 1 y 2 (2) (1) a b = 0 1 2 + ( 1) x = 0 x = 2 (2) a b = a b cos 120 1 2 + ( 1) x = p 1 2 + ( 1) 2 2 2 + x 2 ( 1 2 2 ), 2 x = 4 + x 2 2 ( 2(2 x)) 2 = ( 4 + x 2 ) 2, 2(4 4x + x 2 ) = 4 + x 2, x 2 8x + 4 = 0, x = 4 2 3 x = 4 + 2 3 (3) b = k a (2, x) = k(1, 1), 2 = k x = k x = 2 [7] a = 4, b = 5 a b a b (1) 45 (2) 120 (3) 90 (4) 180 [8] a, b (1) a = (2, 0), b = (2, 1) (2) a = (1, 1), b = (3, 2) (3) a = (k + 2, k 1), b = (2k 4, k + 1) (4) a = (p + q, q), b = (p q, p) [9] a, b (1) a = (1, 3), b = (2, 0) (2) a = (3, 7), b = (2, 5) (3) a = ( 3 1, 3 1), b = (1, 1) (4) a = ( 2, 2), b = ( 3 1, 3 + 1) [10] (1) a = 1, b = 5, 2 a + b = 3 a b (2) a = (3, 7), b = (2, 5) a + t b t 3

4 ABC AB) 3 : 1 D AC 4 : 3 E BE CD P AP BC Q (1) BE, CD AB, AC (2) AP : P Q [11] OAB OA 5 : 2 C OB 3 : 4 D CD M (1) OM OA, OB (2) OM AB N ON : OM, AN : NB [12] ABCDE AB = a, BC = b CD a, b [13] ABC AB = c, BC = a, CA = a AB = c, BC = a, CA = b (1) ABC G AG b c (2) AG 2 a, b, c (3) a = 12, b = 21, c = 3 AGB 4

5 [1] (1) ( 3, 4) v = (1, 2) (2) (1, 3), ( 2, 4) (3) (5, 4) v = (1, 2) [2] (2, 0) ( 1, 3) [14] (1) (, 4) n = (1, 2) (2) (1, 3), ( 2, 4) [15] ( 2, 3) x y 1 = 0 [16] ABC O OA = a, OB = b, OC = c (1) A BC A AA (2) ABC H OH a, b, c 5

6 OABC AB 1 : 2 D CD 3 : 5 E OE 1 : 3 F AF OBC G OA = a, OB = b, OC = c (1) OE a, b, c (2) AG : F G [17] ABCD a, b, c, d 3 : 1 [18] ABCD 3 A(1, 2, 3), B(3, 2, 1), C(6, 4, 4) (1) D (2) E(1, y, 15) ABC y [19] P ABC A P BC H P A = a, P B = b, P C = c (1) a b, b c, c a (2) P H b, c (3) P ABC 6

7 [1] (1) (1, 2, 3) v = (5, 2, 7) (2) (1, 1, 1), ( 2, 1, 3) [2] (1) (1, 2, 3) v = (5, 2, 7) (2) 3 (2, 1, 1), (3, 1, 1), (4, 1, 1) [20] (1) (1, 2, 3) v = (2, 3, 1) (2) ( 1, 5, 2), (3, 4, 1) [21] (1) (4, 2, 1) v = (1, 1, 3) (2) 3 ( 2, 3, 1 ), (2, 2, 3), ( 4, 1, 1) 7

* 8 [1] x 1 3 = y + 1 2 [2] = z + 2, 3x + 2y + z = 1 2 (1, 1, 2), 2x y + z 1 = 0 [22] (1) x + 2 2 = y 3 = y 1, 2x y + 3z + 2 = 0 3 (2) x + 1 2 = y 1 3 [23] (1) (2, 0, 1), x + y 2z + 3 = 0 = z + 2, 2x 3y + z 1 = 0 2 (2) ( 1, 1, 1), 3x y + 2z + 1 = 0 [24] (1, 2, 3) (1) 2x + 3y + z + 1 = 0 (2) x + 2y 2z 3 = 0 8

* 9 (1) a b = b a (2) a ( b + c) = ( a b) + ( a c) (3) (k a) b = k( a b) = a (k b) (4) ( a b) c = ( a c) b ( b c) a (5) ( a b) a ( a b) b (6) ( a b) 2 = ( a a)( b b) ( a b) 2 [25] a b a b (1) a = (1, 1, 1), b = ( 2, 3, 1) (2) a = ( 1, 1, 2), b = (1, 0, 1) [26] a = (1, 1, 3), b = ( 3, 1, 4) [27] a = (1, 0, 2), b = ( 2, k, 4) a b = 0 k [28] 3 a, b, c c ( a b) 9

[1] OABC OAB P, OBC Q OA = a, OB = b, OC = c PQ a, b, c PQ = x a + y b + z c z = AOC = 60, OA = 3, OC = 2 PQ = ( ) [2] OAB OA = 2, OB = 3, AOB = 120 AB M, AM N OA = a, OB = b (1) a b (2) OM, ON a, b (3) OM ON (4) MON = θ cos θ ( ) [3] 2 a = (1, x), b = (2, 1) (1) a + b 2 a 3 b x (2) a + b 2 a 3 b x ( ) [4] ABCD AB // DC, AB = 6, CD = 4 AB, AD M, N MN P MP : PN = 1 : 3 CP AB Q AB = a, AD = b (1) AC, AP a, b (2) CP : PQ (3) AD = 5, BAD = 60 CQ ( ) 10

[5] OABC OA = 3, OB = 4, OC = 5 AOB, AOC, BOC 60 BC O BC OP OB = b, OC = c OP = b + c OP = A OBC AQ OQ = b + c AQ = OABC ( ) [6] 3 a = (cos α, sin α, 0), b = (sin α, cos α, t), c = (sin α, cos α, 0) α, t a, b 0 v v c θ cos θ ( ) [7] (0, 0, 0) O A(1, 0, 0), B(2, 1, 0), C(3, 4, 1) OA = a, OB = b, OC = c 3 A, B, C α (a) P(x, y, z) OP = r a + s b + t c r, s, t x, y, z (b) P α x, y, z (c) (4, 5, 7) D α H DH AB, DH BC H ( ) [8] 3 O(0, 0, 0), A(1, 1, 0), B(1, 0, 1) α (1) α OA (2) α OAB ( ) 11