.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 +
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1 .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 n n A n k n E T ra A 2n A n A 3 A 3 A 2 T raa + detae O A 3 A + T raa 2 T raa + detae + T ra 2 detaa T ra detae {T ra 2 deta}a T ra detae T ra 2 deta A A 3 A 1/12
2 .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 + d 2 a 3 + 2abc + bcd a 2 b + abd + b 2 c + bd 2 a 2 c + bc 2 + acd + cd 2 abc + 2bcd + d 3 a 3 + 2abc + bcd ba 2 + d 2 + ad + bc ca 2 + d 2 + ad + bc abc + 2bcd + d 3 a 3 + 2abc + bcd abc + 2bcd + d 3 a 3 + bc2a + d bca + 2d + d 3 a + d 3 a + d 3 T ra 3 T ra 3 T ra 2 deta A A 3 A A 4 A 4 A 2 A A 5 1 x n x 2 + kx + l x n x 2 + kx + lqx + px + q x 2 + kx + l 2 α, β x n x αx βqx + px + q x α, β α n pα + q β n pβ + q 2/12
3 .1 α n β n pα β α β p αn β n α β βα n pαβ + qβ αβ n pαβ + qα αβ n βα n qα β α β α β q αβn βα n α β x n x α 2 Qx + px α + r 1 x α α n r 1 nx n 1 2x αqx + x α 2 Q x + p x α nα n 1 p r pα 1 nα n x 2 + kx + l α, β α n β n α β x + αβn βα n α β α nα n 1 x + 1 nα n 1 A n A n pa + qe p A n A n p A 2 A 2 A n A A n A 3/12
4 .2.2 A m n > m 3 3 A 3 T raa 2 + aa + be O a, b A 2 A 2 T AT T rat + aa + be O T + aea T rat be T + ae a T + ae T T A T rat be T A S T T A T 1 S A n A A 2 A T a b 1 a b c a c a A 3 T raa {T ra2 T ra 2 }A detae O 1 a b c a 1 a 2 + bc a 2 + bc 2 A λ 1 λ λ 1 λ2 1 λ 1 λ 1 λλ 2 2 λ 3 λ 2 2λ + 2 4/12
5 .2 1 A 3 A 2 2A + 2E O T A T 2A + 2E O T 2EA T 2E A A 2 B 2 A B 1 1 λ λ λ 1 λ1 λ 5 λ λ 1 λ 5 5λ + λ λ 1 2 λ 4 λ 1, 4 λ x y z, x y z x y 2z x k, z l x y z k y 3k 2l l 2 1 k, l λ 1 1 3, 2 1 5/12
6 .2 λ x y z, x y z λ 4 2 C C 1 C 1 AC x y z m , m C 1 B 2 C C 1 BC /12
7 .2 C 1 BC ±1 ±1 ±2 8 B CC 1 BCC 1 ± ± , ± ± ±1 2 ±2 2 6 ±1 + 2 ±2 + 4 ± ±3 2 ±6 6, ± ±3 2 ± Ans.1 C 1 BC ± 1 bc b c 1 bc 2 ± 1 bc b c 1 bc ± 1 bc b c 1 bc 2 B CC 1 BCC ± 1 bc b c 1 bc ± 1 bc b 2 ±3 1 bc 2c 3b ± 2 1 bc c bc bc 3b + 6 ± 1 bc + b 2 ±2 1 bc + 3b bc 9b ±5 1 bc + 3b ±12 1 bc + 9b ±3 1 bc 2c bc + c bc + 2c 4 Ans.2 7/12
8 .2 3 B x A y e A 2 E O O 4 B x y e B x y e E O O 4 B 2 + xy E 1 Bx + xe O 2 yb + ey O 3 yx + e x O y O x O y O y O { B 2 E B ± 2Ex O B B ±2E x O B a b c a x O { a 2 + bc a4 a bc 16 a 2 bc 16 1 x y O x O y O 2 y 3 x { ybx + yxe O 2 ybx + eyx O x 4 B 2 x + xyx Ex 1 B 2 x + x4 e 2 Ex B 2 x + 4 e 2 Ex Ex B 2 x + 3 e 2 Ex O 5 8/12
9 .3 2 B B 2 x + Bxe O B 2 x + ebx O ebx 3 e 2 Ex O {eb 3 e 2 E}x O e ebx 3 e 2 Ex ebx 3 e 2 Ex ebx 3 e 2 Ex O 3 Bx e e x 7 B O 2 7 x O e 3 e e x O y O 2 Ans.1 Ans.2 e e B 2 + xy E 1 Bx O 2 yb O 3 yx 4 4 2,3 yb 2 x O 5 1,5 ye xyx yx yxyx 4 16 e 4 4 k 2.3 A 3 + αa 2 + βa + γe O 2 9/12
10 .4 A 2 T A 3 AT T A AT + βe T + βea αt γe T T + βe t 1, t 2, t 3 i j t i t j A i j a ij * i j a ij t j t i a ij t 1 t 2 t 3 t 1 t 2 t 3 # # # t i t j t i, t j a ij A n A n A 2 T A 2 A gt gt A ht... gt ht T T gt ht 3 ** A.4 A 3 T αa 2 + βa γe T αβa 2 + β 2 A βγe βt 1 A A 4 + αa 3 + βa 2 + γa O 1 2 AT + αt + βa 2 + γa O T A + αt + βa 2 + γa O βa 2 + γa AT αt βa 2 + γa T A αt αβa 2 + βγa αat α 2 T αβa 2 + αγa αt A α 2 T... 2 β 2 βγa αat + α 2 T βγe βt αt A + α 2 T βγe βt AαT ββ γe αt ββ γea α 2 T + βγe + βt 1/12
11 .5 α αt ββ γe ** A α T ra ββ γ A α β 2 α β γ 2 βa βe T A E 1 β T A ** TA AT ATTA T A.6 A Mn; C α 1, α 2,, α n A fx fx x α 1 x α 2 x α n fx A fa A α 1 EA α 2 E A α n E α 1 P 1 AP α 2 *... αn 11/12
12 .6 P P P 1 fap P 1 A α 1 EA α 2 E A α n EP P 1 A α 1 EP P 1 A α 2 EP P 1 A α n EP P 1 AP α 1 EP 1 AP α 2 E P 1 AP α n E α 1 α 1 α 2 * * α 2 *... αn αn O fa O A 12/12
熊本県数学問題正解
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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
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
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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
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高校生の就職への数学II
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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
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[ ] 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 ),
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(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)
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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
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> > <., 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
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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】
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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
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D xy D (x, y) z = f(x, y) f D (2 ) (x, y, z) f R z = 1 x 2 y 2 {(x, y); x 2 +y 2 1} x 2 +y 2 +z 2 = 1 1 z (x, y) R 2 z = x 2 y
5 5. 2 D xy D (x, y z = f(x, y f D (2 (x, y, z f R 2 5.. z = x 2 y 2 {(x, y; x 2 +y 2 } x 2 +y 2 +z 2 = z 5.2. (x, y R 2 z = x 2 y + 3 (2,,, (, 3,, 3 (,, 5.3 (. (3 ( (a, b, c A : (x, y, z P : (x, y, x
(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
3 () 3,,C = a, C = a, C = b, C = θ(0 < θ < π) cos θ = a + (a) b (a) = 5a b 4a b = 5a 4a cos θ b = a 5 4 cos θ a ( b > 0) C C l = a + a + a 5 4 cos θ = a(3 + 5 4 cos θ) C a l = 3 + 5 4 cos θ < cos θ < 4
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