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- あきひろ みしま
- 9 years ago
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3 m n mnm mnn m m m m m m. x x x ax bxc a x x bb ac a fxax bxc fxx x ax bxca b ac x x ax bxca x
4 x x.x x x x x x xxx x x xxx x x xxx x x xx x x x axbcxdacx adbcxbd x xxx
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12 m n mnm mnn m m m m m m m m m m
13 mn m mnn mnm mnn m mnn mn m mn n m mnn mnnmm mnn m mnn m mmm m m mm m m m
14 mn n m m mn n m aaa aa a a a a a a aa a ab ab ab abab ab
15 mabmamb ab a abb ab a abb ababa b xaxbx abxab axbcxdacx adbcxbd ab a a bab b ab a a bab b aba abb a b aba abb a b xxxx xxxx x x xfxgxxfxf x fx x f fxx x f fxxx x fxffx x x fx fxx x f fxxx x x
16 ax by cby cz acz ax babca b c x y z
17 axbyc bycza czaxb axbyczabc cczabc czabc abc z c abc x a abc y x y z c abc a abc b abc c abc
18 xyx y x yxy
19 xy x y xy xy x x yxy y x y xyxy xy xy xy x xyy x y x y x y x y x y xy x y x y x yxy xyx y
20 x x xx x x x x x x x x x x x x x x x x
21 fxx ax bx cxd x xxx xx fx x x PxQx fxxxpxx fxxxqxx x f abcd f abcd x fabcd f abcd abcd abcd abcd abcd ab cd fxx x x
22 ab n n ab n k n k a k b nk a n n a n b n a n b b n
23 abn k k k k +
24 ab a a b b a b
25 a a a aa b b b bb a a b b ab ab ab a a a abab
26 pqr p q r r
27 pp qq r pqp qr pqr pq p qr pq r
28 . x x x x x x x x x x x
29 x x.x x x x x x.x x x x x x x x x x
30 x x x xxx x x xxx x x xxx x x xx x x x x x x ax bxcx bb ac a ax bxc x x x ax bxc b a c a ax bxci i xyx y yfx x pyqyfx pq yfx xyfx yfxy fx yaxbab yxy xy yax
31 x x x axbcxdacx adbcxbd x xxx x f f x b ac b a c a ii xyx y yfxxpyqyfx pq yfx xy fx yfxy fx yaxbab yaxbby y yax
32
33 2012 TOKYO LEGAL MIND K.K., Printed in Japan
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中学校学習指導要領解説数学編
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働く女性の母性健康管理、母性保護に関する法律のあらまし
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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
13 2 13.0 2 ( ) ( ) 2 13.1 ( ) ax 2 + bx + c > 0 ( a, b, c ) ( ) 275 > > 2 2 13.3 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 >
高ゼミサポSelectⅢ数学Ⅰ_解答.indd
数と式 ⑴ 氏点00 次の式を展開せよ ( 各 6 点 ) ⑴ (a-)(a -a+) ⑵ (x+y+)(x+y-5) 次の式を因数分解せよ (⑴⑵ 各 6 点, ⑶⑷ 各 8 点 ) ⑴ x y+x -x-6y ⑵ x -x - ⑶ a +5b ⑷ (x+y+z+)(x+)+yz 数と式 ⑵ 氏点00 次の問いに答えよ ( 各 6 点 ) ⑴ 次の循環小数を分数で表せ. a-5 = ⑵ 次の等式を満たす実数
04年度LS民法Ⅰ教材改訂版.PDF
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熊本県数学問題正解
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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.....................................
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
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 =
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4 4.0 2000 ) 9 81 10 4.1 natural numbers 1, 2, 3, 4, 4.2, 3, 2, 1, 0, 1, 2, 3, integral numbers integers 1, 2, 3,, 3, 2, 1 1 4.3 4.3.1 ( ) m, n m 0 n m 82 rational numbers m 1 ( ) 3 = 3 1 4.3.2 3 5 = 2
EPSON PX-A720 操作ガイド
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26 11 5 1 ( 2 2 2 3 5 3.1...................................... 5 3.2....................................... 5 3.3....................................... 6 3.4....................................... 7
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3 3.0 a n a n ( ) () a m a n = a m+n () (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 ( ) 55 3. (n ) a n n a n a n 3 4 = 8 8 3 ( 3) 4 = 8 3 8 ( ) ( ) 3 = 8 8 ( ) 3 n n 4 n n
5.. z = f(x, y) y y = b f x x g(x) f(x, b) g x ( ) A = lim h 0 g(a + h) g(a) h g(x) a A = g (a) = f x (a, b)
5 partial differentiation (total) differentiation 5. z = f(x, y) (a, b) A = lim h 0 f(a + h, b) f(a, b) h............................................................... ( ) f(x, y) (a, b) x A (a, b) x
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
5.. z = f(x, y) y y = b f x x g(x) f(x, b) g x ( ) A = lim h g(a + h) g(a) h g(x) a A = g (a) = f x (a, b)............................................
5 partial differentiation (total) differentiation 5. z = f(x, y) (a, b) A = lim h f(a + h, b) f(a, b) h........................................................... ( ) f(x, y) (a, b) x A (a, b) x (a, b)
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
LINEAR ALGEBRA I Hiroshi SUZUKI Department of Mathematics International Christian University
LINEAR ALGEBRA I Hiroshi SUZUKI Department of Mathematics International Christian University 2002 2 2 2 2 22 2 3 3 3 3 3 4 4 5 5 6 6 7 7 8 8 9 Cramer 9 0 0 E-mail:hsuzuki@icuacjp 0 3x + y + 2z 4 x + y
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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
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EPSON LP-8900スタートアップガイド
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福岡大学人文論叢47-3
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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 {
04 zz + iz z) + 5 = 0 + i z + i = z i z z z 970 0 y zz + i z z) + 5 = 0 z i) z + i) = 9 5 = 4 z i = i) zz i z z) + = a {zz + i z z) + 4} a ) zz + a + ) z z) + 4a = 0 4a a = 5 a = x i) i) : c Darumafactory
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III No (i) (ii) (iii) (iv) (v) (vi) x 2 3xy + 2. (x,y) (1,0) x 2 + y 2 5x 2 y x 2 + y 2. xy x2 + y 2. 2x + y 3 x 2 + y 2 + 5. sin(x 2 + y 2 ). x 2 + y 2 sin(x 2 y + xy 2 ). xy (i) (ii) (iii) 2xy x 2 +
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取扱説明書[L-02E]
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高校ゼミ 数学A
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[email protected] http://www.image.med.osaka-u.ac.jp/member/yoshi/ II Excel, Mathematica Mathematica Osaka Electro-Communication University (2007 Apr) 09849-31503-64015-30704-18799-390 http://www.image.med.osaka-u.ac.jp/member/yoshi/
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平成9年度水道事業年報 1概況 2施設
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