高考數(shù)學(xué)常用公式及結(jié)論

p9(, 2009 c 6 4 8K 132 153 5 194 AN 245 nn 286 359$ 131 8(1) 8k(5!S5!2) 8x =2UAA,x =2A2. 85(1) ?8gA A(2) 8?88?83) ?f?gf?g?$f?(4) 8 ;f8k 2n n 1 n 1 n 2 3. 8$(1) A(BC) = (AB)CA(BC) = (AB)C(2) A(BC) = (AB)(AC)A(BC) = (AB)(AC)(3) U(AB) = U(AB) = #4. = A,AB = B,A B, ?, B = e#P?8A8?/5. Nn(M) Lk8M B) = n(A) +n(B) n(AB)n(ABC) = n(A) +n(B) +n(C) n(AB) n(BC) n(CA) +n(ABC)6. )8) k(n)8O8n(2) u85Lx9k5P8A = = x; MN = 0;+1)f(x;y) =f(x;y) x; MN =f(0;0);(1;1)g7. op) eK/_K/K/e:pK:K/e:qK:2) KK(K=K(3) 3k/0!/0KK5/=0!/=0(4) p_XKdK=K_KK_Kd14 9$(5) u(XK|dX/p)q,:q):by(6) /=Ky8. q :p p b b b bb b b (/ k k u u u u 1 pq :p:q 3,pq :p:q ? 3,x10. (1) ep)qKpq(2) 7ep)qKqp7(3) ep)qq)pKpq(4) 7,(5) K:B:A,(6) 1;+1)0;+1) 152 1. 5(1) eab!cdKa+cb+da db c(2) eab 0!cd 0bd 3) eab 0 4) 0abK 1a bK 1a 1b(5) ab!1a 1b a 0 b2. (1) ea! 2=a = b/=0(2) ea!Ka+b 2=a = b/=0(3) n/a!4) eab 0!m 0K ( +( (ac+ !xn! !yn!a!b!c!7) ea!8) ea!b! (a+b+c)2 3(ab+ba+=a = b = c/=0(9) ea!b!ab+bc+=a = b = c/=0(10) ea!b! (a+b+c)(c2 ab ac (11) ea!b! 3=a = b = c/=0(12) ea!b!b+3. AC/E|(1) 1n 1n+ 1 = 1n(n+ 1) 2)(2) 1 1pn+1 1)(3) x(1 x)2 = 12 2x(1 x)(1 x) 6 12 2x+ (1 x) + (1 x)33= 427(4) y = x(1 2 1 612233= 427 )y 6 2y = 16 4. (1) L)!(J(2) u(3) f14) |5) (6) !5. x!K(1) pKx = yx+y k 22) ysKx = yxy k 14. )N ( 0bx+)83Abx+c = 0 bx+)83Abx+c = 0 mmx1 x x 08. !5U1.f(x)g(x) 0,f(x) g(x) 0f(x)g(x) 0,8:f(x) g(x) 0;g(x)6= 0:f(x)g(x) 6 0,8:f(x) g(x) 6 0;g(x)6= 0:9. f(x) k)f(x) = M (x x (x p M ,/I0 n S 0 3mSf(x) 03mSf(x) a,xaxg(x),f(x) g(x)(4) )?uuu,=z11. (1) a 1 af(x) ag(x),f(x) g(x)x) x),8:f(x) 0;g(x) 0;f(x) g(x):(2) 0 ag(x),f(x) x),8:f(x) 0;g(x) 0;f(x) ,f(x)t(2) f(x) 6,f(x)t(3) f(x) f(x)t(4) f(x) 6f(x)t(5) f(x) = x) S(6) f(x) = x) 5/(.(x) = c 0 (a 0) ,8080C,2yC B(5) yygl(6) 8Byg,8B(7) )Ku(E,/?#E|!N!n!(8) L|55!3y(9) EyE!/9rK=zlydE|ppK6g(1) d=nzzgz!z(2) a?a?8?n)N3Na?(3) /(k)zmX?(4) g)z)x:KK)mX/I0g 193 51. A) y = f(x+ 1) xx+ 1xXf(x) =Kf(x+ 1) = 2xw,y = f(x+ 1) xx+ 1/y = fg(x)xg(x)(2) f(x) 0;1 of( 0 6 1 (ENf(x) = x); 0;1 f(= f( 1g= 1;1 /f(x) DKy = fg(x) x)23) f(x + 1) = x f(x) t = x + 1f(t) = t2 tf(x) = x2 xp=nf(t) =t2 tf(x) = x2 xnt) = t2 ttxKAKL(4) f(x) = x;gg(x) = x2;gkUAdAKwAKf(x)!g(x)DAg=?kf(x) = g(x)df(x) = x;gg(x) = x2;gAK(5) ey = f(x) 3y = f(x+ 1) y = f 1(x+ 1) y = f(x) y = f 1(x + 1)Uy =f(x+1)y = f 1(x) 1y = f 1(x+1)Ly = f 1(t)t = x+ 1 2. x1!a;bo(1) (x1 f(f( 0,f(f(x2)x1 0,f(x) 3 a;b O(2) (x1 f(f( 0K3m( 1; 1) )( 0;(2) ) y = f(x) :5 1pp(2) f(x)(q 0) =y = 1qf(x) y = f(x) :p5 1q15. y = f(x) 3u5?ykxAN3f(x) = 0316. 5(1) 55(2) f(a) = b,f 1(b) = a(a;b) 3y = f(x) o: (b;a)3y = f 1(x) (3) y = f(x) y = f 1(x) uy = 5y = f(x) x = f 1(y) (4) pk5(5) f(x)ABf 1(x) = x(f 1f(x) = x(ff 1(x)6= f 1f(x)(6) e y = f( b) 3 y = 1kf 1(x) b y =f 1(kx+b)y = f 1(kx+b) y = 1kf(x) b 6Kg(1) )(2) *!_!l!(3) g=kErK=5I(4) ugAn)y = f(x)!y = f 1(x) 55!5!C5Ag)K 0f(n) 0x2(m;n)kf(x) 0=f(x) = 0 3 (m;n) )(2) f(x) = bx+c = 0 (0)kk3( ; )Sdu/f( )f( ) 0 (i)p;qKf(x) f(f(x) p);f(q)g(ii)2p;qKf(x) p);f(q)gf(x) p);f(q)g(2) a+ a 0 f( ) 0f(x) 03p;q ,808:p6 08qf(q) 0a 0 3 p;q ,8 0f(q) 0(3) f(x) 0 ,8 0f(x) : 0mf(m) 0(3) f(x) = 0 m8: 00(4) f(x) = 0 : 0f(m) 0f(n) 0m: 0mf(m) 0(7) f(x) = 0 3m ( 1;n) f(n) : 00(8) f(x) = 03m (m;n) f(m)f(n) :f(m) = 0f(n) 0m:f(n) = 0f(m) 0m: 0f(m) 0f(n) 0m 0;m;n 1(2) a 1a 0;m;n 126 N(3) ar ar+s(a 0;r;(4) (ar)s = a 0;r;(5) (ab)r = a 0;b 0;7. 5(1) ( n = a(2) n an:a a 0;a a 0 L: (1;1)!(0;0)312) 0;1;N 0)(2) N (a 0;1;N 0)11. 0;1;M 0;N 0N 27(1) N) = 2) N = 3) 5Cf(x+y) = f(x) f(y),f(x y) = f(x)f(y)f(x) f(y),f(x y) = f(x) +f(y)12. a 0;1;m 0;1;N 0) nm a 0;a 1;m;n 0;1;1;N 0)13. !5 R R+ R+ 1 a 1 4. f(x) = bx+c)(0)P = ) ef(x) R0 0 0(3) 0 /Iu15. A(1) f(x) = f(x+y) = f(x) +f(y);f(1) = c(2) f(x) = f(x+y) = f(x)f(y);f(1) = 0(3) f(x) = f(= f(x) +f(y);f(a) = 1(a 0;1)(4) f(x) = x f(= f(x)f(y);f(1) = 1(5) uf(x) = f(x+y) +f(x y) = 2f(x)f(y);f(0) = 1(6) u f(x) = u g(x) = f(x y) = f(x)f(y) +g(x)g(y)16. ONOpum xoy N(1 +p)x28 nn5 nn1. !m(1) , k*d 2k (k 2 Z)=2f j = 2k + ;nn(2) m0mkX 2 6 6 2o=h 4;2i(3) 31A 1A(4) ! 2!2 m1K 2 31!n2 31!y1K 2 31!n2 31n!oyK1nK 2 31!o2 31!y1oK 2 31!o2 31n!oyK2. /(/% rK(1) /ll = r (2) /S = 12l = 12r 23. n(1) 2K 4. n) 1(2) 12 + (3) 12 + (4) (5) (6) 1(7) 1(8) 15. u!upw(1) =8:( 1)n2 n = 2k;( 1)n 12 n = 2k 1:(2) =8:( 1)n2 n = 2k;( 1)n 12 n = 2k 1:(nn 29(3) onp(4) p,1 6 ?1m6. (1) ) = (2) ) = (3) ) = (4) b2 + ) 9 3d: (a;b) = 7. (1) = 2 (2) = 2 1 = 1 2 (3) = 2 8. (1) 2 = r1 (2) 2 = (3) 2 = r1 + + 1 9. ) = 2 + (2) = 1 + (3) = 2 10. n(1) = 3 4 4 (2) = 4 3 4 (3) = 3 3 11. (1) + ) ) = (2) + ) ) = 30 nn(3) + 23 ) + + 43 ) = + 23 ) + + 43 ) = 0(4) + 23 ) + + 43 ) = + 23 ) + + 43 ) = 3212. uy = uy = (1) R(2) 1;1 y = x = 2k + 12 (y1x = 2k + 32 (y 1y = x = 2k (y 1x = 2k + (y 1kuK75u3) 5y = y = 2 f(x) = x+) f(x) = x+) T = 2 j!j(4) 55u y = ) % (k ;0) (k 2 Z)x = k + 2 (uy = % (k + 2;0) (x = k (!u.Lp:$:R%x:(5) y = 12 ;2k + 12 i( 12 ;2k + 32 i(y = 2k + i(3 ;2k + 2 i(6) uy = uy = :kIO 0!12 ! !32 !2 :21w5uu3S13. y = 5(1) 12 +k ;kK(2) nn 31(3) 5 y = a:ml (4) 55%0().%ax:,aCx:u!u?(5) 3m12 ;k + 12 (534. y = x+) (A;! 0) 5(1) R(2) A;A(3) Nm2k 12 ! ;2k + 12 !#(O3m2k + 12 ! ;2k + 32 !#(! 0) y = m) y = pI 0m0eXJ k z/(2) y!8B(3) yn!y!|!un9O20. u nn 3321. u) 2RR4(2) un b : c = 2b = 2b = 23) n/)n/e$un22. ) S = 12 12 12ha!hb!La!b!cp(2) S = 12 12 123) S = 2R2 4(4) S = 12r(!B)2 ( ! !(5) S = pp(p a)(p b)(p c) = p rp4=p = 12(a+b+c)r423. n/S +C = ,C = (A+B), 2 A+2C = 2 2(A+B)24. 34(1) 2) 1(3) 2 2 + 2 2 + 2 2 = 1(4) 4 2 2 2(5) 1 + 4 2 2 225. (1) :22i x(2) :2 0; x(3) :22; 2 x26. nnn(1) x) = 2) x) = 3) 2(4) x) = 5) x) = 6) 227. n)(1) a,x = k + ( 1)k )(2) a,x = 2k )(3) a)x = k + k(1) = k + ( 1)k (2) = 2k (3) = k + ( 356 1. an8:S1 n = 1;n 1 n 2:2. (n 1)d = (a1 d)( n( = n(n 1)2 d = (2d)n3. g|#5= . (1) 9 5 a1!d!n!a1!d 5 ? 3 B 2 =/ 3 20(2) $5E|:;a 2d;a d;a;a+d;a+ 2d;:d:;a 3d;a d;a+d;a+ 3d;: 2d5. 5(1) u?n a2 2) (a2 a1)n+ (2a1 = (n m)d(3) q pq = 0 (q)(4) u?p!q!r!sq = r +s 5) u?p!q!rr = 2q 26) u?bKfb (7) e(8) g!:(9) g= ,(10) Y= 3 (m)(11) n)n = 0(12) q pq = (p+q) (q)(13) m!n!n =n (14) 3 2k = (15) 3 2k + 1 = (2k + 1) = k + 1k 36 6. (8:q = 1; q 1:7. O5(1) (3) e 0 1 0(6) 10K(7) eK(8) g!:(9) Sm+n = 10) Y= (m)2 = (2m) 11) 3 2k = (12) 3 2k + 1 2k+1 = ()2k+1(13) qo/0=/q07 3710. b; = d:(1;0) (d b) 11. d# kK l12. 48vk = f(an+k 1;an+k 2;:;48X9k(48485k(1) 8B!8By(2) )n(3) E#5)K(4) S (an )+(