1、null null 5.| 1, 2 null, 9 B Q F , P B 6 B 9 . p H q .(s : 57 429= 6 381 9, 57 186= 6 354 9, 75 249= 8 361 9. )6.| 1, 2 null, 9 B Q F , P B 6 B 8 . p H q .(s : 46 . )7. | 0, 1, null, 9 B Q F T , P T H :! + ! = ! , ! + ! = ! , ! ! = ! ! .(提示: 0 l g . 0 ? T H .s k V 1+ 7= 8, 3+ 6= 9, 4 5= 20. )8. | 0,
2、 1, null, 9V U q 3 , / B T :5 - = ; , P = , , q 3 V B , q 3 V . h s M 3 T .(提示: 0 l g . 0 ? H .s k V 15 4= 60, 29 3=87; 15 4= 60, 39 2= 78. )9. | 0, 1, null, 9 B Q ,F aB , P .(提示:_ 2, V 12 8= 96 . )10. | 0, 1, null, 9 B Q , F B a , P .(提示:_ 2, V 9 4= 36 . )c T 5杨岚清( Z g v , 200011)null null l : 2007
3、- 08- 07null :2007- 11- 02null null (本讲适合高中) B H q / , B c T , p P T | | S # p K , n 5 . s N 5 1 0 H , g( t) max= g( 3) = 3+ 14 ;(3) t 5( a2+ b2+ c2 ) aabac,B i H s Y aabac .( 2002, o 0 X ) : 5 ab+ bc+ ca a2 + b2 + c2 1 “ , k 5, / L k( 6. , k # Z+ , k= 6. K . i i L aabac, ( a2+ b2+ c2 ab+ bc+ ca.B Z
4、 ,k( ab+ bc+ ca) 5( a2+ b2 + c2)5( ab+ bc+ ca) , null k 5. k # Z+ , k 6.6 B Z , y i H 1a1a2 , , 5 i k( 1 1+ 1 2+ 2 1) ( 5( 12+ 12+ 22) , null k ( 6. , k= 6./ k= 6 1 p . ! a ( b ( c. 6( ab+ bc+ ca) 5( a2+ b2+ c2) , 5c2- 6( a+ b) c+ (5a2+ 5b2- 6ab) 0 O d= ( b- c) 2.# 5d- ( a2+ b2+ c2)= 5( b- c) 2- ( a
5、2+ b2+ c2)( 5( b- c) 2- ( 2b- c) 2- b2- c2= - 6bc+ 3c2 ( 0. b a+ c2 ,5 a null( a2+ b2+ c2) + ( 15 - null) a2- ac. , 1 c ( ( 15 - null) a, T l null( a2+ b2+ c2) .y c 0, , 0 H , f ( x) 0- 127( x- a) 3.8 , p nullmax= - 127, O | 1 H q x= 0, = != = 2x, = != 0.说明: ( T p K , 1 B ,= , M % H q , H q , x s u
6、W ) .5 null B X ,y M 1 T 0 “ 9null p K l L M, P L aabac,| ab( a2- b2)+ bc( b2- c2) + ca( c2- a2)|( M ( a2+ b2+ c2) 2.( 47 IMO)92008 M 4 : ! E | T P H ? ( T T 0 , i i | T 0 ( a2 + b2+ c2 ) 2 y “ , 7 p M K l .n 5 I nP( t) = tb( t2- b2) + bc( b2- c2) + ct( c2- t2) . P( b) = P( c) = P( - c- b) = 0.5 | a
7、b( a2- b2) + bc( b2- c2) + ca( c2- a2) |= | P( a) |= | ( b- c) ( a- b) ( a- c) ( a+ b+ c) | . , T N | ( b- c) ( a- b) ( a- c) ( a+ b+ c) |( M ( a2+ b2+ c2 )2 . , ! a ( b ( c,5| ( b- c) ( a- b) | = ( b- a) ( c- b)( ( b- a) + ( c- b)22= ( c- a)24 ,O | 1 H q b- a= c- b,2b= a+ c. i ( b- a) + ( c- b)22(
8、( c- b)2+ ( b- a) 22null3( c- a) 2null ( 2 ( b- a) 2+ ( c- b) 2+ ( c- a) 2 ,O | 1 H q 2b= a+ c.V 7 ,| ( b- c) ( a- b) ( a- c) ( a+ b+ c) |( 14 | ( c- a) 3( a+ b+ c) |= 14 ( c- a) 6( a+ b+ c) 2( 142( b- a)2+ ( c- b)2+ ( c- a)233( a+ b+ c)2( 224( b- a)2+ ( c- b)2+ ( c- a)233( a+ b+ c)22. ( T | ( b- c)
9、 ( a- b) ( a- c) ( a+ b+ c) |( 22( b- a)2+ ( c- b)2+ ( c- a)2+ ( a+ b+ c) 242= 9 232 ( a2+ b2+ c2) 2.# M 9 232 , O | 1 H q 2b= a+ c# null ( b- a)2+ ( c- b) 2+ ( c- a) 23 = ( a+ b+ c)2. 2b= a+ c, ( c- a) 2= 18b2.| b= 1, a= 1- 3 22 , c= 1+ 3 22 .N H , T | .# Mmin= 9 232 .说明: 7 P ( t ) = tb ( t2 - b2 )
10、 + bc ( b2-c2) + ct ( c2- t2) ,P( b) = P( c) = P( - c- b) = 0, | T N | ( b- c) ( a- b) ( a- c) ( a+ b+ c) |( M ( a2+ b2+ c2) 2 1 o ;| | ( b- c) ( a- b)( a- c) ( a+ b+ c) | ( a2+ b2+ c2) 2 y “ , Q ( T , ? M T /F 1 p .练习题1. 1 x T x2 bc, p Kv L k, P ( a2 - bc) 2 k( b2- ca) ( c2- ab) .(提示:5 | b= c= 1, a
11、= 1+ #( # B V i l ) , T kmax= 4, . )5. p K l null, P i n aiabi # 1, 2 ( i = 1, 2, null, n) , O+ni= 1a2i= +ni= 1b2i +ni= 1a3ibi( null+ni= 1a2i.(提示: i ciadi # 1, 2 , 12 (cidi ( 2, (12 di - ci ) ( 2di - ci ) ( 0,5c2i+ d2i (52 cidi .# +ni= 1( c2i + d2i) ( 52 +ni= 1cidi.! aiabi # 1, 2 , 7 ci = ai32bi 12
12、, di =a12i b12i . Q T ,L +ni= 1a3ibi( 1710 +ni= 1a2i, null nullmin = 1710.K / B P nullmin = 1710 L . )命题与解题非线性递归数列化归为线性递归数列的常见技巧邹发明( i g B n , 400030)null null l : 2007- 09- 03null :2007- 12- 17null null , B t B 4 d L B , 5 H | B L B , + Z E p .1null y T s 1 null X an ,a1= 1, a2= 2, a3= - 1,an+ 2 an+ 1= 4an+ 1an- 6an+ 1an- 1+ 9anan- 1- 6a2n.p an .分析: B d L Q B , ? + Z E . i B T = Q Q T , V Y V y T s | B QQ T . :y an+ 2 an+ 1= 4 an+ 1 an- 6 an+ 1 an- 1+ 9 anan- 1- 6 a2n= ( 2an+ 1- 3an ) ( 2an- 3an- 1) ,112008 M 4
