2009 = 7 41 41 n (n−1) 41
n n−1 n≥41 n=41 n n−1
7 n=42 n−1=41 n 7
n 7 n= 42
M E
C A
D
B
M F E
M ABECDF !
√
3
" AB = BE # $ BE = EC # $
AB=BE=EC=CD=DF=FA
∠BAD=∠BED=90 # E$
A, B, E, D BD ABCD "
F % ABECDF !
∠AEB=∠EAB# & E$ =∠ADB# $ %
ABED '
( BM=BD AE
ME=MA=MB=MC=MD,
A, B, E, C, D, F (
' ABEM ) #
AB=BE$ AB ME ( AM=BM ABM
' # BEM$
√
* ) A= (0, 0) B = (a, 0) C= (a, b) D= (0, b) E + , +
EB=EC '
2a b a +b =
b 2,
3a =b =√3
+ 1+2+ +10 = 55,
3 55=165
-17 & ' 17
, x
x+1 x−1 x+2 # ) .$
& 17
5 16+5 17 = 5 33 = 165 ( '
& / & 0 "
!
/1 01 / * x x+1
16 17 x−1
x+2 2 ! # x3$ y
1 y−3 y+3
. # $ " &
# 4 ) $ 4 2 10
5
54
18 18
6 7 P, S, T n.
T −S −2+2P =1/x −2+x = (1/x −1)(1−x )> 0.
T −S −2+2P = (T − −S − −2+2P − ) +
1
x −x −2(1−x )P −
> 1
x −x −2(1−x )
= (1/x −1)(1−x )> 0.
8 9 # '
..:$
5 ℓ
ℓ+1
% - ! .
! k
ℓ < k # $
-ℓ#
$ ( x ,
ℓ+1 x# ℓ $
x y
ℓ+1 ℓ+1 ℓ+2 k
5
• % 1 2 k ! 0 2009−k
1+2+ +k= ( + )
• % 1 2 k ! ! ℓ 0
2010−k ( + )−ℓ
% 2009 = ( + ) − ℓ 0 ≤ ℓ < k ( ( + ) −k = ( − )
'
2009= 63 64
2 −7,
k = 63 ℓ = 7 & 0 2010−63 = 1947
/ ( f(0) = f(0), f(0) =0 f( ) =1.
) B (x) x.
< x < 1, B (x) 1′s, ! x x= +2− +2− y,
m > 1 0 y < 1.
f(x) =1− f(2− + +2− + y) =1−2− + f( + y) =1−2− + +2− f(y).
y=0, f(x) =1−2− .
f( +2− +2− y) =f( +2− ) +f(2− y).
" 5 x=z+2− y, B (z) m
0 y < 1, f(x) =f(z) +2− f(y) =f(z) +f(2− y).
-0 f(y)< 2 x
# $ B (f(x))
# $ 7 B (x) ) #
$ & ) ;
# $ ) < ) #
→ $
# $ &
) u
f(u ).
f(u +u + +u ) =f(u ) +f(u ) + +f(u ).
f( +2− ) =1−2− + = ( +2− ) + (1−2− + ) <
) f(x) x ' 9 < ) f(x) = x.
5
f(x) x, ' B (x) ;
f(x) +f(1−x) x+ (1−x) = .
< '
B (x) B (1−x) , '
B (x) B (1−x) B (x).
' < B (x)
; - 6 B (x)
0= 3=
- & x = 0.00 . . . 3 =3/(4 −1).( 3 4 −1,
x =1/q k.
# $ x= 0,
B (x) ( x, B (x)
B(x) x = y+2− z, B (y)
m 0 z < 1, B (z) p.
z = v+2− z, B (v) ; & 2p B (z),
f(z) =f(v) +2− f(z), f(z) =f(v)/(1−2− ).
B(f(z)) 2p, B (f(v)),