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7/21/2019 Public Page11 19
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Æ
x
102009 + 25
2 − 102009 − 25
2= 10x
9
200
220
12
C D AB
C
A
CB
DA
P
AC
BD Q QP
AB
27 + C 555+ C 1371+C
C
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ABC C BC = a CA = b
a ≥ b ABDE B C FG CAHI
ABC F I EH P F I
DG Q DG EH R
P QR
b
a
1
x − 1 + 1
x − 2 + 1
x − 6 + 1
x − 7 = x2 − 4x − 4
x
102009 + 25
2 − 102009 − 25
2= 10x
9
200
220
12
A C D
AB P CB DA Q
AC BD P Q AB
27 + C 555 + C
1371 + C
C
ABC C BC = a CA = b
a ≥ b
ABC
ABDE
B C FG
CAHI
P Q
R
F I EH F I DG DG EH
b
a P QR
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1
x − 1+
1
x − 2+
1
x − 6+
1
x − 7= x2 − 4x − 4
x2 + bx + c = 0
b c
5 4 c b
2
4
5
4
(x−5)(x−4) = x2−9x + 20 = 0
b
b = −9
2
4
(x − 2)(x − 4) = x2 − 6x + 8 = 0
c
c = 8
x2− 9x + 8 = 0
(x−1)(x−8) = 0
1
8
100
199
1
999
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1xy
1x1
1
9
11y
1
9
1xx
x
1
9
27
100
199
1 999
1
99
9
11 22 . . . 99
100 199 27
27 200
299
900 999
9 + 9 ·27 = 252 1 999
1 999
252
999 =
28
111
ABC
AB = AC
M
BC
P
BM
BC P BA K CA T P K + P T
P
P K + P T
P
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△ABC AB
AC M A ⊥ BC
P T ⊥ BC M A||P T
M A||P K △M BA
△P BK
B
P K
P B =
M A
M B P K =
P B · M A
M B
M A||P T △CP T
△CM A
P T
P C =
M A
M C
P T = P C
·M A
M C
............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
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A
C B
K
P M
T
M B = M C = 1
2BC
P K + P T = P B · M A
M B +
P C · M A
M C =
(P B + P C ) · M A
M B
= BC · M A
M B = 2M A
P K + P T P
M A
ABC B BA BC
3 : 2
BD CA
10 CA
2x
3x
CB AB y
y + 10 CD DA
z
BD
2y + 10
CA
△ABC
(2x)2 + (3x)2 = (2y + 10)2
13x2 = 4y2 + 40y + 100
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C D A
B
z
y y + 10
2x3x
.
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..............................................
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△BDC
△BDA
y2 + z2 = 4x2
z2 + (y + 10)2 = 9x2
z 4x2−y2 = 9x2−(y+10)2
5x2 = (y + 10)2 − y2 = 20y + 100
x2 = 4y + 20
13x2 = 52y + 260
13x2 = 4y2 + 40y + 100
52y + 260 = 4y2 + 40y + 100
4y2 − 12y − 160 = 0
y2 − 3y − 40 = 0
(y + 5)(y − 8) = 0
y > 0
y = 8
CA = 2y + 10 = 26
10
10
1
3
6
5
x y
10
10
10 − x 10 − y
x+6y
3(10−x)+5(10− y) = 80− 3x−5y
x = 9 y = 4
33
33
33
32
64
x + 6y + (80 − 3x − 5y) = 80− 2x + y ≤ 64 2x − y ≥ 16
x y 10
(x, y)
(8, 0) (9, 0) (9, 1)
(9, 2) (10, 0) (10, 1) (10, 2) (10, 3) (10, 4)
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32
x + 6y ≤ 32
80 − 3x − 5y ≤ 32
(10, 4)
3x + 5y ≥ 48
33
f 1 = f 2 = 1
n
≥ 2
f n+1 = f n + f n−1 1 1 2 3 5 8
13 21 34 55 89 144 . . . 100
f 1 = f 2 = 1 f 3 = 2 f n = f n−1 + f n−2 f m
m 3
S n =
nk=1
f 3k
S n = 1
2
n
k=1
(f 3k+f 3k) = 1
2
n
k=1
(f 3k−2+f 3k−1)+f 3k = 1
2
3n
k=1
f k
f 1 = f 3 − f 2 f 2 = f 4 − f 3 f 3 = f 5 − f 4
f r−1 = f r+1 − f r
f r = f r+2 − f r+1
n
r
k=1
f k = f r+2 − f 2 = f r+2 − 1
S n = 1
2(f 3n+2 − 1)
S 100 = 1
2(f 302 − 1)
S 100
f m = 1√ 5
(αm − βm) α = 12
(1 + √ 5) β = 12
(1 + √ 5)
S 100 = 1
2√ 5
α302 − β302
− 1
2
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p
q
p
q
q < 100
3
7
3
7
0.428571 . . .
3
7
q = 99
3
7
42
99 = 0.424242 . . .
43
99 = 0.434343 . . .
3
7
42
99
3
7
42
99
0.004
3
7
14
42
98
1
−1
41
98
43
98
3
7
1
98
0.01
42
99
42
99
3
7
43
99
3
7
p
q
3
7
p
q− 3
7
=
7 p − 3q
7q
= |7 p − 3q|
7q
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0
p
q
3
7
1
p q
p q 1
1
7q q
1
7 p − 3q = ±1
q 7 p = 3q±1
q
p
q = 99
7 p = 3(99)±
1 = 297±
1
7
q = 98
7 p = 3(98) ± 1 = 294± 1
7
q = 97
7 p = 3(97) ± 1 = 291± 1
7
q = 96
7 p = 3(96) ± 1 = 288 ± 1
287
7
q = 96
p = 41
1
4196 − 3
7
= 1
7 · 96 = 1
672
1
2
27 · 99 = 2
693
2
1
672
3
7
p
q =
41
96
sin x
x − 1
6x3 +
1
120x5
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