SO GIAO DUC VA DAO TAO
TAY NII\H
Ky rrrr cHoN Hec srNH Gror Lop 12 THpr voNG riNn
NAvl Hec 2ot3 - zot4
Ngay thi: 25 thing 9 nlnr Z0l3
thi: ToAx - Btroi ttri thri,ntrdt
Tho'i gian: 180 phut (khong kA thdi gian giao
Mdn
DE CHfI\H THTIC
thi
gont
co 0 t trang, thi ,sinh khong phai chep
@A
dA)
di vao giay thi)
Bii 1. (4 dient)
Cho ba s6 ducvnga,b, c tho6 mdn cli6u kiqn
Chirngminhr6ng:
a+b+c
* b * t ,3
-L
dudng ttring dtii xung vdi d qua BC, CA, AB.
Chring rninh rang ba dtrong th[ng
dr, d2, d-, d6ng quy.
Bei s. G aiiim)
tno't minh rf,ng trong I I s5 thuc khirc nhau thu6c croan 10001 o6
the chon cluoc hai
[l;
,
so x vdy sao cho: 0 < x _y 28 thi f(*o)=f(xo -28+28):f(xo-28).f(28)=0.
t(28) = f(l 4 +14): f(l4).f(14) = 0 =+ f(l4) = 0 .
Lim tuong tU nhu trOn thi f(x) LAp lupn tucrng tp thi co:
Vay f(x)
f(x) -
0, Vx > 14
0,
vx >7; f(x)
= 0,
vx
a!
- 0, Vx > 0
trang
)
v6i A = O(0;0).
0, b > 0 vd phuong trinh duong
th[ng AM ld
m*0:
Phuong trinh cludrng thdng AN
Phucmg trinh dudrng thing
h
y=
-l*.
m
Suy ra N(-mb;
b).
MN lA (b - ma)x + (a + mb)y - ab(l + m2) : g
(1)
Phuong trinh dulne th6ng OH le (a + mb)x
GillhA;il""s;'i"h
cdc
- D. Suy ra H eBD.
di6m H la dudrng thang BD.
Gtti chfi: Ndu thl sinh s* dung ki6n ththc itudng thiing Simson
gidc CMN vd di€m A dd gidi thi:
- Phdn thudn: 2il
- PhAn ddo: 2d
diSi
vdi tam
trang 2
Bni 4
G dieln)
Cho itudng *dng d iti,qua trgc tdm H cfia tam gidc ABC. Ggi d1 , dz, dj lhn lwgt
ld gdc itudng thdng ddi x*ng: vt6,i d qua BC, CA; AB. Ch*ng minh rdng ba dwd'ng
thdng & , dz, dj cl6ng quy.
Ggi A', B', I' lAn luqt ld c6c
=j.tr",:"
\
v' netnlth
6,. ffi
{{;
{\
0,5
,,,:,:.: ,11)
r0i do4n c6
rha&U, mo
h0n
[nbr&nEgrnt
thinh11t00 phi
ta I t;;1 0 l
C
lhia
Ill
1
Xlr.1.
.)
I
Chgn
)(=xi,Y=xj thi 0
BAri
3.
6
siS
i
,
Vy e
i
arcm1
Cho hai dudng trdn (O1), (O) cilt nhau tpi hai tti€m A, B vi PrP2 h mgt ti6p tuy6n
chung cira hai ttuimg trdn d6 (P1 thuQc (Or), Pz thuQc (Oz)). Gqi Qr vd Qz lAn luqt li hinh
chi6u vudng g6c cria Pr, Pz l6n dulng thing OrOz. Dulng theng AQr cdt (Or) tAi ili6m
thf hai M1, Dulng thing AQ2 cit (Oz) tai diem thri hai M2. Chimg minh reng ba cli6m M1,
M2, B thing hang.
Bifi 4.
6
aiaml
s
5):64
3y + 3)
: l2+5 lx
(5 diAm)
(1
Cho day sO
(*,)
th6a mdn:
J"'=5
Lr'+1) ' *n = r2Xn*r + ( 4n+2)XnXn+1 Vn > 1, n e N
Tim limxn.
Bei 3. (5 diem)
Chrlng minh rang vdi n nguy6n duong vd n > 3 thi phucrng trinh 4xn +
kh6ng c6 nghiQm nguy6n ducrng (x; y).
Bni 4.
6
Q diem)
Giei h 0 phucrng trinh
I.'(ry
L*y
Vd'i x = 0,
h.6
+ ss) = 64
(y'+ 3y + 3) = lZ + 5lx
ph"""g
Dod 6 hQ tuong duong uoir
f_ __ 64
llv+5s-:'
*3
I
[r'+3
DAt t-
A
),thi
=+
(n
(n + 1)'
Xn+l
Xn
)
0, Vn > l, n e N
+4n+2
-d
xn
-2 thi yr =3 va (r, * t)' (yn*, n2)- n'(y, +z)+ 4niz
-a
xn
* l)'yn+r - n'y n
n2
*Yn+r =-=,Yn
(n+l)
It'
3 n6n
uv+l = x+l = vn -un =(v_u)(vn-l *rn-2, +...+.rn-l) > l+uv +u2 .
v6 lf. vfly phucrng trinh dd cho kh6ng c6 nghiQm nguyEn ducrng.
trang 2
-
a) = (p
= (a+dXa-d):6;t;
-
CD
=a-d; Dp=p_d
d)p . Suy ra a2 = dp
+DB.DC _DP.DM (3)
Tt
(2) vA (3) suy ra: DQ.DR
VOy P, Q, R,
-
DP"DM
M cirng thuQc mOt dudng trdn.
,
......
Bii 1. 6 aiam)
Cho c6c s6 duong
Chimg minh
Beri 2.
6
x,y, zthoi mdn tli6u kiQn x + y * z=1.
L- * -=-l+
,ing
< 3.
e 3xz
- , +y
3z' + x + y
+z 3y' +z+x -;-]-
aieml
f thoe m6n di6u ki0n f(x) +2f (xy)= f(x +Zxy),Vx e
Chimgminhreng f(x+y)=f(x)+f(y),Vxe 1 ,Vye i
Cho hdm
BAri
3.
6
so GIAo Drrc vADAo rAo rAyNIi\H
xi, rnr cHeN Hec srNH cror Lop tz rHpr voNG riNn
NAvr Hgc zot4 -zots
i\gny thi: 25 thfng 9 nIm 2AL4
Mdn thi: TOAX -Budi thi thfr hai
Thbi gian: 180 phrit (khdng kA thdi gian giao dA)
DE CHINHTHTIC
@A SOm cd
0l trang, thi sinh kh\ng phai chdp di vdo gidy
tht)
Bni 1. (5 diAm)
Giai h0 phuo-ng trinh
5)=64
I"'(3y+s
L"V (y'+ 3y + 3) = 12 +5 lx
Bei 2. (5 dihm)
(l
Cho day sO
chdn c6c ducrng cao cta tam gifucABC vE ttr A, B, C xuiSng. c4nh di5i diQn.Ducrng thlng
EF cfu BC t4i P, dudng thang qua D vd song song vdi ef c6t c6c dudng tfrang AC va Ae
tuong img tpi Q vn R. Gqi M ld trung ei6m cpnh BC. Chr?ng minh ring b6n Oi6* P, Q, R,
M ctng thuQc mQt dudng trdn.
I
... H6t...
a
{/ri,
SO GIAO DUC VA DAO TAO TAY NINH
xi. Tm CHQN HQC SINH cror LOP 12 THPT voNG riNn
NAvt Hec 2or4 -zors
TTUONG
oAN
cnAu rm
vrON
roAN - su6r rnr rrrtl NnAr
trang
f (Zxy), Yx e
v
x*0,dflt x-u; y=
2u
e lR'' VY € R
0
= f(2x)
Vai,
f(x +2xY)'Vx
thi f(u + v)
R,Vy e R
- f(u) + f(v), Vu * 0, Vv e R (1)
V6i x =0 thi f(x+y) =f(y)=f(0)+f(y) =f(x)+f(v), Vv e]R' (2)
(1) vd (2) suy ra: f(x + y) - f(x) + f(y), Vx e R, Vy e IR
i;
Bili 3
Q diam)
12
rHpr voNG riNn
Hec zor4 -zots
i\giy thi: 25 thfng 9 nlm 20L4
M6; thi: ToAll - Buoi thi thfr hai
Thli gian: 180 ph fit (kh6rg kA thdi gian giao ai)
DE CHINHTHTTC
@A gom cd 0I trang, thi sinh kh1ng phai chdp di vdo gidy tht)
Bei L. (5 diem)
Giei h0 phuong trinh
[*'(3v
L*V
Bili2.
+
$'+
s
5):64
3y + 3)
Xn
)
0, Vn > l, n e N
+4n+2
-d
xn
-2 thi yr =3 va (r, * t)' (yn*, n2)- n'(y, +z)+ 4niz
-a
xn
* l)'yn+r - n'y n
n2
*Yn+r =-=,Yn
(n+l)
It'
(r,
Do d6
I
_
3 n6n
uv+l = x+l = vn -un =(v_u)(vn-l *rn-2, +...+.rn-l) > l+uv +u2 .
v6 lf. vfly phucrng trinh dd cho kh6ng c6 nghiQm nguyEn ducrng.
trang 2
,?
^1
,'fi|)L i)
a(-
-
CD
=a-d; Dp=p_d
d)p . Suy ra a2 = dp
+DB.DC _DP.DM (3)
Tt
(2) vA (3) suy ra: DQ.DR
VOy P, Q, R,
-
DP"DM
M cirng thuQc mOt dudng trdn.
,
......
HGt
eooooo
trang 3