BENTRE. GQi M Ia giao di8m khac A cua hai dirong tron ngoai tiSp tam giac ABD va AlJ, 0 1fttam dirong tron ngoai ti~p tam giac AlJ.. ----[r]
Trang 1sa GlAo DVC vA DAo T~O
DE cHiNH TBUc
.TRUNG HQC co sa - NAM HQC 2018 - 2019
Mon: TOliN
Ngay thi: 27/02/2019
Thiri gian: 150phut(khong k~ phat d~) ,
CauI (8 di d m)
.a) G'iatp irongtr h Inh': + = + 1· 1 1 . 1
3x-l 2x+4 9x-2' S':"'4x
b)Ruts b'A thu A _ ( x + 2Fx+4 x+2.,1x +I J '(3 1 _2_)
u gon leu ire: - x.,lx _ 8 + x-I +.,Ix _2+ .,Ix +1
voi x ~ 0, x:;C1;x:;C4
c) GiAiMphirong trinh: x2+y
' ~x+y=x -y
Cau 2 (5 didm)
a) Tim cac nghiem nguyen cua phuong trinh: IXYI+ I ~- YI =1
b) Cho cac 8d thuc duong x , y thoa x+ ys;1.Tim gia tri nho nliat cua bi8u thirc:
Cho hinh binh hanh ABeD voi A, C cd dinh va B, D di dQng.Duong phan giac trong cua g6e
BcD cit AB va AD thea thirt1,l taiI va J (Jn~m gitra A va D) GQi M Ia giao di8m khac A cua hai dirong tron ngoai tiSp tam giac ABD va AlJ, 0 1fttam dirong tron ngoai ti~p tam giac AlJ
a) Chimg minh AO la phan giac trong cua g6e.IAJ
-b) Chirng minh OBA=ODJ,
c) Tim ~p hqp di8m M khi cac di8m B, D di dQng
Cau 4 (1 didm)
Cho ham s6 f(x) =(X 3 + 6x-7r 0 19 Tlnh f(a) v&i a==13+:Jfi +13 -10.
Trang 2sa GlAo nyc vA DAo T~O
BENTRE
DK: x:;z!:- x :;z!:-2
O.S
O.S
<=>
(3x -1)(2x +4) =(9x - 2)(S - 4x) (2)
3
(2) <=> 6x 2 +12x-2x-4 =-36x 2 +4Sx+8x-l0 <=> ~
x="6(TM)
b)
O.S
1.0
= -;=-' ;:!::::' ~
c)
kien: x + y > 0
S2+2P _ 2P -1=0
S
O.S
O.S
O.S
Trang 3~ S3+2P-2SP-S = 0 ~ S ( S 2 -1)-2P(S-I) =O~ (S-1)(S 2 + S-2P)= 0
0.5
V6i S =I=> x + y =1 thay vao Izj taduoc: 1=(I- y ) 2 - y ~ y =0 ,y =3 0.5
0.5
V~yMdi'icho co nghiem (x ;y ) =(1;0),-2;3).( 0.5
Ix Y I ~ O Ix -Y I=O
Do =>
I x-YI~ O I xYI =O
I x - Y I=1 (2)
0.5
(1)=> 2 <=> <=> X=Y=±1
( x - y ) =0 x= y
b)
{
x =O
y= ±1
{
x=±1
y =O
0.5
[
x =O
y =O <=>
I x - Y I=1
0.5
(2)=>
0.5
Theo b~t d~ng tlnrcCosi ta co: X 4+ l ~2x2l suyra 8(X4+y4 ) ~ 16 x2 y
16 x y +-=16x y
0.5
Ap dung b~t d~ng thircCosi,ta eo: 16 x2l +_1_+_1_ ~ 3,
4 xy 4 xy
Suyra 16x y +-+-+-~3+2=5.
4 xy 4 xy 2 xy
V~yGTNNcua B la5khivachikhix = y =k.
0 5
0 5
0 5 0.5
Trang 4B
1.0
Vi AIIIDC (do ABCD la hinh binh hanh) nen AIJ=DCJ (so le trong)
Vi AJIIBCnen All = BCJ(d6ngvi)
MelCJ la phan giac g6c BCD nen BCJ = DCJ =>AIJ = AJI => ~ AIJ canaA 0 5
Do 0 la tam duong tron ngoai
giac g6c IAJ
~ AIJ cannenAO la trung tnrc IJ thai lelphan
0.5
-Vi IDI BC nen DJC = JCB = JeD => ~ IDC can tai D 0.5 Suy ra JD = DC = AB (do ABCD lelhinh binh hanh)
Ta c6 OA=OJ ( bang ban kinh (0)) 0 5
Xet ~ OAJ voi g6c ngoai OID c6:
-OJD = AOJ + OAJ = 2AIJ + OAJ = 2DCJ + OAJ
- - -
0.5
Xet ~ OAB vel~ OID c6:
{
OA=OJ(cmt)
6AB =(jjjj(cmt)=> ~OAB =~OJD(c.g.c)
AB=JD(cmt)
0.5
GQi0' la tam duong tron ngoai tam giac ABD
GQiK la giao BD velAC =>K leltrung diSm BD velAC=>K E00'
3
Trang 5· '
Vi OA = OM,0' A = O'M nen 00' la trung tnrc cua AM
~ M -thuQcdirong tron tam K ban kinh KA, hay dirong tron duong kinh AC
0.5
V~y khi B, D thay d6i, M luon n~m tren dirongtron duong kinh AC
4)
<;:::;> a3 +6a~6 = 0