(1,0 • I A = 21Bc:> l A - 2 1 8 = 0 ( I ) 0,25
diem)
. 3JA + 2JC = 0 <=> 3 ( I A - IJ) + 2(IC - IJ) = 6
» 3 I A + 2IC = 5IJ (2)
0,25
L a y (2) irir (1) theo ve ta dUdc:
0,25
2 ( I A + I B + I C ) = 5IJ 6IG = 5IJ. 0,25
Suy ra I , J, G thang hiing nen dicing thang IJ di qua trong tam
G eiia tam giac A B C . 0,25
D E SO 1 8 D E THLT SLTC HOC K l I M O N T O A N L6P 1 0
Thdi gian lam bai: 90 phut
C a u I . (1,0diem) Cho ham so r(x) = x / l - x ^ + 7 3 - x . 1. Tim tap xac djnh A eiia ham só ('(x).
2. Gia su- B = | x e !R I - 2 < X < 2 | . Hay xac djnh cac tap hcJp A n B , A .
C a u I I . (2,0 diem) Cho ham so y = - x " - 4x + 6 c6 do thj la parabol (P).
1. T i m toa do dinh va phiTdng Irinh triie doi xufng cua (P).
2. V e do thi ciia (P).
3. Difa vao do t h i , hay cho bict tap hdp cac gia trj cua x sao cho y > 0.
c a u I I I . (2,0 diem)
1. T i m ciic gia trj cíia m de phiTdng trinh x " - 2 ( m - l ) x + m ^ - 3m = Oco hai nghiem x , , x-, thoa man he ihiifc x^ + x^ = 8.
x - 1 3x - 1
2. Giiii phúctng trinh
3 - 2 x x + 1
3. G i a i phúdng trinh x(x - 1) - 27x'^ - x + 3 = 0.
C S u I V . (2,0 diem). Trong msll phiing Oxy cho cac d i e m A ( l ; - 2 ) , B ( - l ; 2) va
C( 2 7 3 ; 7 3 ) .
1. Cho biet hinh dang eiia lam giac A B C . ' 2. T i m do dai dirdng cao cua lam giiic A B C va tam d\Xl1ng Iron noi liep lam
giac A B C .
Cfiu V . (2,0 diem)
1. G i i i i he nlufttng l i i n h P ^ " '
[ ( x - l ) ( y - l ) = 18 (adsbygoogle = window.adsbygoogle || []).push({});
2. C h u a, h, c, d la cac so'diCiJiig. ChiVng m i n h rllng: a + b + c a + b + d a + c + d b + c + d
d c
C'duVl. (1,0 diem).
•> 12.
Cho hinli binh haiih A B C D c6 A B = 3a, A D = 5a va B A D = 120". 1. T i n h cac lich \ hififng A B . A D ; A C B D .
2. T i n h do d a i doan B D va ban i<inh dirr<ng Iron ngoai t i c p l a m giac A B C .
D A P A N T H A M K H A O
Can Dap an Diem
I
(1,0 diem) diem)
Ị (0,5 diem) T i m tap xac dinh A ciia h a m s6'f(x).
I
(1,0
diem) H a m so IX\) = \ / l - x " + N/B - x xac dinh ichi | ' " ^ ^ [ 3 - x > 0 0,25 I (1,0 diem) f x - < l | - l < x < I <:> < <=>< < t = > - l< X< l . | x < 3 [ x < 3
V a y lap xac dinh ciia ham so ('(x) la A = [ - 1 ; 1].
0,25
I
(1,0 diem) diem)
2. (0,5 diem) H a y xac djnh c a c tap h(/p...
I (1,0 (1,0 diem) B - ( - 2 ; 2), A n B = l - i ; 1|, A \ = 0 . 0,5 I I (2,0 diem)
1. (0,5 diem) Vim toa do dinh va phifdng trinh true dói xrfng...
I I
(2,0 diem) diem)
D i n h I ( - 2 ; lOj.PhiTctng irinh Iriic d o i xrfngx = - 2 . 0,5
I I
(2,0 diem) diem)
2. (1,0 diem) V e do thi ciia (P) (doc gia l i f v c ) .
I I
(2,0 diem) diem)
3. (0,5 diem) D i / a vao do thj, hay cho biet tap h<fp... (adsbygoogle = window.adsbygoogle || []).push({});
I I
(2,0 diem) diem)
Tir do Ihi la COy > 0 o - 2 - N / I O < x < - 2 + NAỌ 0,5
I I I
(2,0 diem) diem)
1. (1,0 diem) Tim c a c «ia trj ciia m de phif(/ng trinh...
I I I
(2,0 diem) diem)
Dc phi^^ng trinh c6 hai nghiC-m x , . X T thi
Á = m + l > ( ) o m > - l (*} 0,25 I I I (2,0 diem) X , + X T = 2(m - 1) T h c o dinh l i Vi-c:t la c6: <^ " ^ X J X T = m " - 3m 0,25 76 T h c o dc la c6: xj^ + x ; = 8 <=> ( x , + x-, i"^ - 2X | X T = 8 o 4 ( m " l ) - - 2 ( m - - 3 m ) = 8 c : > m - - m - 2 = () 0,25
<=> m = - 1 hoac m = 2 (ihoa man ( * } ) . 0,25
2. (0,5diem) G i a i phi/t/n^ trinh . . .
F)icu k i c n X
Ncu X > - 1 : PhifiJng trinh da cho lr('< thiinh X - - 1 - - 6 x - + l l x - 3 o 7x- - l l x + 2 = 0
l ± x / 6 5 < = > X -
14 (ihoa man (*)).
. N c u X < -1 : PhiÂng Irlnh da cho trc'l lhanh 1 - X - = - 6 x - + 1 Ix - 3 <=> 5x- - 1 Ix + 4 = 0
I1 ±N/ 4T
C=> X = •
10 (khong ihoa man (*)).
Vay phifinig trinh da cho c6 hai nghicMn x = 1 ± V65 14
3. (0,5 diem) G i a i phif</nfj trinh . . .
D a l I = \ / x - - X + 3 (t > 0) la c6 phifiJng Irinh: t = - 1 (loai)
1 = 3 (nhan). I - - 2 l - 3 = ( ) o
Vfn t = 3 thi - x + 3 = 3 o X " - X - 6 = 0 o
Vay phifc^ng trinh da cho c6 hai nghiC'iii x = - 2 , x = 3 x = 3 x = - 2 '
I V
(2,0 diem) diem)
1. (1,0 diem) C h o biet hinh dan>; ciia tam giac A B C .
A B = ( - 2 ; 4) 3 > A B = ^ ( - 2 ) ^ + 4 " = ^ 2 0 = 2V5. (adsbygoogle = window.adsbygoogle || []).push({});
A C = ( - I + 2 V 3 ; 2 + 7 3 ) ^ A C = 2V5.
B C = (2Ny3 + l ; V 3 - 2 ) = > B C = 2V5. V I A B = B C = C A ncn tam giac A B C d c u . V I A B = B C = C A ncn tam giac A B C d c u .
V
(2,0 diem) diem)
2. (1,0 diem) T i m do dai 6i(iinn cao cíia tarn }»iac A B C . . .
V i tarn giac A B C d c u ncn do dai difc^ng cao cua l a m giac A B C
A B V 3 2j5.^l3
la A H = = ^yl5.
Do lam giac A B C d c u nC-n loa do Irong l a m ciia l a m giac la lam dúclng iron noi l i c p lam giac A B C la c6:
+ X H + x c _ 1 - 1 + 2 7 3 _ 2 V 3 —»^ I
„ y A+ y n + yc _ -2 + 2 + ^(3 _ 7 3
y . = ^ ^ -Y
X i =
1. (1,0 diem) G i a i he phifc^n^ trinh
, X " + y" = 65 a CO < (x + y ) - - 2 x y = 65 (1) ( x - l ) ( y - l ) = 18 [xy = 17 + (x + y ) (2) Thay (2) vao (1) la c6 (x + y ) " - 2(x + y) - 9 9 = 0 o x + y = 11 x + y = - 9 • V d i X + y = 11 ihi xy = 28 . Do do x, y la n g h i c m cua phiTdng i n n h X - - l l X + 28 = ()c:>X = 4hoacX = 7.
• V('ii X + y = - 9 ihi xy = 8 . Do do x, y la n g h i c m cua phiTdng i r i n h X - + 9 X + 8 = 0 <r> X = - I hoac X = - 8 .
V a y he da cho c6 n g h i c m la (4; 7).(7; 4 ) . ( - l ; - 8 ) , ( - 8 ; - 1 ) . 2. (1,0 diem) ChuTng minh rans...
^ . a + b + c a + b + d a + c + d b + c + d Dal A = + + + A = a d f b d^ ^ • d^ a c f b c^ f a b] — + — + — + — + h — + — + — + — + — + — + — i d a ; [ d b , Vd c , [ b a j Ap dung B D T Co si la c6: 'd a V d b Vd c Vc a Vc b Vb a
Dilng IhiJc xiiy ra <=> a = b = c = d.