(Đ thi c 01 trang) ! "#$%! %! % #&'()*+, / 0!1*23 4+ 5( 60789:/ !" 2 3 3 2 2 : 1 (víi 0, 9) 9 3 3 3 x x x x x x x x x x + − + − − ≥ ≠ ÷ ÷ ÷ ÷ − + − − #$%! &' x !" − 4+%5( 60789:/ '() =y x *+ y m x m = + − + , /0'()* 1,22 3*4'567 4+;5( 60789:/ 8975:;<& = y x y y x y − = + + = + 4+<5( 60789:/ 5:;<& x mx + + = >&' m ?" x x x x − + − 4@<01AB&'@<01AC B x x 27'567 D 4+05(;789:/ E:<F6'G:H!IJ<KE:<F2A' I 1; !IBL,MN5N+O:<F6'G 3!I4P>QRS +C +O!IS ∈ !IBRIQ+C +OPQ ∈ PJS 3IQ4T> 'I256@ DC PT 'IQUIPVT 'IS>!P"!S>IP 4+=( 789:/ '<W > > a b b a + + + > ÷ ÷ B+O B a b > XXXXXXXXXXXXSYXXXXXXXXXXXX H(:Z (E,['@H\L N7 S%KH(]]]]]]]]]]]]]]]>^)@,]]]]]] > ? # 5 %5 -O B _x x ≥ ≠ C + − + − − ÷ ÷ ÷ ÷ − + − − − + + − − − − + = + − − 2 3 3 2 2 A= : 1 9 3 3 3 2 ( 3) ( 3) 3 3 2 2 3 : ( 3)( 3) 3 x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x − + + − − + − − − = = × + − − + − + − + − = = + + + &' x !" − !" − _ x x x x − − ⇔ = ⇔ + = ⇔ = ⇔ = + 1'` B _x x ≥ ≠ -a!" − x = > /0* =y x C9@<0 x X X X y _ b b _ *:;<&c' D*+, x m x m x m x m= + − + ⇔ − + + − = "J" m − + J " m − C [ ] b>> _ b m m m m m ∆ = − + − − = + + − + " + + > ∀ 2 ( 1) 20 0 víi m m ⇒ *:;<& C7'567 ⇒ ,22 3*4'56 ;5 <5 05 7 − = + + = + 2 2 2 2 10 5 1 1 (I) 20 3 11 1 y x y y x y de = 2 x u ≥ 0u + = + 2 10 1 y v y S7f<g − = − = = = ⇔ ⇔ ⇔ + = + = − = = 5 1 10 2 2 13 13 1 3 2 11 3 2 11 5 1 4 u v u v u u u v u v u v v -O = ⇒ = ⇔ = ± 2 1 1 1u x x -O = = ⇒ = ⇔ − + = ⇔ + = 2 2 2 10 4 4 4 10 4 0 1 1 2 y y v y y y y E24A7f$+O = ± = = 1 1; 2 hoÆc 2 x y y -a7f Cb7'JJJ 1 2 JXJJXJ 1 2 *:;<& + + = 2 2 1 0 (1)x mx C ∆ = − ' 2 1m d5:;<& C7'567 1 2 , x x & < − ∆ > ⇔ − > ⇔ > 2 1 ' 0 1 0 1 m m m \-\ C + = − = 1 2 1 2 2 (I) 1 x x m x x \ C?" − + − = − + − 2 2 2 2 4 2 4 2 1 1 2 2 1 1 2 2 ( 2012) ( 2012) 2012 2012x x x x x x x x = + − − + = + − − − + − 2 2 2 2 2 2 2 1 2 1 2 1 2 2 2 2 2 1 2 1 2 1 2 1 2 1 2 ( ) 2 2012( ) ( ) 2 2( ) 2012 ( ) 2 x x x x x x x x x x x x x x x x 7 f+ ? C ?" − − − − 2 2 2 (4 2) 2012(4 2) 2m m " − − − + − − 2 2 2 2 2 (4 2) 2.(4 2).1006 1006 1006 2m m " − − − + ≥ + 2 2 2 2 (4 2) 1006 (1006 2) -(1006 2)m ?4@<01A − − = ⇔ = ⇔ = 2 2 2 4 2 1006 0 4 1008 252m m m = ⇔ = − 6 7 6 7 m m 175:;<& C7' QC'?"X U @/ !"#$%&' GJJ PN5NJP 3!I4P S IQBSIQ4T I256@ D IQUIPVT IS>!P"!S>IP =5 C ã ã ằ DCB CAB (cùng chắn BC)= ã ã BCE CAB (góc có cạnh t ơng ứng vuông góc)= PCI256@ DC PT () *+ &,' ?hiPT C EK CD (BK CD) B là trực tâm của CDE DH CE (CH AB) CB DE tại F I2: DiPT>jI256@ DC PTKiPT 64 ã ã CED CDE = je@ ả à 1 1 D E (góc có cạnh t ơng ứng vuông góc)= PCiIPT 64I BD = BE IPUIQ"ITUIQ"TQ <'@ QT+4Q CTQVT 42OA IQUIPVT ) -./&0/-. & ?h'@ !I C ã 0 ACB 90 (góc nội tiếp chắn nửa đ ờng tròn)= 2 BH . BA = BC 7 + 4+: <'@ + 24 C BH BC BHC BFD (g-g) BH . BD = BC . BF BF BD = ~ BH.(BA+BD) = BC.(BC + BF) BH . AD = BC . CF (1) je@ C!PT k+C +Ol ả ã ã ã 2 0 D CAB (so le trong) AH AC ACH DF BD mà AHC DFB 90 = = = = ~ DBF (g-g) AH . BD = DF . AC (2) je@ AC CF ABC CDF (g -g) BC . CF = DF . AC (3) BC DF = ~ mJ+(<IS>!P"!S>IP n C + + + = + + + ữ ữ 1 1 21 3 21. 3. 21 3a b a b b a b a -O > , 0a b >o5,[Ap (B:Z + ì = 3 3 21 2 21 6 7a a a a + ì = 21 21 3 2 3 6 7b b b b cm+N D+:Z ì + + ì + ữ ữ 1 1 21 3 12 7a b a b j = = 12 7 144.7 1008 J = = 2 31 31 961 > 12 7 31 ì + + ì + ữ ữ 1 1 21 3 > 31a b a b 5 ' XXXXXXXXSYXXXXXXXXXXX 1232(45657 896-:5;<="=*1$'$62.(>$.4 ã ã DCB BCE = ã ả 2 2 D E = . 2 2 (4 2) 2.(4 2) .100 6 100 6 100 6 2m m " − − − + ≥ + 2 2 2 2 (4 2) 100 6 (100 6 2) - (100 6 2)m ?4@<01A − − = ⇔ = ⇔ = 2 2 2 4 2 100 6 0 4 100 8 252m m m = ⇔ =. = + 2 2 2 2 10 5 1 1 (I) 20 3 11 1 y x y y x y de = 2 x u ≥ 0u + = + 2 10 1 y v y S7f<g − = − = = = ⇔ ⇔ ⇔ + = + = − = = 5 1 10 2 2 13 13 1 3. v u u u v u v u v v -O = ⇒ = ⇔ = ± 2 1 1 1u x x -O = = ⇒ = ⇔ − + = ⇔ + = 2 2 2 10 4 4 4 10 4 0 1 1 2 y y v y y y y E24A7f$+O = ± = = 1 1; 2 hoÆc 2 x y y -a7f