Tài liệu đề thi học sinh giỏi toán lớp 9 p6

®Ò thi häc sinh giái cÊp huyÖn Trêng THCS Yªn trung M«n: To¸n (Thêi gian lµm bµi: 150 phót) Líp 9 §Ò bµi C©u1: (4.0 ®iÓm) Cho biÓu thøc A=  x x 1   x 1   x 1   : x 1    x    x 1  x a) T×m §KX§ cña A. Rót gän A b) T×m gi¸ trÞ cña x ®Ó A = 3 Câu 2: (5.0 điểm) Trªn mÆt ph¼ng täa ®é cho c¸c ®êng th¼ng (d): 3x – 2y + 3 = 0 vµ (d') : 3x + 2y – 9 = 0 c¾t nhau t¹i C vµ lÇn lît c¾t trôc Ox t¹i A, B. a) T×m täa ®é cña c¸c ®iÓm A, B, C. b) T×m diÖn tÝch vµ chu vi cña tam gi¸c ABC biÕt ®¬n vÞ ®o ®é dµi trªn c¸c trôc lµ cm. C©u 3:(4.0 ®iÓm). a) Cho biÓu thøc : M  x 2  5 x  y 2  xy  4 y  2014 . Víi gi¸ trÞ nµo cña x, y th× M ®¹t gi¸ trÞ nhá nhÊt ? T×m gi¸ trÞ nhá nhÊt ®ã b) Gi¶i hÖ ph¬ng tr×nh :  x 2  y 2  x  y  18    x  x  1 . y  y  1  72 C©u 4 (5.5®): Cho tam gi¸c ABC. Ph©n gi¸c AD (D  BC) vÏ ®êng trßn t©m O qua A vµ D ®ång thêi tiÕp xóc víi BC t¹i D. §êng trßn nµy c¾t AB vµ AC lÇn lît t¹i E vµ F. Chøng minh a) EF // BC b) C¸c tam gi¸c AED vµ ADC; AFD vµ ABD lµ c¸c tam gi¸c ®ång d¹ng. c) AE.AC = AF.AB = AD2 C©u 5 (1,5 ®iÓm).Cho a, b lµ c¸c sè thùc d¬ng. Chøng minh r»ng :  a  b 2  ab  2a b  2b a 2 §¸p ¸n vµ híng dÉn chÊm C©u1: (4®iÓm) a)  §KX§: x > 0 vµ x  1  Ta cã: A=  x x 1   x 1   (0.5®) x 1   : x 1    x    x 1  x  ( x  1)( x  x  1)    ( x  1)( x  1)  = x 1   : x 1    x( x  1) x 1    x 1  x (0.5®)  x  x 1    x 1  = = x =  2 x x => x 1 x 1 x 1 x 2 x 1 : (0.5®) x : (0.5®) x 1 x = x 1  x 2 x 1  x 1 x (0.75®) 2 x x = b) A = 3 x 1   x  x  x  :   x 1  x 1    =3 (0.5®) => 3x + x -2=0 (0.25) => x = 2/3 (0,5®) Câu 2: (5,0 điểm)  C lµ giao ®iÓm cña d vµ d/ nªn täa ®é cña C tháa m·n hÖ :  2y  3x  3   2y  9 - 3x   2y  3x  3   4y  12 x  1 y  3  VËy C(1 ; 3) (1.0®)  Ph¬ng tr×nh trôc Ox lµ y = 0 nªn täa ®é A tháa m·n hÖ :  2y  3x  3   y0 x  - 1  y  0 VËy A(- 1; 0) (0.5®) x  3  y  0  VËy B(3 ; 0) (0.5®) täa ®é B tháa m·n hÖ :  2y  9 - 3x   y0   Gäi H lµ h×nh chiÕu cña C trªn trôc Ox th× CH lµ ®êng cao cña tam gi¸c CAB vµ CH = 3 cm ( tung ®é cña ®iÓm C) ; c¹nh ®¸y AB = AO + OB = 1 + 3 = 4 (cm) . y y= 3 3x+3 2 C 1 1  dt(ABC) = 2 AB.CH = 2 .4.3 = 6 (cm2) (1.5®)  HA = HO + OA = 1 + 1 = 2 (cm)  HB = AB - AH = 2 (cm)  HA = HB = 2(cm)  tam gi¸c CAB c©n t¹i C (CH võa lµ ®êng cao võa lµ trung tuyÕn) ; tam gi¸c vu«ng HCA cã : 2 2 2 2 CA  AH  HC  2  3  13 (cm)  chu vi ABC lµ : AB + BC + CA = 4  2 13 (cm) (1.5®) C©u 3: (4.0 ®iÓm) Ta cã : -1 1 A O H x 3 B y= 9-3x 2     M  x 2  4 x  4  y 2  2 y  1   xy  x  2 y  2   2007 (0,25®) M   x  2    y  1   x  2   y  1  2007  (0,5®) 2 2 2 1 2   3  M   x  2    y  1    y  1  2007 2   4 (0,25®) 2 Do  y  1 2  0 vµ  x  2   1  y  1   0 x, y   2   (0,25®)  M (0,25®). (0,5®). 2007  M min  2007  x  2; y  1  u  x  x  1   v  y  y  1  §Æt :  (0,25®).  u  v  18  u ; v lµ nghiÖm cña ph¬ng tr×nh :  uv  72 Ta cã :  (0.25®) (0,5®). X 2  18 X  72  0  X 1  12; X 2  6  u  12 ;  u  6    v  6  v  12   x  x  1  12    y  y  1  6 (0,25®).  x  x  1  6 ;   (0,25®).  y  y  1  12 Gi¶i hai hÖ trªn ta ®îc : NghiÖm cña hÖ lµ : (3 ; 2) ; (-4 ; 2) ; (3 ; -3) ; (-4 ; -3) vµ c¸c ho¸n vÞ. 1 � � � C©u 4: (5.5 ®iÓm) a) EAD  EFD( sd ED) (0,5®) 2 1 � � � FAD  FDC( sd FD ) 2 � � � � mµ EDA  FAD  EFD  FDC (0,5®) (0,5®) (0,5®)  EF // BC (2 gãc so le trong b»ng nhau) � � b) AD lµ ph©n gi¸c gãc BAC nªn DE  DF 1 (0,5®). A 1 � � � � � s® ACD  s®( AED  DF ) = s® AE = s® ADE (0.5®) 2 2 � � � � do ®ã ACD  ADE vµ EAD  DAC E  DADC (g.g) (0,5®) T¬ng tù: 1 � 1 � � � s® ADF  sd AF  sd ( AFD  DF ) 2 2 1 � � � = (sd AFD  DE )  sd ABD 2 � �  ADF  ABD do ®ã AFD ~ d(g.g) c) Theo trªn: + AED ~ DB F B (0.25®) (0.25®) (0,5®) D C AE AD  hay AD2 = AE.AC (1) AD AC AD AF  + ADF ~ ABD  AB AD   AD2 = AB.AF (2) Tõ (1) vµ (2) ta cã AD2 = AE.AC = AB.AF 2 2 C©u 5: (1,5 ®iÓm) Ta cã :  a  1   0;  b  1   0     2 2   (0,5®) (0.25®) (0.25®) (0,5®)  a,b>0 1 1  0; b  b   0 4 4 1 1  ( a  a  )  (b  b  )  0  a , b > 0 4 4 1  ab  a  b  0 2  a a  MÆt kh¸c a  b  2 ab  0   a  b 2 2  (0,25®) (0,25®) (0,25®) 1 Nh©n tõng vÕ ta cã :  a  b   a  b     2 ab  a  b     (0,25®)  a  b 2  2a b  2b a (0,25®) (0,25®)