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De thi vao lop 10

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  • 1. http://w ebdethi.net http://webdethi.neth h GIAÛI ÑEÀ THI VAØO LÔÙP 10 MOÂN TOAÙN CHUNG TRÖØÔØNG THPT CHUYEÂN LEÂ QUYÙ ÑOÂN BÌNH ÑÒNH NAÊM HOÏC 2008 – 2009 – Ngaøy: 17/06/2008 Thôøi gian laøm baøi: 150 phuùt http://webdethi.net Caâu 1.(1 ñieåm) Ruùt goïn: A = a a 1 a a 1 a a a a      (a > 0, a  1) =         3 3 a 1 a 1 a a 1 a a 1 a aa a 1 a a 1            = a a 1 a a 1 2 a 2 a a        (a > 0, a  1) Caâu 2.(2 ñieåm) a) Haøm soá y =  1 3 x – 1 ñoàng bieán treân R vì coù heä soá a =  1 3 < 0. b) Khi x = 1 3 thì y =   1 3 1 3 1   = 1 – 3 – 1 = - 3. Caâu 3.(3 ñieåm) a) Phöông trình x2 – 4x + m + 1 = 0 Ta coù bieät soá ’ = 4 – (m + 1) = 3 – m. Ñieàu kieän ñeå phöông trình coù hai nghieäm phaân bieät laø: ’ > 0  3 – m > 0  m < 3. b) Khi m= 0 thì phöông trình ñaõ cho trôû thaønh: x2 – 4x + 1 = 0 ’ = 4 – 1 = 3 > 0 Phöông trình coù hai nghieäm phaân bieät: x1 = 2 - 3 , x2 = 2 + 3 . Caâu 4.(3 ñieåm) A N B M C P O 1 2 2 1 1 2 2 1 1 2
  • 2. http://w ebdethi.net http://webdethi.neth h a) Chöùng minh O laø taâm ñöôøng troøn ngoaïi tieáp MNP Ta coù: O laø giao ñieåm ba ñöôøng phaân giaùc cuûa ABC neân töø ñieàu kieän giaû thieát suy ra: OBM = OMN (c.g.c)  OM = ON (1) OCM = OCP (c.g.c)  OM = OP (2) Töø (1), (2) suy ra OM = ON = OP. Vaäy O laø taâm ñöôøng troøn ngoaïi tieáp MNP. b) Chöùng minh töù giaùc ANOP noäi tieáp Ta coù OBM = OMN  1 1 M N , OCM = OCP  2 2 P M Maët khaùc 0 1 2 1 2 P P 180 M M    (keà buø)  1 1 P M  1 1 P N Vì 1 2 N N = 1800 neân 1 2 P N = 1800 . Vaây töù giaùc ANOP noäi tieáp ñöôøng troøn. Caâu 5. (1 ñieåm) Chöùng minh tam giaùc ñeàu Ta coù: 2x2 + 3y2 + 2z2 – 4xy + 2xz – 20 = 0 (1) Vì x, y, z  N* neân töø (1) suy ra y laø soá chaün. Ñaët y = 2k (k  N* ), thay vaøo (1): 2x2 + 12k2 + 2z2 – 8xk + 2xz – 20 = 0  x2 + 6k2 + z2 – 4xk + xz – 10 = 0  x2 – x(4k – z) + (6k2 + z2 – 10) = 0 (2) Xem (2) laø phöông trình baäc hai theo aån x. Ta coù:  = (4k – z)2 – 4(6k2 + z2 – 10) = 16k2 – 8kz + z2 – 24k2 – 4z2 + 40 = = - 8k2 – 8kz – 3z2 + 40 Neáu k  2, thì do z  1 suy ra  < 0: phöông trình (2) voâ nghieäm. Do ñoù k = 1, suy ra y = 2. Thay k = 1 vaøo bieät thöùc :  = - 8 – 8z – 3z2 + 40 = - 3z2 – 8z + 32 Neáu z  3 thì  < 0: phöông trình (2) voâ nghieäm. Do ñoù z = 1, hoaëc 2. Neâu z = 1 thì  = - 3 – 8 + 32 = 21: khoâng chính phöông, suy ra phöông trình (2) khoâng coù nghieäm nguyeân. Do ñoù z = 2. Thay z = 2, k = 1 vaøo phöông trình (2): x2 – 2x + (6 + 4 – 10) = 0  x2 – 2x = 0  x(x – 2) = 0  x = 2 (x > 0) Suy ra x = y = z = 2. Vaäy tam giaùc ñaõ cho laø tam giaùc ñeàu.