Bai TUYtN SINH V Ao LOP 10 THPT NAM HQC 2018 - 2019 y DiEm N~idung 1,S 1.a Bieu thuc A c6 nghia 2x -1 > ~x~2· 0,25 ½thi bi~u thuc A c6 nghia V~y x l.b 0,25 B = /3( ,./3 0,25 )= /3(313-2.2 /3 +413) 0,25 =13(313)=9 V~y B=9 1.c C +.,&_-~} a- /a · ,a+ /a-1 =( t a -1 , J r.+1 I -(_E_ a-1 - /a-1 Ja( /a-1) ( /a-1)( /a+l) Jal )( fa - 1) = /a -1 V~y C = fa -1 a -1 0,25 1,S 2.a 2.b E>~t t=x ~o 0,25 Phucmgtrinh viitl~i: t +3t-4=0 (1) Giai PT(l), ta c6 hai nghi~m t1 = (tmdk); t2 =-4 (khong tmdk:) 0,25 V6i t=l, ta duqcx= ± V~y phucmg trinh da cho c6 nghi~m la x = ± 0,25 E>ucmgthing(d)diqua A(l;-1) nen -l=m-l+n~m+n=0 (1) 0,25 E>ucmgthingdc6h~s6g6cbing -3 nen m-1=-3~m=-2 (2) 0,25 0,25 1,0 Tir(l)va(2),tac6: m=-2 va n=2 0,25 GQi x la s6 chiec n6n la ma co so d6 d\l kien lam m6i nouven ducmg) Theo d\l ki8n s6 co so d6 phai lam la oo (ngay) (x X 0,25 0,25 Th\lc te m6i lam duqc nhieu ban chiec nen theo th\lc te, s6 300 co so d6 dil lam la - - (ngay) x+ Vi co so dil hoan truoc nen ta c6 phuang trlnh: 300 _ 300 =3 x x+5 (=> oo(x+ )- 00x =3 (=> 1500=3x(x+5) (vi x nguyen duong) x(x+5) (=> 3x + 15x -1500 = Giai phuang trinh, ta duqc x = 20 (tmctk), ·x = -25 (khong tmctk) viv theo d\l ki&i ban ddu, mlii co su d6 lam 20 chi@c n6n la 4.a PT(l) c6 hai nghi~m phan bi~t (=> !! ' = m2 - ( m + m) > o Vai m < thi PT( 1) c6 hai nghi~m phan bi~t X1 + X2 = -2m va X1 X2 0,25 2,0 0,25 0,25 0,25 (=>m x= · 4.b - Xi ' X2 0,25 = m2 + m Dod6: (*)(=>(x1 -xi}2(x1 +.xi)=32(::>[(x1 +xi}2 - 4x1x ]{x1 +Xi)=32 (=> 8m2 = 32 (=> m = ±2 0,25 0,25 Ma mNAD=MND 0,25 ABC+ NM) = 9b (1) Nill+ MNfj = 90° Tir (I), (2), (3) suy ABC ==fiiJ Ma S.c • (2) • 0,25 (3) 0,25 Suy tu giac BNDI n9i ti8p (dpcm) 0,25 Theo kBt qua cau b) ta c6\.'tu giac BNDI n(>i tiBp, suy NiiB =NiB =90° " _ 0,25 DB l AB t{li diSm B nen dm'mg thAng BD cd djnh 0,25 MJt khac diSm D nim tren dm'mg phan giac AD cua g6c BAG (cd djnh) nen ducmg thAng AD cfJ dinh, suy diSm D cd djnh Vfy ducmg tr