... C = ( -2 ; -1 0) hc C = (1 ; -1 ) VI.a 0 ,25 0 ,25 Gäi G(t;3t-8) lµ träng t©m tam gi¸c ABC th× d(G, AB)= 0 ,25 0,5 0 ,25 0 ,25 ( ⇔ 2+ Bpt ( t = 2+ ) x2 − x ) x2 − x ( + 2 ) x2 − x BPTTT : (t > 0) 0 ,25 ... y>0 VIb (1 ) ⇔ 22 log3 xy − 2log3 xy − = 0,5 0 ,25 x 2 (2 ) ⇔ log 4(4 x +4y ) = log 4(2 x +6xy) ⇔ x2+ 2y2 = ⇔log3xy = ⇔ xy = 3⇔y= Kết hợp (1 ), (2 ) ta nghiệm hệ: ( ; ) ( ; S M A D H B C ) 0 ,25 ... VP 2 DÊu ®¼ng thøc x¶y a=b=c=1/3 V.a 5 , trung ®iĨm M ( ; − ), 2 pt (AB): x – y – = 3 S ∆ABC = d(C, AB).AB = ⇒ d(C, AB)= 2 Ta cã: AB = ⇒ d(G, AB)= t − (3 t − 8) − ⇒ G(1; - 5) hc G (2 ; - 2) = 0 ,25 ...