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1、p20172018c1“p()YK 0,2(a + 1) = 4,)?a = 1.(9)na 1a = 1. . (10)18. ):(1)?CPlRu,.(2) kCP=65 20= 12, l?k = 2, . (4) l : y = 2x + 5 . (6)(2) CM CP,;,CP?. . (9)?%CP:(2,6),|CP|2=52.(11)(x + 1)2+ (y 11 2)2=5 4.(12)19.):(1)?AC?BDu:O,KBD ACqAA1 BD,AA1AC = A, BD AA1C, BD A1C,.(2)A1B1BB1C1C, A1B1 BE,q BE B1C,A1
2、B1 B1C = B1, BE A1C,.(4)qBD BE = B, A1C BDE.(6)(2)PA1C ?BDEu:O0, 4BCB1 4BCE, |CE| = 1, . (7)VCDEB=1 3O0C S4DEB=1 3CE S4CDB,.(9) O0C =63. A1O0= A1C CO0= 26 63=56 3, . (11) A1?BDE?ld =56 3. . (12)20. ):(1)?M(x0,y0),P(x,y),dKx =x0+4 2,y0 2, x0= 2x 4,y0= 2y, . (2) M3?x2+ y2= 4, (2x 4)2+ (2y)2= 4, . (4);
3、,(x 2)2+ y2= 1. . (6)(2) CR QR, |PQ|2= |CQ|2 |CR|2= |CQ|2 1,.(8)q:C?l?ldmin=|32026|32+42= 4, . (10) S4QRC=1 2|CR|PQ| =1 2p|CQ|2 1 =152. . (11)K?152.(12)21. ):(1)a = 1,f(x) =x2+ x + 1,x 0,x2 x + 1,x 0, . (2) f(x)?NOm:(12,0),(1 2,+).(4) (2)?x 1,2,f(x) = ax2 x + 2a 1. . (5)?ea = 0,f(x) = x 131,2N4,g(a)
4、 = f(2) = 3, . (6)p()o211?a 0,f(x)3m1,2,g(a) = f(2) = 6a 3,.(7)?0 a 1 4,f(x)3m1,2N4,g(a) = f(2) = 6a 3, . (8) ?14 a 1 2,g(a) = f(1 2a) = 2a 1 4a 1,.(10)n,g(a) =6a 3,a 1 4,2a 1 4a 1,1 4 a 1 2.(12)22. ):(1)dK?f(x),1+ax x1=x1 1ax,.(2) )?a = 1.(3)u?a = 1.(4)(2)d(1)?f(x) = log1 2x+1 x1,3m(1,+)S4O. . (5) y:-g(x) =x+1 x1= 1 +2 x1,duy =2 x13m(1,+)S4.g(x)3 m(1,+)SN4, . (7)?f(x)3m(1,+)4O. . (8)(3)-h(x) = f(x) (12)x,Kd(2)?h(x)33,4N4O,.(10)g(x)min= 98,.(11) m 98.(12)p()o212
