18、
????解：?1?a?b?5,OQ?221a2?b2???????????????2?OC?OA?OB??a＝??－1,2?,??0,C点在圆O上，且?－1,2?在圆O上，??＝1，C?－1,2?，设A?x1,y1?,B?x2,y2?,?x1?x2??1,????????????????????11又OC?OA?OB，OA＝OB?AB?OC，kAB???,kOC21?设l的方程为y=x?b代入圆的方程得5x2?4bx?4b2?20?024b515?x1?x2????1,?b?,?l的方程为y=x?.542419、
5.511解：fx?fx,?gx???f?x1??f?x2???1?设g?x??f?x?????????12?1?,2?2?211g?x2???fx?fx,?gx?gx??fx?fx???????????????21?1212??0,2?4?又?g?x?也是二次函数，由二次函数的图象特征知，g?x?在?x1,x2?上必有一个根，且g?x?＝0有两个不相等的实数根。?f?x??1f?x1??f?x2??有两个不相等的实数根,且在?x1,x2?上必有一个根.???2b11122fx?fx?ax?x?b?x1?x2??c??2??x0??,f?m??am2?bm?c???????12?12??2a22211b12?am2?bm?a??x1?x2??2x1x2??b?2m?1??x0????m2?2m??x1x2?22?2a2212m?1?11???m2?2m??x1?2m?1?x1????x1??m??m??m2,?22?44?1?上式等号成立时,m?,x1?0,?x1?x2?2m?1?0?x2?0与x1?x2矛盾。2?x0?m2.
6