1 ··································································································································· 5分 x?2213········································································ 7分 ?当x??时，原式的值为? ·
23??24311122． 解：摸到绿球的概率为：1??? ···································································· 1分
3261则袋中原有三种球共 3??18 （个） ············································································· 3分
61所以袋中原有红球 ··········································································· 5分 ?18?6 （个） ·
31袋中原有黄球 ·················································································· 7分 ?18?9 （个） ·
2??y?x,?x1?1?x2??1?,?23．解：（1）解方程组?得， ··············· 2分 ?1y?1y??1y??1?2?x?所以A、B两点的坐标分别为：A（1，1）、B（－1，－1） ······· 4分
（2）根据图象知，当?1?x?0或x?1时，正比例函数值大于反比例函数值 24． 证明：（1）?四边形ABCD和四边形DEFG都是正方形
7分
?AD?CD,DE?DG,?ADC??EDG?90?,
······················· 3分 ??ADE??CDG,?△ADE≌△CDG，······································································· 4分 ?AE?CG ·
（2）由（1）得 ?ADE??CDG,??DAE??DCG,又?ANM??CND，
?ANMN?，即AN?DN?CN?MNCNDN ······················································· 7分
∴?AMN∽?CDN ····································································································· 6分 25解：（1）由平移的性质得
AF//BC且AF?BC，△EFA≌△ABC，?四边形AFBC为平行四边形，?S?EFA?S?BAF?S?ABC?3，
?四边形EFBC的面积为9． ······························································································ 3分
（2）BE?AF．证明如下：由（1）知四边形AFBC为平行四边形
?BF//AC且BF?AC，又AE?CA，?BF//AE且BF?AE，?四边形EFBA为平行四边形又已知AB?AC，?AB?AE，
?平行四边形EFBA为菱形，?BE?AF ········································································· 5分
(3)作BD?AC于D，??BEC?15?，AE?AB，??EBA??BEC?15?，??BAC?2?BEC?30?,?在Rt?BAD中,AB?2BD.设BD?x,则AC?AB?2x,?S?ABC?3,且S?ABC?11AC?BD??2x?x?x2,?x2?3,?x为正数,?x?3,?AC?23......................7分22
BE9?，?设BE?9k，AE?5k?k为正数?，AE55则在Rt?ABE中，?BEA?90?，AB?106，AB2?BE2?AE2，.....................................2分226.解:?1??i?5522?5?即?106???9k???5k?，解得k?，?BE?9??22.5?m?.22?2?故改造前坡顶与地面的距离BE的长为22.5米....................................................................4分27
FH?tan?FAH,?2?由?1?得AE?12.5,设BF?xm,作FH?AD于H,则AH22.5由题意得?tan45?,即x?10.x?12.5?坡顶B沿BC至少削进10m,才能确保安全.........................................................................7分．解： (1)因为租用甲种汽车为x辆，则租用乙种汽车?8?x?辆．
2??4x?2?8?x?≥30,由题意，得? ·················································································· 2分
??3x?8?8?x?≥20.44解之，得7?x?···································································································· 3分 . ·
5即共有两种租车方案：
第一种是租用甲种汽车7辆，乙种汽车1辆；
第二种是全部租用甲种汽车8辆 ···················································································· 5分 （2）第一种租车方案的费用为7?8000?1?6000?62000元 ·································· 6分 第二种租车方案的费用为8?8000?64000元 ······························································· 7分 所以第一种租车方案最省钱······························································································ 8分 28．解：（1）设AB的函数表达式为y?kx?b.
3??0??8k?b,?k??,∵A??8,0?,B?0,?6?,∴?∴?4
?6?b.???b??6.∴直线AB的函数表达式为y??3········································································ 3分 x?6． ·
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