学 海 无 涯 例：当a?0时，如a?6则a?6?6，故此时a的绝对值是它本身 当a?0时，a?0，故此时a的绝对值是零 当a?0时，如a??6则a??6?6??(?6)，故此时a的绝对值是它的相反数 ?综合起来一个数的绝对值要分三种情况，即 当a?0?a?a??0当a?0 ??a当a?0?这种分析方法涌透了数学的分类讨论思想． 问：（1）请仿照例中的分类讨论的方法，分析二次根式a2的各种展开的情况． （2）猜想a2与a的大小关系． 25．（9分）如图，在△ABC中?ACB?90，D是AB的中点，以DC为直径的O交△ABC的三边，交点分别是G，F，E点．GE，CD的交点为M，且ME?46，MD:CO?2:5． （1）求证：?GEF??A． （2）求O的直径CD的长． （3）若cos?B?0.6，以C为坐标原点，CA，CB所在的直线分别为X轴和Y轴，建立平面直角坐标系，求直线AB的函数表达式． B G F M O C E 第25题图 A D 学 海 无 涯 2008年凉山州初中毕业、高中阶段招生统一考试 数学参考答案及评分标准 一、选择题（每小题3分，共30分） 1~5：CDCBC 6~10：BBADA 二、填空题（每小题3分，共12分） 11．a(b?a)或a(a?b) 2212．甲 13．x??4 14．47 三、（15题18分，16、17各6分，共30分） 15．解答下列各题（每小题6分，共18分） （1）计算： ?1?解：?2?(tan60?1)?3?????(??)0?2?3 ?2?2?2??4?(3?1)3?4?1?2?3 ····································································· 3分 ??4?3?3?4?1?2?3 ··········································································· 5分 ············································································································ 6分 ?2 ·（2）解： 1?x2?1? ?1????x?2?2x?4x?1x2?1?? ···························································································· 1分 x?22x?4?x?12(x?2)? ···················································································· 3分 x?2(x?1)(x?1)2 ········································································································ 4分 x?121当x?3时，原式?············································· 6分 ? ·3?12① ?（3）解：①众数为9，中位数为8 ······································· 2分 ②平均分?25% 5?10?8?9?4?8?3?7?8.75分 ······················ 4分 2020% 40% 第15－3题图 ③圆心角度数?(1?25％?40％?20％)?360?54 ··············· 6分 16．（6分） 顶点a 边数b 区域c 第1排从左至右为：12 18 7 ··············································· 1分 第3排从左至右为：20 30 11 ············································· 2分 第4排从左至右为：24 36 13 ············································· 3分 学 海 无 涯 规律：b?a?c?1或各种正确的等式 ································································· 6分 17．（6分） 四、（18题6分、19题6分，共12分） 18．（6分） （1）添加条件：BE?DF或?BAE??DAF或?BAF??DAE等 ······················ 1分 （2）证明： 四边形ABCD是菱形 ································································································· 2分 ?AB?AD ······································································ 3分 ?B??D ·A 在△ABE和△ADF中 C 第18题图 ······················································································ 5分 ?△ABE≌△ADF ·································································································· 6分 ?AE?AF ·注：其它合理的推理参照评分． 19．（6分） （1） 1 2 3 1 （2，1） 2 3 （2，3） （1，2） （1，3） ?AB?AD???B??D ?BE?DF?B F D （3，1） （3，2） 开始 1 2 3 1 2 3 1 3 2 ?能组成的两位数有21，31，12，32，13，23 ············································································· 3分 能组成的两位数有21，31，12，32，13，23 ························································· 5分 （2）P(学号12)?1． ························································································ 6分 6学 海 无 涯 五、（20题8分、21题8分，共16分） 20．（8分） （1）设A，B两处粮仓原有存粮x，y吨 ?x?y?450?根据题意得：??································································· 2分 3??2? ·1?x?1?y????5?5?????解得：??x?270 ?y?18035答：A，B两处粮仓原有存粮分别是270，180吨． ··············································· 3分 （2）A粮仓支援C粮仓的粮食是?270?162（吨） 2B粮仓支援C粮仓的粮食是?180?72（吨） 5······································ 4分 A，B两粮仓合计共支援C粮仓粮食为162?72?234吨 ·234?200 ?此次调拨能满足C粮仓需求． ········································································ 5分 （3）根据题意知： ?A?26，AB?180千米，?ACB?90 ························································· 6分 在Rt△ABC中， BC，?BC?ABsin?BAC?180?0.44?79.2 ······························ 7分 AB此车最多可行驶4?35?140（千米）?2?79.2 ?小王途中须加油才能安全回到B地． ······························································· 8分 sin?BAC?（若用时间比较，可参考评分） 21．（8分） ①由题意得y与x之间的函数关系式y?x?30（1≤x≤160，且x整数） ············· 2分 （不写取值范围不扣分） ②由题意得P与x之间的函数关系式P?(x?30)(1000?3x)??3x?910x?30000 ·· 4分 ③由题意得W?(?3x?910x?30000)?30?1000?310x 22??3(x?100)2?30000 ·················································································· 6分 ?当?100时，W最大?30000 ·········································································· 7分 100天?160天 ?存放100天后出售这批野生菌可获得最大利润30000元． ···································· 8分 （用抛物线的顶点坐标公式求最值可参照给分） 六、（22题3分、23题3分，共6分） 学 海 无 涯 22．（3分） 83cm2 43cm ······················································································ 3分 23．（3分）7或8 ··························································································· 3分 七、（24小题5分，25小题9分，共14分） 24．（5分） （1）写出类似例的文字描述 ············································································· 2分 当a?0?a?a2??0当a?0 ····················································································· 4分 ??a当a?0?（2）a?a ······························································································ 5分 25．（9分） （1）连接DF 2CD是圆直径，??CFD?90，即DF?BC ?ACB?90，?DF∥AC． ······································································ 1分 ······················ 2分 ??BDF??A．在O中?BDF??GEF，??GEF??A． ·（2）D是Rt△ABC斜边AB的中点，?DC?DA，??DCA??A， 又由（1）知?GEF??A，??DCA??GEF． 又?OME??EMC，?············································· 3分 △OME与△EMC相似 ·OMME2 ?ME?OM?MC ·································································· 4分 ??MEMC又ME?46，?OM?MC?(46)2?96 ································ 5分 MD:CO?2:5，?OM:MD?3:2，?OM:MC?3:8·设OM?3x，MC?8x，?3x?8x?96，?x?2 ?直径CD?10x?20． ················································································· 6分 （3）Rt△ABC斜边上中线CD?20，?AB?40 BC在Rt△ABC中cos?B?0.6?，?BC?24，?AC?32 ·························· 7分 ABB 设直线AB的函数表达式为y?kx?b， G 0)，B(0，24) 根据题意得A(32，3??0?k?b?24?k?? 解得???4 32?k?b?0???b?24F M O C E D A 3?直线AB的函数解析式为y??x?24（其他方法参照评分） ····························· 9分 4 第25题图
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