广西钦州市

26．（本题满分10分）

32

x＋bx＋c与坐标轴交于A、B、C三点， A点的坐标43为（－1，0），过点C的直线y＝x－3与x轴交于点Q，点P是线段BC上的一个动点，

4t如图，已知抛物线y＝

过P作PH⊥OB于点H．若PB＝5t，且0＜t＜1．

（1）填空：点C的坐标是_▲_，b＝_▲_，c＝_▲_； （2）求线段QH的长（用含t的式子表示）；

（3）依点P的变化，是否存在t的值，使以P、H、Q为顶点的三角形与△COQ相似？若存在，求出所有t的值；若不存在，说明理由．

yQ AOBxP

C

附加题：（本题满分10分，每小题5分）

请你把上面的解答再认真地检查一遍，别留下什么遗憾，并估算一下成绩是否达到了80分，如果你的全卷得分低于80分，则本题的得分将计入全卷总分，但计入后全卷总分最多不超过80分；如果你全卷得分已经达到或超过80分，则本题的得分不计入全卷总分．

（1）计算2 3的结果是_▲_；

（2）一组数据1，2，3，它的平均数是_▲_．

祝贺你，你真棒!但还是请你再检查一遍！ H钦州市2009年初中毕业升学考试参考答案及评分标准 数 学

评卷说明：

1．填空题和选择题中的每小题，只有满分和零分两个评分档，不给中间分．

2．解答题每小题的解答中所对应的分数，是指考生正确解答到该步骤所应得的累计分数．考生的其他解法，请参照评分意见进行评分．

3．如果考生在解答的中间过程出现计算错误，但并没有改变试题的实质和难度，其后续部分酌情给分，但最多不超过正确解答分数的一半；若出现严重的逻辑错误，后续部分就不再给分．

一、填空题：（每小题2分，共24分） 1．a（a＋2）

2．60

3．3.17×105 8．－2

4．主视图 9．4

5．黄

n6．90

7．y＝kx＋2（k＞0即可）

a3n?110．(?1)

n二、选择题：（每小题3分，共24分）

题号 答案 11 B 12 C 13 D 14 B 15 B 16 A 17 A 18 B 三、解答题：（本大题共8小题，共76分．解答应写出文字说明或演算步骤）

19．解：（1）去分母，移项，得 x＜3．····························································· 3分

这个不等式的解集在数轴上表示如下：

?03 ·································································· 5分

（2）两边都乘以x＋1，得

2＝x＋1． ··················································································· 7分 移项，合并同类项，得 x＝1． ························································································ 8分 当x＝1时， x＋1＝2≠0， ····························································· 9分 ∴原方程的根是：x＝1． ····························································· 10分

20．解：（1）∵b≠0时，∴b＞0或b＜0． ························································· 1分

当b＞0时，1＋b＞1， ·································································· 3分 当b＜0时，1＋b＜1； ·································································· 5分

2aa2?1（2）原式＝ ····································································· 6分 ?a?1a＝

2a(a?1)(a?1) ···························································· 7分 ?a?1a＝2（a－1）． ······································································· 8分 ∵a＝7＋1， ∴原式＝2（a－1）

＝2（7＋1－1） ······························································· 9分

＝27≈5.29． ······························································· 10分

21．（1）证明：∵AF＝BE，EF＝EF，∴AE＝BF． ···················· 1分

∵四边形ABCD是矩形，

∴∠A＝∠B＝90°，AD＝BC． ···························· 3分 ∴△DAE≌△CBF． ········································· 4分 ∴DE＝CF； ··················································· 5分

（2）解：过点O1作O1C⊥AB，垂足为C，

则有AC＝BC． ·················································· 6分 由A（1，0）、B（5，0），得AB＝4，∴AC＝2． ··· 7分 在Rt△AO1C中，∵O1的纵坐标为5，

∴O1C＝5． ··················································· 9分

AOA A F E B C D yO1O C B xB图2 ∴⊙O1的半径O1A＝O1C2?AC2?(5)2?22＝3． ······················· 10分

22．解：（1）树状图为：

小王 小李 正面 正面 反面 开始 反面 正面 反面 小林 正面 反面 正面 反面 正面 反面 正面 反面 不确确确确确确不结果 确定定定定定定确定定 （答对一组得1分）； ··································································· 4分 （2）由（1）中的树状图可知：

P（一个回合能确定两人先上场）＝

23．解：（1）地面总面积为：（6x＋2y＋18）m2； ·············································· 4分

?6x?2y?21,（2）由题意，得? ················································ 6分

6x?2y?18?15?2y.??x?4,?解之，得?········································································ 8分 3 ·

y?.??2∴地面总面积为：6x＋2y＋18＝6×4＋2×∵铺1m2地砖的平均费用为80元，

∴铺地砖的总费用为：45×80＝3600（元）． ···································· 10分

63＝． ····································· 8分 843＋18＝45（m2）． ············· 9分 2 24．解：（1）根据题意，2009年一季度全区生产总值为1552.38亿元，

设2008年一季度全区生产总值为x亿元，则

1552.38-x＝12.9%． ········ 2分 x解之，得x≈1375.00（亿元）． ······················································ 3分 答：2008年一季度全区生产总值约是1375.00亿元； ··························· 4分 （2）能推算出2007年一季度全区生产总值． ············································ 5分

设2007年一季度全区生产总值为y亿元，同理，由（1）得

1375.00-y＝11.3%． ··································································· 6分 y解之，得y≈1235.40（亿元）．

所以2007年一季度全区生产总值约是1235.40亿元； ·························· 7分 （3）近三年广西区生产总值均为正增长；2008年1季度增长率较2007年同期增长

率有较大幅度下降；2009年1季度增长率较2008年同期增长率有所上升，经济发展有所回暖；2007年广西经济飞速发展；…．等等，只要能有自己的观点即可给分． ·············································································· 8分

25．解：（1）∵∠ABC＝90°，

∴OB⊥BC． ·················································· 1分 ∵OB是⊙O的半径，

∴CB为⊙O的切线． ······································· 2分 又∵CD切⊙O于点D，

∴BC＝CD； ·················································· 3分 （2）∵BE是⊙O的直径，

∴∠BDE＝90°．

∴∠ADE＋∠CDB ＝90°． ······························· 4分 又∵∠ABC＝90°，

∴∠ABD＋∠CBD＝90°． ······························································· 5分 由（1）得BC＝CD，∴∠CDB ＝∠CBD．

∴∠ADE＝∠ABD； ······································································ 6分 （3）由（2）得，∠ADE＝∠ABD，∠A＝∠A．

∴△ADE∽△ABD． ······································································ 7分 ∴∴

CDAEO?BADAE＝． ············································································ 8分 ABAD21＝，∴BE＝3，······························································ 9分 1?BE2∴所求⊙O的直径长为3． ·························································· 10分

26．解：（1）（0，－3），b＝－

9，c＝－3． ··················································· 3分 439（2）由（1），得y＝x2－x－3，它与x轴交于A，B两点，得B（4，0）．

44y ∴OB＝4，又∵OC＝3，∴BC＝5．

QH由题意，得△BHP∽△BOC，

AOBx∵OC∶OB∶BC＝3∶4∶5， P∴HP∶HB∶BP＝3∶4∶5， ∵PB＝5t，∴HB＝4t，HP＝3t． C∴OH＝OB－HB＝4－4t．

3由y＝x－3与x轴交于点Q，得Q（4t，0）．

4t∴OQ＝4t． ·············································································· 4分 ①当H在Q、B之间时， QH＝OH－OQ

＝（4－4t）－4t＝4－8t． ······················································· 5分 ②当H在O、Q之间时， QH＝OQ－OH

＝4t－（4－4t）＝8t－4． ······················································· 6分 综合①，②得QH＝｜4－8t｜； ······················································ 6分 （3）存在t的值，使以P、H、Q为顶点的三角形与△COQ相似． ················ 7分

①当H在Q、B之间时，QH＝4－8t，

4?8t3t若△QHP∽△COQ，则QH∶CO＝HP∶OQ，得＝，

34t7∴t＝． ··············································································· 7分

323t4?8t若△PHQ∽△COQ，则PH∶CO＝HQ∶OQ，得＝，

4t3即t2＋2t－1＝0．

∴t1＝2－1，t2＝－2－1（舍去）． ········································· 8分 ②当H在O、Q之间时，QH＝8t－4．

8t?43t若△QHP∽△COQ，则QH∶CO＝HP∶OQ，得＝，

34t25∴t＝． ··············································································· 9分

323t8t?4若△PHQ∽△COQ，则PH∶CO＝HQ∶OQ，得＝，

4t3即t2－2t＋1＝0．

∴t1＝t2＝1（舍去）． ······························································ 10分

725综上所述，存在t的值，t1＝2－1，t2＝，t3＝． ···················· 10分

3232附加题：解：（1）8； ···················································································· 5分

（2）2． ···················································································· 10分