（2）解：去分母得：2?x?1??x?3 ·············································································· 1分 解得x??1 ········································································································ 2分 检验x??1是原方程的解 ················································································· 3分 所以，原方程的解为x??1 ············································································· 4分 19．（本小题满分7分）
（1）证明：∵四边形ABCD是平行四边形，
∴AD?BC，AD∥BC． ∴∠ADE?∠FBC ······················································································ 1分 在△ADE和△CBF中，
∵AD?BC，∠ADE?∠FBC，DE?BF ∴△ADE≌△CBF ···················································································· 2分 ∴AE?CF ····································································································· 3分
A
A D
F O E C D E B C B
（第19题图 ①） （第19题图②）
（2）解：∵DE是?O的直径
∴∠DBE?90? ······························································································ 1分 ∵∠ABD?30?
∴∠EBO?∠DBE?∠ABD?90??30??60? ········································· 2分 ∵AC是?O的切线
∴∠CAO?90? ······························································································ 3分 又∠AOC?2∠ABD?60?
∴∠C?180??∠AOC?∠CAO?180??60??90??30? ······················· 4分
20．（本小题满分8分） 解：（1）k为负数的概率是 （2）画树状图
第一次
第二次
或用列表法： 第二次 第一次 2 ··································································································· 3分 3开始
?1 ?2?23 ?1 3
3?2 1
?1 （?2，?1） （3，?1） ?2 （?1，?2） （3，?2） 3 （?1，3） （?2，3） ?1 ?2 3 ·········································································· 5分
26
共有6种情况，其中满足一次函数y?kx?b经过第二、三、四象限，
即k?0，b?0的情况有2种 ··························································································· 6分 所以一次函数y?kx?b经过第二、三、四象限的概率为
21? ··································· 8分 6321．（本小题满分8分）
解：（1）设职工的月基本保障工资为x元，销售每件产品的奖励金额为y元 ················· 1分
由题意得??x?200y?1800 ·························································································· 3分
?x?180y?1700?x?800 ······························································································ 4分
y?5?解这个方程组得?答：职工月基本保障工资为800元，销售每件产品的奖励金额5元. ································· 5分
（2）设该公司职工丙六月份生产z件产品 ··········································································· 6分
由题意得800?5z≥2000 ·························································································· 7分 解这个不等式得z≥240
答：该公司职工丙六月至少生产240件产品 ········································································· 8分 22．解：（1）将A?3，2?分别代入y? ∴k?6，a?kk，y?ax中，得2?，3a?2 x32 ······································································································ 2分 36
∴反比例函数的表达式为：y? ········································································· 3分
x2 正比例函数的表达式为y?x ··········································································· 4分
3y （2）观察图象，得在第一象限内，
当0?x?3时，反比例函数的值大 于正比例函数的值．
B M D A OC x ·························· 6分 （第22题图） ·
（3）BM?DM ···································································································· 7分 理由：∵S△OMB?S△OAC?1?k?3 2 ∴S矩形OBDC?S四边形OADM?S△OMB?S△OAC?3?3?6?12
OB?12 即OC? ∵OC?3 ∴OB?4 ················································································································ 8分 即n?4
27
63? n2333∴MB?，MD?3??
222∴MB?MD ··········································································································· 9分
∴m?23.（本小题满分9分） 解：（1）如图①，过A、D分别作AK?BC于K，DH?BC于H，则四边形ADHK是矩形
∴KH?AD?3． ······································································································ 1分
在Rt△ABK中，AK?AB?sin45??42．2?4 2BK?AB?cos45??42?2········································································· 2分 ?4 ·
2在Rt△CDH中，由勾股定理得，HC?52?42?3
∴BC?BK?KH?HC?4?3?3?10 ······························································ 3分 A D A D
N
B C B C K H G M
（第23题图①） （第23题图②） ADGB（2）如图②，过D作DG∥AB交BC于G点，则四边形是平行四边形
∵MN∥AB ∴MN∥DG ∴BG?AD?3 ∴GC?10?3?7 ·································································································· 4分 由题意知，当M、N运动到t秒时，CN?t，CM?10?2t． ∵DG∥MN
∴∠NMC?∠DGC 又∠C?∠C
∴△MNC∽△GDC
CNCM? ·········································································································· 5分 CDCGt10?2t即? 5750解得，t? ·········································································································· 6分
17∴
（3）分三种情况讨论：
①当NC?MC时，如图③，即t?10?2t
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∴t?10 ·················································································································· 7分 3A D A D
N N
B B C M H E M
（第23题图④） （第23题图③）
②当MN?NC时，如图④，过N作NE?MC于E 解法一：
由等腰三角形三线合一性质得EC?C
11MC??10?2t??5?t 22EC5?t?在Rt△CEN中，cosc? NCtCH3? 又在Rt△DHC中，cosc?CD55?t3? ∴t525解得t? ·············································································································· 8分
8解法二：
∵∠C?∠C，?DHC??NEC?90? ∴△NEC∽△DHC
NCEC? DCHCt5?t即? 5325∴t? ·················································································································· 8分
811③当MN?MC时，如图⑤，过M作MF?CN于F点.FC?NC?t
22∴
解法一：（方法同②中解法一）
1tFC3cosC??2?
MC10?2t560解得t?
17解法二：
∵∠C?∠C，?MFC??DHC?90? ∴△MFC∽△DHC ∴
B
A D
N F C
H M
（第23题图⑤）
FCMC? HCDC29
1t10?2t即2?
3560∴t?
17102560综上所述，当t?、t?或t?时，△MNC为等腰三角形 ···················· 9分
817324.（本小题满分9分）
?b?2a?1??解：（1）由题意得?9a?3b?c?0 ············································································· 2分
????c??22?a??3?4?解得?b?
3??c??2??224x?x?2 ······························································ 3分 33（2）连结AC、BC.因为BC的长度一定，所以△PBC周长最小，就是使
PC?PB最小.B点关于对称轴的对称点是A点，AC与对称轴x??1的交点即为所求的点P.
∴此抛物线的解析式为y?设直线AC的表达式为y?kx?b
y E A P C O B D x ??3k?b?0，则?
b??2? ···························································· 4分
2??k??解得?3
??b??2∴此直线的表达式为y??把x??1代入得y??∴P点的坐标为??1，?（第24题图）
2x?2． ········································································ 5分 34 3??4?···················································································· 6分 ? ·3?（3）S存在最大值 ································································································· 7分
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