21．解：依题意得，直线的解析式为．······················································· 2分

　　因为在直线上，

　　则．································································································ 3分

　　即．

　　又因为在的图象上，

　　可求得．························································································· 4分

　　所以反比例函数的解析式为．························································ 5分

20．解：(1)(万人)．···························································· 1分

　故从2000年到2005年北京市常住人口增加了154万人．

　(2)(万人)．

　故2005年北京市常住人口中，少儿(岁)人口约为157万人．············· 3分

　(3)例如：依数据可得，2000年受大学教育的人口比例为，2005年受大学教育的人口比例为．可知，受大学教育的人口比例明显增加，教育水平有所提高．·········································· 5分

19．解：(1)证明：如图，连结．

　因为，

　所以．

　故．································· 1分

　又，

　所以是等边三角形．

　故．··························································································· 2分

　因为，

　所以．

　所以是的切线．················································································ 3分

(2)解：因为，

　所以垂直平分．

　则．·························································································· 4分

　所以．································································································ 5分

　在中，，

　由正切定义，有．

　所以．··························································································· 6分

18．解：如图，过点作交于点．············································· 1分

　因为，

　所以四边形是平行四边形．································································· 2分

　所以．

　由，

　得．

　在中，，，

　由，

　求得．································································································ 3分

　所以．·············································································· 4分

　在中，，

　　．

　求得．·························································································· 5分

17．解：

　　··········································································· 2分

　　．····························································································· 3分

　　当时，原式．························· 5分

16．证明：因为，

则．····················································································· 1分

又，

则．····················································································· 2分

在与中，

·················································································· 3分

所以．···································································· 4分

所以．·················································································· 5分

15．解：．······················································· 2分

　　．···································································· 3分

　　　　　　．··········································································· 4分　

　　经检验是原方程的解．

　　所以原方程的解是．········································································ 5分

14．解：由不等式解得　．························································ 2分

　　由不等式解得　．····················································· 4分

　　则不等式组的解集为　．····················································· 5分

13．解：

　························································································· 4分

　．···································································································· 5分

第II卷(非机读卷　共88分)