2012年初中毕业学业考试参考答案.doc

2012年永兴场中学初中毕业学业考试模拟试卷
参考答案及评分标准
一、1．2； 2．(2)(2)x x +-； 3．6
1.8310⨯；
4．1x ≥；
5．2x <； 6．500； 7．ABC DCB ∠=∠或AC DB =均可； 8．90；
9．你；
10．5.5．
三、21．解：原式1322
=+-⨯
····················································· 3分（每个1分） 41=- 3= ····················································································· 5分 22．解：方程两边同乘以(1)x x + ······································································ 1分
得：2
212(1)x x x x ++=+ ········································································ 2分 解方程得：1x = ······················································································ 3分 检验：当1x =时，(1)0x x +≠ ··································································· 4分 所以，原分式方程的解是1x = ···································································· 5分
（1）P （石，石）9
=
··················································································· 3分 （2）P （不同手势）62
93
== ·········································································· 6分
24．答：同意．······························································································ 1分 理由如下：连结AC ，BD ························ 2分 因为在梯形ABCD 中，AD BC ∥，AB CD = 所以AC BD = ········································ 3分
又因为E F G H ，，，分别为AB BC CD
DA ，，，的中点，
所以在ABC △，ACD △，DAB △，BCD △中分别有：
12EF GH AC ==
，1
2
EH FG BD == ····· 5分 所以EF GH EH FG ===
所以四边形EFGH 是菱形． ······················ 6分
25．解：（1）2550%50÷=（人） ············ 2分 （2）如图 ·············································· 5分 （3）60040%240⨯=
26y 亩， ·
··································· 1分
根据题意有：4200800
20%30%800
x y x y +=-⎧⎨
+=⎩································································ 5分
解得：22001200x y =⎧⎨=⎩
··························································································· 9分
答：该村去年种植烟叶和蔬菜的面积各是2200亩，1200亩． ································ 10分
附加题：
解：（1）由表中信息可知点(22)，，(30)，在直线a 上，描点连线得直线a 的图象，1分 由待定系数法可求得直线a 的解析式为26y x =-+ ··············································· 3分 点(1010)-，的坐标不满足26y x =-+
所以点(1010)-，不在直线a 图象上 ···································································· 4分
（2）解方程组26y x y x =⎧⎨=-+⎩
······················ 6分
得2x y == 故点C 的坐标为(22)，
..8分 （3）当02m <≤时，2
12S m =
·············· 11分 当23m <<时，1132(3)(26)22
S m m =⨯⨯---+2
m =-分
（4）若有这样的P 点，使直线l 平分OBC △分
由于OBC △面积等于3，故当l 平分OBC △面积时，3
2
S =
213
22
m =∴ 解得m =故存在这样的P 点，使l 平分OBC △的面积．点P 的坐标为． ··················· 20分
10% 类别 其它 职高 普高。