2024北京市大兴区初三一模数学（图片版含答案）

大兴区 2023～2024 学年度第二学期初三期中检测 数学参考答案及评分标准 一、选择题（共 16分，每题 2分） 题号 1 2 3 4 5 6 7 8 答案 D B C B C D A D 二、填空题（共 16分，每题 2分） 题 9 10 11 12 13 14 15 16 号 答 x≥3 a (x + 2) (x 2) x = 1 -5 45 1 240 60，30 案 三、解答题（共 68 分，第 17-20 题，每题 5 分，第 21 题 6 分，第 22-23 题，每题 5 分，第 24-26 题，每题 6分，第 27，28题，每题 7分）解答应写出文字说明、演算步骤或证明过程. 2 17．解：原式= 3 + 1 + 2 2 2 ························································· 4 分 2 = 4 + 2 ． ··········································································· 5 分 4x-1≥2x+5,① 18. 解： 2x-1 ＜x. ② 3 解不等式①，得 x≥3. ································································· 2 分 解不等式②，得 x＞-1. ······························································· 4 分 所以不等式组的解集为 x≥3. ························································ 5 分 19.解： (a+1)2 + a(a+ 4) 2 = a2 + 2a +1+ a2 + 4a 2 ··························································· 2 分 = 2a2 + 6a 1 . ·········································································· 3 分 ∵a 2 +3a 1= 0 ， a2∴ +3a =1 . ············································································· 4 分 ∴2a 2 + 6a = 2 . 2 ∴原式 = 2a + 6a 1 =2-1 =1. ················································································ 5 分 初三数学参考答案及评分标准 第1页（共 8 页） 20.解：设每本 A 书籍厚度为 x cm，桌子高度为 y cm. ····································· 1 分 由题意可得 3x + y = 79, 6 ····································································· 3 分 5 x + y = 82. 5 x = 1, 解得 ············································································· 4 分 y = 76. 答：每本 A 书籍厚度为 1cm. ···································································· 5 分 21. (1)证明：∵四边形 ABCD 是正方形， ∴AD∥BC，AD=BC. …………………………1 分 ∵BE=FD， ∴AD-FD=BC-BE. 即 AF=CE . …………………………2 分 又∵AF∥CE， ∴ 四边形 AECF 是平行四边形. ……………………………………3 分 (2)解：∵四边形 ABCD 是正方形， ∴AB∥CD，∠BCD=∠D=90°，AD=CD. ………………………… ... ...