习题练习

[{"id":2021,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生在整理班级数学测验成绩时，发现一组数据的平均数为85分，后来发现漏记了一个成绩90分。将这个成绩加入后，新的平均数变为85.5分。请问原来这组数据共有多少个成绩？","answer":"A","explanation":"设原来有n个成绩，则原来总分是85n。加入90分后，总人数变为n+1，总分变为85n + 90，新的平均数为85.5。根据平均数公式列出方程：(85n + 90) \/ (n + 1) = 85.5。两边同乘(n + 1)得：85n + 90 = 85.5(n + 1) = 85.5n + 85.5。移项整理：85n - 85.5n = 85.5 - 90 → -0.5n = -4.5 → n = 9。因此原来有9个成绩，正确答案是A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 10:31:38","updated_at":"2026-01-09 10:31:38","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"9","is_correct":1},{"id":"B","content":"10","is_correct":0},{"id":"C","content":"11","is_correct":0},{"id":"D","content":"12","is_correct":0}]},{"id":2404,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某校八年级开展了一次数学实践活动，要求学生测量校园内一个不规则四边形花坛ABCD的边长与角度。已知AB = 5 m，BC = 12 m，CD = 9 m，DA = 8 m，且对角线AC将四边形分成两个直角三角形△ABC和△ADC，其中∠ABC = 90°，∠ADC = 90°。若一名学生想计算该花坛的面积，以下哪个选项是正确的？","answer":"A","explanation":"题目中给出四边形ABCD被对角线AC分成两个直角三角形：△ABC和△ADC，且∠ABC = 90°，∠ADC = 90°。因此，可以分别计算两个直角三角形的面积，再相加得到整个四边形的面积。\n\n在△ABC中，AB = 5 m，BC = 12 m，∠ABC = 90°，所以面积为：\n(1\/2) × AB × BC = (1\/2) × 5 × 12 = 30 m²。\n\n在△ADC中，AD = 8 m，DC = 9 m，∠ADC = 90°，所以面积为：\n(1\/2) × AD × DC = (1\/2) × 8 × 9 = 36 m²。\n\n因此，花坛总面积为：30 + 36 = 66 m²。\n\n本题综合考查了勾股定理的应用背景（直角三角形识别）、三角形面积计算以及实际问题中的几何建模能力，符合八年级学生知识水平。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 12:09:17","updated_at":"2026-01-10 12:09:17","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"66 m²","is_correct":1},{"id":"B","content":"72 m²","is_correct":0},{"id":"C","content":"78 m²","is_correct":0},{"id":"D","content":"84 m²","is_correct":0}]},{"id":2247,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学生在一次数学实践活动中，记录了一周内某城市每日的气温变化情况。规定：气温上升记为正，下降记为负。已知这七天的气温变化依次为：+3℃，-2℃，+5℃，-4℃，+1℃，-6℃，+2℃。若第一天的起始气温为-1℃，请回答以下问题：经过这七天的连续变化后，最终气温是多少摄氏度？并判断最终气温比起始气温是升高了还是降低了，变化了多少摄氏度？","answer":"最终气温是-2℃，比起始气温降低了1℃。","explanation":"本题综合考查正负数在连续变化中的加减运算，要求学生理解正负数表示相反意义的量，并能进行多步有理数加法运算。题目设置了真实情境（气温变化），避免机械计算，强调过程推理。通过逐日累加变化量，最终得出结果，并比较起始与结束状态的差异，体现了正负数在实际问题中的应用，符合七年级课程标准中‘有理数运算’与‘实际问题建模’的要求。","solution_steps":"1. 起始气温为-1℃。\n2. 第一天变化：-1 + (+3) = 2℃\n3. 第二天变化：2 + (-2) = 0℃\n4. 第三天变化：0 + (+5) = 5℃\n5. 第四天变化：5 + (-4) = 1℃\n6. 第五天变化：1 + (+1) = 2℃\n7. 第六天变化：2 + (-6) = -4℃\n8. 第七天变化：-4 + (+2) = -2℃\n9. 最终气温为-2℃。\n10. 比起始气温-1℃的变化量：-2 - (-1) = -1℃，即降低了1℃。","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 14:44:04","updated_at":"2026-01-09 14:44:04","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":516,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"72°","answer":"待完善","explanation":"解析待完善","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:20:15","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":1529,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学校组织七年级学生进行校园绿化活动，计划在矩形花坛中种植两种花卉：玫瑰和郁金香。花坛的长比宽多6米，面积为91平方米。现需在花坛四周铺设一条宽度相同的步行道，铺设后整个区域（包括花坛和步行道）的总面积为195平方米。已知铺设步行道的费用为每平方米80元，且预算不超过8000元。问：(1) 花坛原来的长和宽分别是多少米？(2) 步行道的宽度最多为多少米？（结果保留一位小数）(3) 若实际铺设时步行道宽度取最大值，总费用是否在预算范围内？请说明理由。","answer":"(1) 设花坛的宽为x米，则长为(x + 6)米。\n根据题意，花坛面积为91平方米，得方程：\nx(x + 6) = 91\nx² + 6x - 91 = 0\n解这个一元二次方程：\n判别式 Δ = 6² - 4×1×(-91) = 36 + 364 = 400\nx = [-6 ± √400] \/ 2 = [-6 ± 20] \/ 2\nx = 7 或 x = -13（舍去负值）\n所以花坛的宽为7米，长为7 + 6 = 13米。\n\n(2) 设步行道的宽度为y米。\n铺设步行道后，整个区域的长为(13 + 2y)米，宽为(7 + 2y)米。\n总面积为195平方米，得方程：\n(13 + 2y)(7 + 2y) = 195\n展开得：91 + 26y + 14y + 4y² = 195\n4y² + 40y + 91 = 195\n4y² + 40y - 104 = 0\n两边同时除以4：y² + 10y - 26 = 0\n解这个方程：\nΔ = 10² - 4×1×(-26) = 100 + 104 = 204\ny = [-10 ± √204] \/ 2 ≈ [-10 ± 14.28] \/ 2\n取正值：y ≈ (4.28) \/ 2 ≈ 2.14\n保留一位小数，步行道宽度最多为2.1米。\n\n(3) 步行道面积 = 总面积 - 花坛面积 = 195 - 91 = 104（平方米）\n总费用 = 104 × 80 = 8320（元）\n由于8320 > 8000，超出预算。\n因此，即使取最大宽度2.1米，总费用仍超过预算，不在预算范围内。","explanation":"本题综合考查了一元二次方程、面积计算、不等式思想及实际应用能力。第(1)问通过设未知数建立一元二次方程求解花坛尺寸，需注意舍去不符合实际的负解；第(2)问引入新变量表示步行道宽度，利用整体面积建立方程，解出合理范围并按要求保留小数；第(3)问结合费用计算与预算比较，体现数学建模与决策能力。题目融合了代数运算、几何图形初步和一元二次方程的应用，情境真实，思维层次丰富，符合困难难度要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 12:15:16","updated_at":"2026-01-06 12:15:16","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":214,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"一个长方形的长是8厘米，宽是5厘米，它的周长是_厘米。","answer":"26","explanation":"长方形的周长计算公式是：周长 = 2 × (长 + 宽)。将长8厘米和宽5厘米代入公式，得到：2 × (8 + 5) = 2 × 13 = 26。因此，这个长方形的周长是26厘米。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:40:05","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":891,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级环保活动中，某学生收集了若干个废旧电池。他将这些电池分成两类：可回收的和不可回收的。已知可回收电池的数量比不可回收的多6个，两类电池总数为24个。设不可回收电池的数量为x，则可列出方程：x + (x + 6) = 24。解这个方程，不可回收电池有___个。","answer":"9","explanation":"根据题意，设不可回收电池数量为x，则可回收电池数量为x + 6。两类电池总数为24，因此方程为x + (x + 6) = 24。化简得2x + 6 = 24，两边减去6得2x = 18，再除以2得x = 9。所以不可回收电池有9个。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 02:08:37","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":2182,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"在一次数学测验中，某学生需要计算三个有理数的和：-2.5，3\/4，以及比-1.2大0.8的数。该学生列式如下：(-2.5) + (3\/4) + (-1.2 + 0.8)。请问这个算式的正确结果是多少？","answer":"B","explanation":"首先计算比-1.2大0.8的数：-1.2 + 0.8 = -0.4。然后将三个数相加：-2.5 + 0.75 + (-0.4)。先算-2.5 + 0.75 = -1.75，再算-1.75 + (-0.4) = -2.15。因此正确答案是B。本题综合考查了有理数的加减运算、小数与分数的转换以及运算顺序，符合七年级有理数运算的教学要求。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-09 14:21:04","updated_at":"2026-01-09 14:21:04","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"-2.65","is_correct":0},{"id":"B","content":"-2.15","is_correct":1},{"id":"C","content":"-1.95","is_correct":0},{"id":"D","content":"-1.75","is_correct":0}]},{"id":1906,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某班级组织了一次环保知识竞赛，参赛学生需完成一份包含10道选择题的试卷。每答对一题得5分，答错或不答扣2分。一名学生最终得分为29分，请问这名学生答对了多少道题？","answer":"B","explanation":"设这名学生答对了x道题，则答错或不答的题目数为(10 - x)道。根据得分规则：每答对一题得5分，答错或不答扣2分，总得分为29分，可列出一元一次方程：5x - 2(10 - x) = 29。展开并化简：5x - 20 + 2x = 29 → 7x = 49 → x = 7。因此，这名学生答对了7道题。验证：7×5 = 35分，答错3题扣3×2 = 6分，35 - 6 = 29分，符合题意。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-07 13:10:44","updated_at":"2026-01-07 13:10:44","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"6道","is_correct":0},{"id":"B","content":"7道","is_correct":1},{"id":"C","content":"8道","is_correct":0},{"id":"D","content":"9道","is_correct":0}]},{"id":2363,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生在研究一次函数与几何图形的综合问题时，绘制了平面直角坐标系中的两个点A(0, 4)和B(6, 0)，并连接AB构成线段。若点P(x, y)是线段AB上的一点，且满足AP : PB = 2 : 1，则点P的坐标是下列哪一个？","answer":"B","explanation":"本题考查一次函数背景下的线段定比分点问题，结合坐标几何与比例关系。已知A(0, 4)，B(6, 0)，点P在线段AB上且AP : PB = 2 : 1，说明P将AB分为2:1的内分点。使用定比分点公式：P的横坐标x = (1×0 + 2×6)\/(2+1) = 12\/3 = 4；纵坐标y = (1×4 + 2×0)\/(2+1) = 4\/3。因此P(4, 4\/3)。也可通过向量法验证：向量AB = (6, -4)，AP = (2\/3)AB = (4, -8\/3)，故P = A + AP = (0+4, 4−8\/3) = (4, 4\/3)。选项B正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 11:13:53","updated_at":"2026-01-10 11:13:53","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"(2, 8\/3)","is_correct":0},{"id":"B","content":"(4, 4\/3)","is_correct":1},{"id":"C","content":"(3, 2)","is_correct":0},{"id":"D","content":"(5, 2\/3)","is_correct":0}]}]