习题练习

[{"id":1836,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"如图，在平面直角坐标系中，点A(0, 4)、B(3, 0)、C(-3, 0)构成△ABC。若点D是线段BC上的一点，且△ABD与△ACD的周长相等，则点D的横坐标为多少？","answer":"B","explanation":"由题意，点B(3,0)、C(-3,0)，所以线段BC在x轴上，中点为原点O(0,0)。因为△ABD与△ACD的周长相等，即AB + BD + AD = AC + CD + AD。两边同时减去AD，得AB + BD = AC + CD。计算AB和AC的长度：AB = √[(3-0)² + (0-4)²] = √(9+16) = 5；AC = √[(-3-0)² + (0-4)²] = √(9+16) = 5。所以AB = AC，代入得BD = CD。因此D是BC的中点，坐标为(0,0)，横坐标为0。故选B。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-06 16:49:42","updated_at":"2026-01-06 16:49:42","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":"-1","is_correct":0},{"id":"B","content":"0","is_correct":1},{"id":"C","content":"1","is_correct":0},{"id":"D","content":"2","is_correct":0}]},{"id":1933,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"在平面直角坐标系中，点A(2, 3)、点B(6, 7)，点C在x轴上，且△ABC是以AB为斜边的等腰直角三角形，则点C的横坐标为___。","answer":"4","explanation":"由AB中点M(4,5)为直角顶点对称中心，C在x轴上且满足AC=BC，利用距离公式列方程解得C横坐标为4。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 14:10:20","updated_at":"2026-01-07 14:10:20","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":438,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某班级在一次数学测验中，收集了20名学生的成绩（单位：分），数据如下：68, 72, 75, 76, 78, 79, 80, 82, 82, 83, 85, 86, 87, 88, 89, 90, 91, 92, 94, 98。如果将这些成绩按从小到大的顺序排列，那么中位数是多少？","answer":"B","explanation":"中位数是指将一组数据按从小到大（或从大到小）的顺序排列后，处于中间位置的数。如果数据个数为偶数，则中位数是中间两个数的平均数。本题共有20个数据，是偶数个，因此中位数是第10个和第11个数据的平均数。将数据排序后，第10个数是83，第11个数是85。计算中位数：(83 + 85) ÷ 2 = 168 ÷ 2 = 84。因此，中位数是84分。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:40:10","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":"A","content":"83分","is_correct":0},{"id":"B","content":"84分","is_correct":1},{"id":"C","content":"85分","is_correct":0},{"id":"D","content":"86分","is_correct":0}]},{"id":1415,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市为了优化公交线路，对一条主干道的车流量进行了为期一周的观测。观测数据如下：周一至周五每天的车流量分别为 1200、1350、1280、1420、1300 辆；周六和周日分别为 980 和 860 辆。交通部门计划在车流量超过平均日流量的日子增加临时班次。已知每增加一个临时班次可多运送 50 名乘客，且每名乘客的平均票价为 2 元。若临时班次的运营成本为每班次 80 元，问：在一周中，交通部门因增加临时班次总共能获得多少净利润？（净利润 = 总收入 - 总成本）","answer":"第一步：计算一周的总车流量。\n1200 + 1350 + 1280 + 1420 + 1300 + 980 + 860 = 8390（辆）\n\n第二步：计算平均日车流量。\n8390 ÷ 7 ≈ 1198.57（辆\/天）\n\n第三步：找出车流量超过平均日流量的天数。\n比较每天车流量与 1198.57：\n- 周一：1200 > 1198.57 → 超过\n- 周二：1350 > 1198.57 → 超过\n- 周三：1280 > 1198.57 → 超过\n- 周四：1420 > 1198.57 → 超过\n- 周五：1300 > 1198.57 → 超过\n- 周六：980 < 1198.57 → 未超过\n- 周日：860 < 1198.57 → 未超过\n\n因此，有 5 天需要增加临时班次。\n\n第四步：计算每天增加的临时班次数。\n题目未直接给出班次数，但说明“每增加一个临时班次可多运送 50 名乘客”，我们假设交通部门根据超出部分合理配置班次，但题目未给出具体配置规则。然而，结合问题目标（求净利润），需明确班次数。\n\n重新审题：题目隐含条件是“在车流量超过平均的日子增加临时班次”，但未说明增加几个。考虑到七年级知识范围，应理解为：只要超过，就增加一个临时班次（标准做法）。否则无法计算。\n\n因此，每天超过平均流量的日子增加 1 个临时班次，共 5 天 → 共增加 5 个临时班次。\n\n第五步：计算总收入。\n每班次多运送 50 名乘客，每名乘客票价 2 元：\n每班次收入 = 50 × 2 = 100（元）\n5 个班次总收入 = 5 × 100 = 500（元）\n\n第六步：计算总成本。\n每班次成本 80 元，5 个班次总成本 = 5 × 80 = 400（元）\n\n第七步：计算净利润。\n净利润 = 总收入 - 总成本 = 500 - 400 = 100（元）\n\n答：交通部门因增加临时班次总共能获得 100 元的净利润。","explanation":"本题综合考查了数据的收集、整理与描述（计算平均数、比较数据大小）、有理数的运算（加减乘除）、以及实际问题的建模能力。解题关键在于理解“平均日流量”的计算方法，并据此判断哪些天需要增加班次。题目设置了真实情境——城市公交调度，要求学生在处理实际数据的基础上进行逻辑推理和数学计算。难点在于学生需自主判断“增加临时班次”的具体数量，结合七年级认知水平，合理假设为每天增加一个班次，使问题可解。同时涉及收入、成本、利润等经济概念，体现了数学在生活中的应用，符合新课标对数学建模能力的要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:29:46","updated_at":"2026-01-06 11:29:46","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":652,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次班级大扫除中，某学生负责统计各小组清理的垃圾袋数量。已知第一组清理了3袋，第二组清理了5袋，第三组清理了x袋，三组共清理了12袋垃圾。根据题意列出的一元一次方程是：3 + 5 + x = ___","answer":"12","explanation":"题目中明确指出三组共清理了12袋垃圾，而第一组清理3袋，第二组清理5袋，第三组清理x袋，因此总数量为3 + 5 + x。根据总数量等于12，可得方程：3 + 5 + x = 12。空白处应填写总数12，这是建立一元一次方程的关键步骤，考查学生将实际问题转化为数学表达式的能力。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:11:40","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":723,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次班级图书角整理活动中，某学生统计了上周同学们借阅图书的天数，发现借阅天数最多的为7天，最少的为2天。如果将每位同学的借阅天数都减去3天，则新的数据中，最大值与最小值的差是___天。","answer":"5","explanation":"原数据中最大值为7天，最小值为2天，它们的差是7 - 2 = 5天。当每个数据都减去同一个数（这里是3）时，数据之间的差距（即极差）不会改变。因此，新的最大值是7 - 3 = 4，新的最小值是2 - 3 = -1，它们的差仍然是4 - (-1) = 5天。所以答案是5。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:57:40","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":470,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生调查了班级同学每天用于课外阅读的时间（单位：分钟），并将数据整理如下：15, 20, 25, 30, 35, 40, 45。如果再加入一个数据后，这组数据的中位数变为30，那么加入的这个数据可能是多少？","answer":"B","explanation":"原数据有7个数，按从小到大排列为：15, 20, 25, 30, 35, 40, 45。中位数是第4个数，即30。加入一个数据后，总共有8个数，中位数是第4个和第5个数的平均数。要使中位数为30，则第4个和第5个数的平均数必须为30。若加入30，则新数据为：15, 20, 25, 30, 30, 35, 40, 45，此时第4个数是30，第5个数也是30，中位数为(30+30)÷2=30，符合条件。其他选项加入后，中位数均不等于30。因此正确答案是B。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:53:54","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":"A","content":"25","is_correct":0},{"id":"B","content":"30","is_correct":1},{"id":"C","content":"35","is_correct":0},{"id":"D","content":"40","is_correct":0}]},{"id":787,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次班级数学测验成绩整理中，某学生将10名同学的成绩按从小到大的顺序排列，得到的数据为：72，75，78，80，82，85，88，90，93，96。这组数据的中位数是____。","answer":"83.5","explanation":"中位数是指将一组数据按大小顺序排列后，处于中间位置的数。当数据个数为偶数时，中位数是中间两个数的平均值。本题中有10个数据（偶数个），因此中位数是第5个和第6个数据的平均数。第5个数是82，第6个数是85，所以中位数为 (82 + 85) ÷ 2 = 167 ÷ 2 = 83.5。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 00:06:37","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":1516,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市地铁线路规划部门正在设计一条新的地铁线路，线路在平面直角坐标系中表示为一条直线 L。已知该线路经过站点 A(2, 5) 和站点 B(6, 1)。为优化换乘，需在站点 C(4, 3) 处设置一个换乘枢纽。经测量，换乘枢纽 C 到线路 L 的垂直距离为 d。现计划在线路 L 上新建一个临时施工点 P，使得点 P 到点 C 的距离等于 d，且点 P 位于线段 AB 上（包括端点）。若存在多个满足条件的点 P，取横坐标较小的一个。求点 P 的坐标。","answer":"解：\n\n第一步：求直线 L 的方程\n已知直线 L 经过点 A(2, 5) 和 B(6, 1)，先求斜率 k：\nk = (1 - 5) \/ (6 - 2) = (-4) \/ 4 = -1\n\n设直线方程为 y = -x + b，代入点 A(2, 5)：\n5 = -2 + b ⇒ b = 7\n所以直线 L 的方程为：y = -x + 7\n\n第二步：求点 C(4, 3) 到直线 L 的距离 d\n点到直线的距离公式：对于直线 ax + by + c = 0，点 (x₀, y₀) 到直线的距离为\n|ax₀ + by₀ + c| \/ √(a² + b²)\n\n将 y = -x + 7 化为标准形式：x + y - 7 = 0，即 a = 1, b = 1, c = -7\n代入点 C(4, 3)：\nd = |1×4 + 1×3 - 7| \/ √(1² + 1²) = |4 + 3 - 7| \/ √2 = |0| \/ √2 = 0\n\n发现点 C(4, 3) 在直线 L 上！因为当 x = 4 时，y = -4 + 7 = 3，确实在直线上。\n因此 d = 0，即点 C 到直线 L 的距离为 0。\n\n第三步：找点 P，使 P 在线段 AB 上，且 |PC| = d = 0\n|PC| = 0 意味着 P 与 C 重合，即 P = C\n\n检查点 C(4, 3) 是否在线段 AB 上：\n参数法判断：设线段 AB 上任意点可表示为：\n(x, y) = (1 - t)(2, 5) + t(6, 1) = (2 + 4t, 5 - 4t)，其中 t ∈ [0, 1]\n令 x = 4：2 + 4t = 4 ⇒ 4t = 2 ⇒ t = 0.5 ∈ [0, 1]\n此时 y = 5 - 4×0.5 = 5 - 2 = 3，正好是点 C(4, 3)\n所以点 C 在线段 AB 上\n\n因此，满足条件的点 P 就是 C(4, 3)\n题目要求若存在多个点取横坐标较小者，此处仅有一个点\n\n最终答案：点 P 的坐标为 (4, 3)","explanation":"本题综合考查了平面直角坐标系、直线方程、点到直线的距离公式以及线段上的点参数表示等多个知识点。解题关键在于发现点 C 恰好落在直线 AB 上，从而得出距离 d 为 0，进而推出点 P 必须与 C 重合。通过参数法验证点 C 是否在线段 AB 上是关键步骤，体现了数形结合思想。题目设计巧妙，表面看似复杂，实则通过计算揭示几何本质，考查学生逻辑推理与计算能力，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 12:10:08","updated_at":"2026-01-06 12:10:08","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":2436,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某公园内有一个矩形花坛ABCD，长为12米，宽为8米。现计划在花坛内部修建一条宽度相同的十字形步道（步道沿花坛中心对称分布，将花坛分为四个面积相等的小矩形区域），使得剩余绿化区域的面积为60平方米。设步道宽度为x米，则可列方程为：","answer":"B","explanation":"花坛总面积为12×8=96平方米。十字形步道由一条水平步道和一条垂直步道组成，宽度均为x米。水平步道面积为12x，垂直步道面积为8x，但两者在中心重叠了一个x×x的正方形区域，因此被重复计算了一次，实际步道总面积为12x + 8x - x² = 20x - x²。剩余绿化面积为总面积减去步道面积：96 - (20x - x²) = 96 - 20x + x²。根据题意，该面积等于60，即96 - (12x + 8x - x²) = 60，整理得12×8 - (12x + 8x - x²) = 60，对应选项B。选项A和D错误地将整个花坛视为减去一圈边框，不符合十字形步道结构；选项C未扣除重叠部分，导致多减面积。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 13:11:18","updated_at":"2026-01-10 13:11:18","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":"(12 - x)(8 - x) = 60","is_correct":0},{"id":"B","content":"12×8 - (12x + 8x - x²) = 60","is_correct":1},{"id":"C","content":"12×8 - 2×(12x + 8x) = 60","is_correct":0},{"id":"D","content":"(12 - 2x)(8 - 2x) = 60","is_correct":0}]}]