[{"id":277,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在平面直角坐标系中描出三个点：A(2, 3)、B(2, -1)、C(-4, -1)。这三个点构成的三角形是什么类型的三角形？","answer":"C","explanation":"首先观察三个点的坐标：A(2, 3)、B(2, -1)、C(-4, -1)。点A和点B的横坐标相同，说明AB是一条垂直于x轴的线段，长度为|3 - (-1)| = 4。点B和点C的纵坐标相同，说明BC是一条平行于x轴的线段，长度为|2 - (-4)| = 6。因此，AB与BC互相垂直，夹角为90度。根据勾股定理，若一个三角形中两条边互相垂直，则该三角形为直角三角形。所以，△ABC是以B为直角顶点的直角三角形。正确答案是C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:30:57","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":"等边三角形","is_correct":0},{"id":"B","content":"等腰三角形","is_correct":0},{"id":"C","content":"直角三角形","is_correct":1},{"id":"D","content":"钝角三角形","is_correct":0}]},{"id":769,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次班级环保活动中，某学生收集了若干个塑料瓶。若每3个塑料瓶可以兑换1支铅笔，且该学生最终兑换了___支铅笔后，还剩下2个塑料瓶。已知他最初收集的塑料瓶总数不超过20个，且兑换过程没有浪费，则他最初至少收集了___个塑料瓶。","answer":"6；20","explanation":"设该学生兑换了x支铅笔，则他用于兑换的塑料瓶数量为3x个，加上剩下的2个，总瓶数为3x + 2。根据题意，总瓶数不超过20，即3x + 2 ≤ 20，解得x ≤ 6。要使最初收集的瓶数最少，应使x尽可能小，但题目问的是“至少收集了多少个”，结合“兑换了___支铅笔”这一空，需满足兑换后剩2个且总数不超过20。当x = 6时，总瓶数为3×6 + 2 = 20，符合“不超过20”且为最大可能值，但题目要求“至少收集”，需反向思考：若兑换6支铅笔，则必须至少有18个用于兑换，加上剩余2个，共20个，这是满足条件的最小总数（因为若总数少于20，则无法兑换6支）。因此，第一个空填6（兑换铅笔数），第二个空填20（最初至少收集的瓶数）。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 23:47:55","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":620,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"72度","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 21:45:21","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":2298,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生测量了一个等腰三角形的底边长为8 cm，腰长为5 cm。若该三角形的一条对称轴将其分成两个全等直角三角形，则每个直角三角形的斜边长为多少？","answer":"A","explanation":"等腰三角形的对称轴是从顶角垂直平分底边的高，它将原三角形分成两个全等的直角三角形。每个直角三角形的底边为原底边的一半，即8 ÷ 2 = 4 cm，一条直角边为高（未知），另一条直角边为4 cm，斜边即为原等腰三角形的腰长，为5 cm。因此，每个直角三角形的斜边长为5 cm。选项A正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 10:43:17","updated_at":"2026-01-10 10:43:17","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":"5 cm","is_correct":1},{"id":"B","content":"6 cm","is_correct":0},{"id":"C","content":"8 cm","is_correct":0},{"id":"D","content":"10 cm","is_correct":0}]},{"id":285,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在整理班级同学的课外阅读时间数据时，记录了5名同学每天阅读的分钟数分别为：15、20、25、20、30。这组数据的众数和中位数分别是多少？","answer":"A","explanation":"首先将数据按从小到大的顺序排列：15、20、20、25、30。众数是出现次数最多的数，其中20出现了两次，其他数各出现一次，因此众数是20。中位数是位于中间位置的数，由于共有5个数据，中间位置是第3个数，即20，因此中位数也是20。所以正确答案是A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:31:49","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":"众数是20，中位数是20","is_correct":1},{"id":"B","content":"众数是20，中位数是25","is_correct":0},{"id":"C","content":"众数是25，中位数是20","is_correct":0},{"id":"D","content":"众数是15，中位数是25","is_correct":0}]},{"id":1475,"subject":"数学","grade":"七年级","stage":"小学","type":"解答题","content":"某学生在研究平面直角坐标系中的点与图形关系时，设计了如下实验：在坐标系中，点A的坐标为(2, 3)，点B位于x轴上，且线段AB的长度为5。点C是线段AB的中点，点D在y轴上，且满足CD的长度等于AB长度的一半。已知点D位于y轴正半轴，求点D的坐标。","answer":"解题步骤如下：\n\n1. 设点B的坐标为(x, 0)，因为点B在x轴上。\n\n2. 根据两点间距离公式，AB的长度为：\n AB = √[(x - 2)² + (0 - 3)²] = 5\n 即：(x - 2)² + 9 = 25\n (x - 2)² = 16\n x - 2 = ±4\n 所以x = 6 或 x = -2\n 因此点B有两个可能位置：(6, 0) 或 (-2, 0)\n\n3. 分别求两种情况下点C的坐标（AB中点）：\n - 若B为(6, 0)，则C = ((2+6)\/2, (3+0)\/2) = (4, 1.5)\n - 若B为(-2, 0)，则C = ((2-2)\/2, (3+0)\/2) = (0, 1.5)\n\n4. 点D在y轴上，设其坐标为(0, y)，且y > 0（因在正半轴）\n 已知CD = AB \/ 2 = 5 \/ 2 = 2.5\n\n5. 分情况讨论CD的距离：\n\n 情况一：C为(4, 1.5)\n CD = √[(0 - 4)² + (y - 1.5)²] = 2.5\n 16 + (y - 1.5)² = 6.25\n (y - 1.5)² = -9.75 → 无实数解（舍去）\n\n 情况二：C为(0, 1.5)\n CD = √[(0 - 0)² + (y - 1.5)²] = |y - 1.5| = 2.5\n 所以 y - 1.5 = 2.5 或 y - 1.5 = -2.5\n 解得 y = 4 或 y = -1\n 但y > 0，故y = 4\n\n6. 因此点D的坐标为(0, 4)\n\n答案：点D的坐标是(0, 4)","explanation":"本题综合考查了平面直角坐标系、两点间距离公式、中点坐标公式以及实数运算。解题关键在于分类讨论点B的两种可能位置，并通过距离条件排除不符合的情况。特别需要注意的是，当点C在y轴上时，CD的距离计算简化为纵坐标差的绝对值，这是解题的突破口。同时，题目设置了无解情况以检验学生对方程解的合理性判断能力，体现了对数学严谨性的考查。整个过程涉及代数运算、几何直观和逻辑推理，属于较高难度的综合题。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:53:23","updated_at":"2026-01-06 11:53:23","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":1701,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市地铁系统正在进行客流数据分析。已知在早高峰时段，A站和B站之间的乘客流动情况如下：从A站上车、B站下车的乘客人数为x人，从B站上车、A站下车的乘客人数为y人。调查发现，若将A站到B站的乘客人数增加20%，B站到A站的乘客人数减少10%，则总单向流动人数（即A到B与B到A之和）将增加8人。另外，若A站到B站的乘客人数减少10人，B站到A站的乘客人数增加15人，则两者人数相等。现需根据以上信息建立方程组，并求解x和y的值。进一步地，若该线路单程票价为3元，求调整后（即第一种变化情况）该区间一天的票务收入增加了多少元？","answer":"设从A站到B站的乘客人数为x人，从B站到A站的乘客人数为y人。\n\n根据题意，第一种变化情况：\nA到B人数增加20% → 变为1.2x\nB到A人数减少10% → 变为0.9y\n总单向流动人数增加8人：\n1.2x + 0.9y = x + y + 8\n化简得：\n1.2x + 0.9y - x - y = 8\n0.2x - 0.1y = 8 → 方程①\n\n第二种变化情况：\nA到B减少10人 → x - 10\nB到A增加15人 → y + 15\n两者人数相等：\nx - 10 = y + 15 → 方程②\n\n由方程②得：x = y + 25\n代入方程①：\n0.2(y + 25) - 0.1y = 8\n0.2y + 5 - 0.1y = 8\n0.1y + 5 = 8\n0.1y = 3\ny = 30\n代入x = y + 25得：x = 55\n\n所以，原来A到B有55人，B到A有30人。\n\n调整后人数：\nA到B：1.2 × 55 = 66（人）\nB到A：0.9 × 30 = 27（人）\n总人数：66 + 27 = 93（人）\n原来总人数：55 + 30 = 85（人）\n增加人数：93 - 85 = 8（人），符合题意。\n\n票务收入增加计算：\n每张票3元，总人数增加8人，因此收入增加：\n8 × 3 = 24（元）\n\n答：x = 55，y = 30；调整后一天的票务收入增加了24元。","explanation":"本题综合考查二元一次方程组的建立与求解，并结合实际情境进行数据分析。首先根据文字描述提取两个等量关系，列出方程组。第一个关系涉及百分数变化后的总量变化，需将百分数转化为小数参与运算；第二个关系是人数调整后的相等关系，可直接列式。通过代入法求解方程组，得到原始人数。最后结合票价计算收入变化，体现数学在现实问题中的应用。题目融合了二元一次方程组、有理数运算和实际问题建模，思维层次较高，属于困难难度。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 13:42:13","updated_at":"2026-01-06 13:42:13","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":195,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"小明买了3支铅笔和2本笔记本，共花费18元。已知每本笔记本比每支铅笔贵3元，设每支铅笔的价格为x元，则下列方程正确的是（ ）。","answer":"A","explanation":"设每支铅笔的价格为x元，根据题意，每本笔记本比每支铅笔贵3元，因此每本笔记本的价格为(x + 3)元。小明买了3支铅笔，总价为3x元；买了2本笔记本，总价为2(x + 3)元。两者相加等于总花费18元，因此方程为：3x + 2(x + 3) = 18。选项A正确。其他选项中，B错误地将笔记本价格设为比铅笔便宜，C和D则颠倒了铅笔和笔记本的数量与单价对应关系，均不符合题意。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:04:01","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":"3x + 2(x + 3) = 18","is_correct":1},{"id":"B","content":"3x + 2(x - 3) = 18","is_correct":0},{"id":"C","content":"3(x + 3) + 2x = 18","is_correct":0},{"id":"D","content":"3(x - 3) + 2x = 18","is_correct":0}]},{"id":1907,"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活动，收集废旧纸张和塑料瓶。已知收集的废旧纸张总重量比塑料瓶多12千克，且两种物品的总重量为48千克。设塑料瓶的重量为x千克，则根据题意列出的方程是：","answer":"B","explanation":"根据题意，塑料瓶重量为x千克，废旧纸张比塑料瓶多12千克，因此纸张重量为(x + 12)千克。两者总重量为48千克，所以方程为：x + (x + 12) = 48。选项B正确表达了这一数量关系。选项A错误地将纸张表示为比塑料瓶少；选项C的减法不符合实际意义；选项D错误地将12与x相乘，而非相加。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-07 13:11:04","updated_at":"2026-01-07 13:11:04","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":"x + (x - 12) = 48","is_correct":0},{"id":"B","content":"x + (x + 12) = 48","is_correct":1},{"id":"C","content":"x - (x + 12) = 48","is_correct":0},{"id":"D","content":"x + 12x = 48","is_correct":0}]},{"id":2152,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在解方程时，将方程 3(x - 2) = 2x + 1 的解题步骤写成了：第一步：3x - 6 = 2x + 1；第二步：3x - 2x = 1 + 6；第三步：x = 7。该学生在哪一步开始出现错误？","answer":"D","explanation":"该学生的解题过程完全正确：第一步去括号得 3x - 6 = 2x + 1，正确；第二步移项得 3x - 2x = 1 + 6，正确；第三步合并同类项得 x = 7，正确。因此整个解答过程无误。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 13:00:46","updated_at":"2026-01-09 13:00:46","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":"A","content":"第一步","is_correct":0},{"id":"B","content":"第二步","is_correct":0},{"id":"C","content":"第三步","is_correct":0},{"id":"D","content":"没有错误，解答正确","is_correct":1}]}]