习题练习

[{"id":618,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"3.42元","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 21:44:59","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":2312,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"一个等腰三角形的两条边长分别为5和11，则这个三角形的周长是（ ）","answer":"B","explanation":"本题考查三角形三边关系及等腰三角形的性质。等腰三角形有两条边相等，已知两边分别为5和11，需分两种情况讨论：\n\n情况一：腰为5，底为11。此时三边为5、5、11。但5 + 5 = 10 < 11，不满足三角形两边之和大于第三边的条件，不能构成三角形。\n\n情况二：腰为11，底为5。此时三边为11、11、5。检查三边关系：11 + 11 > 5，11 + 5 > 11，均成立，可以构成三角形。\n\n因此，唯一可能的三边为11、11、5，周长为11 + 11 + 5 = 27。\n\n故正确答案为B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 10:45:50","updated_at":"2026-01-10 10:45:50","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":"21","is_correct":0},{"id":"B","content":"27","is_correct":1},{"id":"C","content":"21或27","is_correct":0},{"id":"D","content":"无法确定","is_correct":0}]},{"id":2139,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在解方程 3(x - 2) = 2x + 1 时，第一步去括号得到 3x - 6 = 2x + 1，第二步移项得到 3x - 2x = 1 + 6，第三步合并同类项得到 x = 7。该学生解题过程中哪一步开始出错？","answer":"D","explanation":"该学生解题过程完全正确：第一步去括号正确，3(x - 2) 展开为 3x - 6；第二步移项正确，将 2x 移到左边变为 -2x，将 -6 移到右边变为 +6；第三步合并同类项，3x - 2x = x，1 + 6 = 7，得到 x = 7，符合解一元一次方程的步骤和规则，因此整个过程没有出错。","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":"没有出错","is_correct":1}]},{"id":1859,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市地铁线路规划图中，两条平行轨道AB和CD被一条斜向联络线EF所截，形成多个角。已知∠1与∠2是同旁内角，且∠1的度数是∠2的2倍少30°。同时，在平面直角坐标系中，点A的坐标为(2, 3)，点B在x轴正方向上，且AB的长度为5个单位。若将线段AB向右平移3个单位，再向下平移2个单位得到线段A'B'，求：(1) ∠1和∠2的度数；(2) 点A'的坐标；(3) 若点C是线段A'B'的中点，求点C的坐标。","answer":"(1) 设∠2的度数为x°，则∠1 = (2x - 30)°。\n因为AB∥CD，EF为截线，∠1与∠2是同旁内角，所以∠1 + ∠2 = 180°。\n列方程：(2x - 30) + x = 180\n3x - 30 = 180\n3x = 210\nx = 70\n所以∠2 = 70°，∠1 = 2×70 - 30 = 110°。\n\n(2) 点A(2, 3)向右平移3个单位，横坐标加3，得(5, 3)；再向下平移2个单位，纵坐标减2，得(5, 1)。\n所以点A'的坐标为(5, 1)。\n\n(3) 点B在x轴正方向上，且AB = 5，A(2, 3)，设B(x, 0)，由距离公式：\n√[(x - 2)² + (0 - 3)²] = 5\n(x - 2)² + 9 = 25\n(x - 2)² = 16\nx - 2 = ±4\nx = 6 或 x = -2\n因为B在x轴正方向上，且从A向右延伸更合理（结合平移方向），取x = 6，即B(6, 0)。\n将B(6, 0)同样平移：向右3单位得(9, 0)，向下2单位得(9, -2)，即B'(9, -2)。\n点C是A'B'的中点，A'(5, 1)，B'(9,...","explanation":"本题综合考查平行线性质、一元一次方程、平面直角坐标系中的平移与坐标计算、中点公式。第(1)问利用同旁内角互补建立方程求解角度；第(2)问考查图形平移对坐标的影响；第(3)问需先通过距离公式确定点B坐标，再经平移得B'，最后用中点公式求C。关键步骤是正确理解几何关系与坐标变换规则，并准确进行代数运算。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 09:39:21","updated_at":"2026-01-07 09:39:21","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":128,"subject":"数学","grade":"初一","stage":"初中","type":"解答题","content":"某文具店出售一种笔记本，每本售价5元。小明购买了若干本这种笔记本，共花费了35元。请问小明买了多少本笔记本？","answer":"7本","explanation":"本题考查一元一次方程的实际应用。根据题意，每本笔记本5元，小明共花费35元，设他买了x本笔记本，则可列出方程：5x = 35。解这个方程即可求出x的值。这是初一学生应掌握的基础代数问题，涉及设未知数、列方程和简单求解。","solution_steps":"Array","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-24 08:54:36","updated_at":"2025-12-24 08:54:36","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":2770,"subject":"历史","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在参观博物馆时看到一件唐代的陶俑，其服饰风格融合了中亚地区的特点，面部轮廓立体，手持胡琴。这件文物最能反映唐代哪一方面的历史特征？","answer":"C","explanation":"题目中的陶俑具有中亚服饰特征和胡琴等外来文化元素，说明唐代社会受到外来文化的影响。唐朝国力强盛，对外交通发达，通过丝绸之路与中亚、西亚等地频繁交流，吸收了大量外来艺术、音乐和服饰文化。因此，这件文物最能体现唐代中外文化交流频繁的特点。选项A与题干无关；选项B错误，唐代是开放的朝代；选项D不符合史实，佛教虽盛行但并未取代本土信仰。故正确答案为C。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-12 10:41:04","updated_at":"2026-01-12 10:41:04","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":"唐代农业技术高度发达","is_correct":0},{"id":"B","content":"唐代实行严格的闭关锁国政策","is_correct":0},{"id":"C","content":"唐代中外文化交流频繁","is_correct":1},{"id":"D","content":"唐代佛教完全取代了本土信仰","is_correct":0}]},{"id":446,"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时，收集了10名同学每天阅读的分钟数：25，30，35，30，40，35，30，45，35，30。如果将这些数据按从小到大的顺序排列，那么位于中间两个数的平均数是多少？","answer":"B","explanation":"首先将数据从小到大排序：25，30，30，30，30，35，35，35，40，45。共有10个数据（偶数个），因此中位数是中间两个数的平均数，即第5个和第6个数的平均值。第5个数是30，第6个数是35，所以中位数为 (30 + 35) ÷ 2 = 65 ÷ 2 = 32.5。本题考查数据的整理与描述中的中位数概念，属于简单难度，符合七年级数学课程标准要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:44:01","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":"30","is_correct":0},{"id":"B","content":"32.5","is_correct":1},{"id":"C","content":"35","is_correct":0},{"id":"D","content":"37.5","is_correct":0}]},{"id":2301,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"在一次班级数学测验中，某学生记录了5名同学完成一次函数应用题所用的时间（单位：分钟），分别为：8、10、12、10、15。这组数据的中位数和众数分别是多少？","answer":"A","explanation":"首先将数据从小到大排列：8、10、10、12、15。共有5个数据，奇数个，因此中位数是中间的那个数，即第3个数，为10。众数是出现次数最多的数，10出现了两次，其余数各出现一次，因此众数是10。所以正确答案是A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 10:44:00","updated_at":"2026-01-10 10:44:00","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":"中位数是10，众数是10","is_correct":1},{"id":"B","content":"中位数是12，众数是10","is_correct":0},{"id":"C","content":"中位数是10，众数是12","is_correct":0},{"id":"D","content":"中位数是11，众数是10","is_correct":0}]},{"id":1983,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生在纸上画了一个边长为12 cm的正方形，并在正方形内部画了一个以正方形中心为圆心、半径为6 cm的圆。若将该圆绕其圆心逆时针旋转45°，则旋转前后两个圆重叠部分的面积占原圆面积的多少？","answer":"D","explanation":"本题考查旋转与圆的综合应用。圆具有任意角度的旋转对称性，即绕其圆心旋转任意角度后，图形都与原图形完全重合。题目中圆绕其圆心逆时针旋转45°，由于圆上每一点到圆心的距离不变，且旋转不改变圆的形状和大小，因此旋转后的圆与原圆完全重合。所以，旋转前后两个圆的重叠部分就是整个圆本身，重叠面积等于原圆面积，占比为1。故正确答案为D。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-07 15:03:01","updated_at":"2026-01-07 15:03:01","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\/4","is_correct":0},{"id":"B","content":"1\/2","is_correct":0},{"id":"C","content":"3\/4","is_correct":0},{"id":"D","content":"1","is_correct":1}]},{"id":1009,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生在整理班级同学的课外阅读时间时，将一周内每天阅读超过30分钟的人数记录如下：周一5人，周二7人，周三6人，周四8人，周五4人，周六9人，周日10人。若该学生想计算这周平均每天有多少人阅读超过30分钟，则计算结果为___人。","answer":"7","explanation":"本题考查数据的收集、整理与描述中的平均数计算。首先将每天的人数相加：5 + 7 + 6 + 8 + 4 + 9 + 10 = 49，共有7天，因此平均每天人数为49 ÷ 7 = 7（人）。计算过程简单，符合七年级学生对平均数概念的理解和应用能力。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 05:14:13","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":[]}]