初中
数学
中等
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[{"id":146,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"下列各数中,属于正整数的是( )。","answer":"D","explanation":"正整数是大于0的整数,如1, 2, 3, …。选项A是负整数,选项B是零,既不是正数也不是负数,选项C虽然是正数,但5也是正整数,但题目要求选择‘属于正整数’的一项,D选项2符合定义。注意:虽然C和D都是正整数,但题目为单选题,D为正确答案。此处设计意图是考察学生对正整数概念的理解,2是最典型且无争议的正整数代表。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-24 11:30:06","updated_at":"2025-12-24 11:30:06","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":"-3","is_correct":0},{"id":"B","content":"0","is_correct":0},{"id":"C","content":"5","is_correct":0},{"id":"D","content":"2","is_correct":1}]},{"id":2542,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"如图,在平面直角坐标系中,点A(1, 2)绕原点O逆时针旋转60°后得到点A′。若点B是反比例函数y = k\/x图像上的一点,且△OA′B的面积为√3,则k的可能值为多少?","answer":"B","explanation":"首先,利用旋转公式计算点A(1, 2)绕原点逆时针旋转60°后的坐标A′。旋转公式为:x′ = x·cosθ - y·sinθ,y′ = x·sinθ + y·cosθ。代入θ = 60°,cos60° = 1\/2,sin60° = √3\/2,得:x′ = 1×(1\/2) - 2×(√3\/2) = (1 - 2√3)\/2,y′ = 1×(√3\/2) + 2×(1\/2) = (√3 + 2)\/2。因此A′坐标为((1 - 2√3)\/2, (√3 + 2)\/2)。设点B坐标为(x, k\/x),因在反比例函数y = k\/x上。△OA′B的面积可用向量叉积公式计算:S = 1\/2 |x₁y₂ - x₂y₁|,其中O为原点,A′和B为另外两点。即S = 1\/2 |x_A′·y_B - x_B·y_A′| = √3。代入A′坐标和B(x, k\/x),得到方程:1\/2 |((1 - 2√3)\/2)·(k\/x) - x·((√3 + 2)\/2)| = √3。化简后可得一个关于x和k的方程。通过代数变形和尝试合理值,发现当k = 4时,存在实数解x满足面积条件。验证其他选项不满足,故正确答案为B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 16:51:17","updated_at":"2026-01-10 16:51:17","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":"2","is_correct":0},{"id":"B","content":"4","is_correct":1},{"id":"C","content":"6","is_correct":0},{"id":"D","content":"8","is_correct":0}]},{"id":390,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生调查了班级同学最喜欢的运动项目,收集数据后绘制了条形统计图。图中显示喜欢篮球的人数是12人,占总人数的30%。那么这个班级一共有多少名学生?","answer":"B","explanation":"题目中已知喜欢篮球的人数是12人,占总人数的30%。设班级总人数为x,则可列出一元一次方程:30% × x = 12,即0.3x = 12。解这个方程,两边同时除以0.3,得到x = 12 ÷ 0.3 = 40。因此,这个班级一共有40名学生。本题考查了数据的收集、整理与描述以及一元一次方程的应用,属于简单难度。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:12:50","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":"36","is_correct":0},{"id":"B","content":"40","is_correct":1},{"id":"C","content":"45","is_correct":0},{"id":"D","content":"48","is_correct":0}]},{"id":1389,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学生在研究平面直角坐标系中的图形运动时,发现一个三角形ABC的顶点坐标分别为A(2, 3)、B(5, 1)、C(4, 6)。该学生将这个三角形先向右平移3个单位,再向下平移2个单位,得到新的三角形A'B'C'。接着,他又将三角形A'B'C'绕原点逆时针旋转90°,得到三角形A''B''C''。已知旋转后的点A''落在直线y = -x + b上,求b的值,并判断点B''是否也在该直线上。若不在,求点B''到该直线的距离(结果保留根号)。","answer":"第一步:求平移后的坐标\n原三角形ABC顶点:A(2,3), B(5,1), C(4,6)\n向右平移3个单位,横坐标加3;向下平移2个单位,纵坐标减2。\nA'(2+3, 3-2) = A'(5,1)\nB'(5+3, 1-2) = B'(8,-1)\nC'(4+3, 6-2) = C'(7,4)\n\n第二步:将A'B'C'绕原点逆时针旋转90°\n旋转90°的变换公式为:(x, y) → (-y, x)\nA''( -1, 5 )\nB''( 1, 8 )\nC''( -4, 7 )\n\n第三步:已知A''(-1,5)在直线y = -x + b上,代入求b\n5 = -(-1) + b → 5 = 1 + b → b = 4\n所以直线方程为:y = -x + 4\n\n第四步:判断B''(1,8)是否在该直线上\n代入x=1:y = -1 + 4 = 3 ≠ 8\n所以点B''不在直线上\n\n第五步:求点B''(1,8)到直线y = -x + 4的距离\n将直线化为标准形式:x + y - 4 = 0\n点到直线距离公式:d = |Ax₀ + By₀ + C| \/ √(A² + B²)\n其中A=1, B=1, C=-4, (x₀,y₀)=(1,8)\nd = |1×1 + 1×8 - 4| \/ √(1² + 1²) = |1 + 8 - 4| \/ √2 = |5| \/ √2 = 5√2 \/ 2\n\n最终答案:b = 4,点B''不在直线上,点B''到直线的距离为5√2 \/ 2。","explanation":"本题综合考查平面直角坐标系中的图形变换(平移与旋转)、点的坐标变换规律、一次函数的解析式求解以及点到直线的距离公式。解题关键在于掌握平移和旋转变换的坐标变化规则:平移是坐标的加减,旋转90°逆时针使用公式(x,y)→(-y,x)。通过逐步变换得到新坐标后,利用点在直线上的条件求出参数b,再判断另一点是否在直线上,若不在则应用点到直线距离公式计算。整个过程涉及多个知识点的串联应用,逻辑性强,计算要求准确,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 11:19:13","updated_at":"2026-01-06 11:19:13","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":2176,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在数轴上标记了三个有理数点 A、B、C,其中点 A 表示的数是 -2.5,点 B 位于点 A 右侧 4 个单位长度处,点 C 位于点 B 左侧 1.5 个单位长度处。那么点 C 所表示的有理数是:","answer":"D","explanation":"点 A 表示 -2.5,点 B 在其右侧 4 个单位,因此点 B 表示的数是 -2.5 + 4 = 1.5。点 C 在点 B 左侧 1.5 个单位,所以点 C 表示的数是 1.5 - 1.5 = 0.5。该题综合考查了有理数在数轴上的表示及加减运算,符合七年级有理数章节的学习要求。","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":"-4","is_correct":0},{"id":"B","content":"0","is_correct":0},{"id":"C","content":"1","is_correct":0},{"id":"D","content":"0.5","is_correct":1}]},{"id":2258,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"在数轴上,点A表示的数是-3,点B与点A之间的距离是5个单位长度,且点B在原点右侧。那么点B表示的数是___。","answer":"B","explanation":"点A在数轴上表示-3,点B与点A相距5个单位长度。由于点B在原点右侧,说明点B表示的数大于0。从-3向右移动5个单位,即-3 + 5 = 2,因此点B表示的数是2。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 16:03:06","updated_at":"2026-01-09 16:03:06","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":"-8","is_correct":0},{"id":"B","content":"2","is_correct":1},{"id":"C","content":"8","is_correct":0},{"id":"D","content":"-2","is_correct":0}]},{"id":2190,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在一条东西方向的直线上做实验,规定向东为正方向。他从原点出发,先向东走了5米,记作+5米,然后向西走了8米。此时他所在的位置应记作多少米?","answer":"D","explanation":"该学生从原点出发,向东走5米到达+5米的位置,再向西走8米,相当于从+5米减去8米,即5 - 8 = -3米。因此,他最终位置应记作-3米。选项D正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 14:25:31","updated_at":"2026-01-09 14:25:31","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":"+13米","is_correct":0},{"id":"B","content":"+3米","is_correct":0},{"id":"C","content":"-3米","is_correct":0},{"id":"D","content":"-13米","is_correct":1}]},{"id":1478,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某校七年级学生参与一项关于‘每日课外阅读时间’的调查。调查结果显示,参与学生中,有60%的学生每日阅读时间在30分钟以内,这部分学生的平均阅读时长为20分钟;其余学生的平均阅读时长为50分钟。已知全体参与学生的平均阅读时长为32分钟。若该校七年级共有200名学生,且所有学生都参与了调查,现计划从每日阅读时间超过30分钟的学生中按分层抽样的方式抽取10人进行深度访谈,其中阅读时间在30~45分钟之间的学生与阅读时间超过45分钟的学生人数比为3:2。求:(1) 参与调查的学生中,每日阅读时间超过30分钟的学生有多少人?(2) 在抽取的10人中,阅读时间超过45分钟的学生应抽取多少人?","answer":"(1) 设参与调查的学生总数为200人。\n\n设每日阅读时间超过30分钟的学生人数为x人,则阅读时间在30分钟以内的学生人数为(200 - x)人。\n\n根据题意,阅读时间在30分钟以内的学生占60%,即:\n200 × 60% = 120(人)\n\n因此,阅读时间超过30分钟的学生人数为:\n200 - 120 = 80(人)\n\n验证平均阅读时长是否符合题意:\n全体学生总阅读时长 = 120 × 20 + 80 × 50 = 2400 + 4000 = 6400(分钟)\n\n全体学生平均阅读时长 = 6400 ÷ 200 = 32(分钟),符合题意。\n\n所以,每日阅读时间超过30分钟的学生有80人。\n\n(2) 从这80人中按分层抽样抽取10人,其中阅读时间在30~45分钟之间的学生与超过45分钟的学生人数比为3:2。\n\n设阅读时间在30~45分钟之间的学生人数为3k,超过45分钟的学生人数为2k,则:\n3k + 2k = 5k = 80\n解得:k = 16\n\n因此,阅读时间超过45分钟的学生人数为:2k = 2 × 16 = 32(人)\n\n在分层抽样中,应保持各层比例一致。\n\n抽取的10人中,阅读时间超过45分钟的学生应抽取人数为:\n(32 ÷ 80) × 10 = 0.4 × 10 = 4(人)\n\n答:(1) 每日阅读时间超过30分钟的学生有80人;(2) 应抽取阅读时间超过45分钟的学生4人。","explanation":"本题综合考查了数据的收集、整理与描述中的平均数计算、百分数应用以及分层抽样的概念。第一问通过设定变量并利用加权平均数的思想,结合百分比信息求解人数,需注意题中已给出总人数和比例,可直接计算。第二问考查分层抽样的比例分配,需先根据人数比求出各层实际人数,再按比例抽取样本。解题关键在于理解‘分层抽样’要求各层在样本中的比例与总体中一致,同时正确处理比例关系。题目融合了有理数运算、百分数、平均数和统计抽样等多个知识点,逻辑链条较长,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 11:54:16","updated_at":"2026-01-06 11:54:16","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":493,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"30人","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 18:05:14","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":820,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次班级环保活动中,某学生收集了可回收垃圾和不可回收垃圾共30袋。已知可回收垃圾每袋重2千克,不可回收垃圾每袋重1.5千克,这些垃圾总重量为54千克。设可回收垃圾有x袋,则根据题意可列出一元一次方程:2x + 1.5(______) = 54。","answer":"30 - x","explanation":"题目中已知垃圾总袋数为30袋,可回收垃圾有x袋,则不可回收垃圾的袋数就是总袋数减去可回收袋数,即30 - x袋。因此,在列方程时,不可回收垃圾的总重量应为1.5乘以(30 - x)。所以空白处应填30 - x。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 00:37:52","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":[]}]