习题练习

[{"id":2220,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生在记录一周内每天气温变化时，发现某天的气温比前一天上升了3℃，记作+3℃；那么如果某天的气温比前一天下降了2℃，应记作____℃。","answer":"-2","explanation":"根据正数和负数表示相反意义的量的概念，气温上升用正数表示，气温下降则用负数表示。因此，下降2℃应记作-2℃。这符合七年级数学中正负数在实际生活中的应用要求。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 14:27:19","updated_at":"2026-01-09 14:27:19","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":218,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生计算一个数的相反数时，将原数5写成了____，这个相反数是-5。","answer":"5","explanation":"相反数的定义是：一个数与它的相反数相加等于0。已知相反数是-5，那么原数就是5，因为5 + (-5) = 0。题目中说某学生计算的是这个数的相反数，并得到-5，因此原数应为5。空白处应填写原数5。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:40:16","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":911,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次环保活动中，某学生收集了不同种类的垃圾，其中可回收垃圾占总量的3\/8，厨余垃圾占总量的1\/4，有害垃圾占0.125，其余为其他垃圾。如果其他垃圾的重量是2.5千克，那么这次收集垃圾的总重量是___千克。","answer":"10","explanation":"首先将各部分垃圾所占比例统一为分数形式：可回收垃圾占3\/8，厨余垃圾占1\/4 = 2\/8，有害垃圾占0.125 = 1\/8。将这些比例相加：3\/8 + 2\/8 + 1\/8 = 6\/8 = 3\/4。因此，其他垃圾占总量的1 - 3\/4 = 1\/4。已知其他垃圾为2.5千克，设总重量为x千克，则有(1\/4)x = 2.5，解得x = 2.5 × 4 = 10。所以总重量是10千克。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 02:32:30","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":1726,"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开展‘校园植物分布调查’活动，要求将校园平面图绘制在平面直角坐标系中。已知校园主干道AB为一条直线，其两端点A和B的坐标分别为(-6, 0)和(4, 0)。校园内有一条与主干道AB垂直的小路CD，且小路CD经过点P(1, 5)。现需在小路CD上设置一个垃圾分类回收站Q，使得Q到主干道AB的距离为4个单位长度。同时，为了便于管理，要求回收站Q到点P的距离不超过3个单位长度。问：满足上述所有条件的回收站Q的坐标可能有哪些？请写出所有符合条件的点Q的坐标。","answer":"解题步骤如下：\n\n第一步：确定主干道AB所在直线的位置。\n已知A(-6, 0)，B(4, 0)，两点纵坐标均为0，说明AB是x轴上的一条线段，因此主干道AB所在的直线为y = 0。\n\n第二步：确定小路CD的方程。\n小路CD与AB垂直，AB是水平的（斜率为0），所以CD是竖直的，即斜率不存在，应为一条竖直线。\n但注意：若AB是水平线，则与之垂直的直线应为竖直线（即平行于y轴）。然而题目说CD经过点P(1, 5)，且与AB垂直，因此CD是过点(1, 5)且垂直于x轴的直线，即x = 1。\n\n第三步：确定点Q的位置。\n点Q在小路CD上，即Q的横坐标为1，设Q的坐标为(1, y)。\n\n第四步：Q到主干道AB的距离为4个单位长度。\n主干道AB在直线y = 0上，点Q(1, y)到直线y = 0的距离为|y - 0| = |y|。\n根据题意，|y| = 4，解得y = 4 或 y = -4。\n因此，可能的点Q有两个：(1, 4) 和 (1, -4)。\n\n第五步：筛选满足到点P(1, 5)距离不超过3的点。\n计算(1, 4)到P(1, 5)的距离：\n√[(1-1)² + (4-5)²] = √[0 + 1] = 1 ≤ 3，满足条件。\n\n计算(1, -4)到P(1, 5)的距离：\n√[(1-1)² + (-4-5)²] = √[0 + 81] = 9 > 3，不满足条件。\n\n第六步：得出结论。\n只有点(1, 4)同时满足：\n① 在小路CD上（x=1）；\n② 到主干道AB的距离为4；\n③ 到点P的距离不超过3。\n\n因此，符合条件的回收站Q的坐标只有一个：(1, 4)。","explanation":"本题综合考查了平面直角坐标系、点到直线的距离、两点间距离公式以及不等式的应用。解题关键在于理解几何关系：AB在x轴上，CD与之垂直，故CD为竖直线x=1。点Q在CD上，故横坐标为1。利用点到直线的距离公式确定纵坐标的可能值，再结合两点间距离公式和不等式条件进行筛选。题目融合了坐标几何与实际情境，要求学生具备较强的空间想象能力和代数运算能力，属于综合性较强的困难题。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 14:15:49","updated_at":"2026-01-06 14:15:49","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":1023,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某学生在整理班级同学的身高数据时，将150厘米到160厘米之间的身高记录为一个区间。如果一名学生的身高是155.3厘米，那么这个数据应被归入该区间的第___个十分位段（将150到160平均分成10段，每段为1厘米）。","answer":"6","explanation":"将150厘米到160厘米的区间平均分成10段，每段为1厘米，分别对应第1段（150≤身高<151）、第2段（151≤身高<152）……第6段（155≤身高<156）。因为155.3厘米满足155 ≤ 155.3 < 156，所以它属于第6个十分位段。本题考查数据的收集与整理中对数据区间的划分与归类，属于‘数据的收集、整理与描述’知识点，符合七年级数学课程内容。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 05:40:10","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":734,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生测量了教室里一盏灯到地面的垂直距离为2.8米，灯正下方地面上有一张课桌，课桌的高度为0.75米，那么灯到课桌桌面的垂直距离是______米。","answer":"2.05","explanation":"灯到地面的距离是2.8米，课桌高度为0.75米，课桌桌面距离地面0.75米。因此灯到桌面的垂直距离为2.8减去0.75，即2.8 - 0.75 = 2.05（米）。本题考查有理数的减法在实际生活中的应用，属于简单难度的计算题。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 23:06:44","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":1368,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市地铁线路规划中，需确定两个站点A和B之间的最短运行时间。已知列车在平直轨道上的平均速度为每小时60千米，但在弯道处需减速至每小时40千米。线路设计图显示，从A站到B站的总轨道长度为12千米，其中包含一段弯道。若列车全程运行时间不超过15分钟，且弯道长度至少为2千米，试求弯道长度的可能取值范围。假设列车在直道和弯道上均以恒定速度行驶，且不考虑停站和加减速时间。","answer":"解：\n设弯道长度为x千米，则直道长度为(12 - x)千米。\n根据题意，弯道长度至少为2千米，即：\nx ≥ 2。\n列车在弯道上的速度为40千米\/小时，行驶时间为：\n弯道时间 = x \/ 40 小时。\n列车在直道上的速度为60千米\/小时，行驶时间为：\n直道时间 = (12 - x) \/ 60 小时。\n总运行时间为两者之和，且不超过15分钟，即15\/60 = 0.25小时。\n因此，建立不等式：\nx \/ 40 + (12 - x) \/ 60 ≤ 0.25。\n为消去分母，两边同乘以120（40和60的最小公倍数）：\n120 × (x \/ 40) + 120 × ((12 - x) \/ 60) ≤ 120 × 0.25\n3x + 2(12 - x) ≤ 30\n3x + 24 - 2x ≤ 30\nx + 24 ≤ 30\nx ≤ 6\n结合弯道长度至少为2千米的条件，得：\n2 ≤ x ≤ 6\n因此，弯道长度的可能取值范围是大于等于2千米且小于等于6千米。\n答：弯道长度的取值范围是2千米到6千米（含端点）。","explanation":"本题综合考查了一元一次不等式的建立与求解，以及实际问题的数学建模能力。首先根据题意设定未知数x表示弯道长度，利用速度、时间与路程的关系分别表示直道和弯道的行驶时间，再根据总时间不超过15分钟（即0.25小时）建立不等式。通过通分消去分母，化简不等式得到x ≤ 6，再结合题设中弯道长度至少为2千米的条件，最终确定x的取值范围为2 ≤ x ≤ 6。解题过程中需注意单位统一（时间换算为小时），并合理运用不等式的性质进行变形。本题背景新颖，贴近现实，考查学生将实际问题转化为数学表达式的能力，属于困难难度的综合性应用题。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:11:20","updated_at":"2026-01-06 11:11:20","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":644,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级图书捐赠活动中，某学生捐出的图书数量比全班平均每人捐书数量的2倍少3本。已知该学生捐了7本书，那么全班平均每人捐书____本。","answer":"5","explanation":"设全班平均每人捐书 x 本。根据题意，该学生捐出的图书数量为 2x - 3 本，而实际捐了7本，因此可列方程：2x - 3 = 7。解这个一元一次方程：两边同时加3，得 2x = 10；再两边同时除以2，得 x = 5。所以全班平均每人捐书5本。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 22:09:30","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":134,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"下列各数中，最小的数是（ ）。","answer":"D","explanation":"在有理数中，负数小于0，0小于正数。比较负数时，绝对值越大的负数越小。-5 比 -3 更小，因此 -5 是四个选项中最小的数。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-24 09:40:59","updated_at":"2025-12-24 09:40:59","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":"1","is_correct":0},{"id":"D","content":"-5","is_correct":1}]},{"id":2246,"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，记录了一周内每天的温度变化情况。以某基准温度0℃为标准，高于0℃记为正，低于0℃记为负。已知这一周七天的温度变化值分别为：+3，-2，+5，-4，+1，-6，+2（单位：℃）。该学生发现，若将其中连续三天的温度变化值相加，可以得到一个最大的正数和最小的负数。请找出这个最大的正数和最小的负数，并说明是由哪连续三天得到的。","answer":"最大的正数是6，由第1天、第2天和第3天的温度变化值（+3，-2，+5）相加得到；最小的负数是-9，由第4天、第5天和第6天的温度变化值（-4，+1，-6）相加得到。","explanation":"本题考查正负数的加减运算及在实际情境中的应用，要求学生在多个连续数据中寻找极值组合，涉及枚举、计算与比较，符合七年级学生对正负数运算的综合运用能力要求。题目设计结合生活情境，避免机械重复，强调逻辑推理与系统分析，难度较高，适合用于提升学生的数学思维能力。","solution_steps":"1. 列出七天的温度变化值：第1天：+3，第2天：-2，第3天：+5，第4天：-4，第5天：+1，第6天：-6，第7天：+2。\n2. 找出所有可能的连续三天组合，共5组：\n - 第1-3天：+3 + (-2) + (+5) = 3 - 2 + 5 = 6\n - 第2-4天：-2 + (+5) + (-4) = -2 + 5 - 4 = -1\n - 第3-5天：+5 + (-4) + (+1) = 5 - 4 + 1 = 2\n - 第4-6天：-4 + (+1) + (-6) = -4 + 1 - 6 = -9\n - 第5-7天：+1 + (-6) + (+2) = 1 - 6 + 2 = -3\n3. 比较所有结果：6，-1，2，-9，-3。\n4. 其中最大的正数是6，最小的负数是-9。\n5. 确定对应的连续三天：最大正数6来自第1-3天，最小负数-9来自第4-6天。","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":[]}]