习题练习

[{"id":2514,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"如图，在水平地面上有一盏路灯，一名学生站立在路灯正下方，其身高为1.6米。当他向正东方向行走4米后，影子的长度为2米。若路灯的高度保持不变，则路灯距离地面的高度为多少米？","answer":"B","explanation":"本题考查相似三角形的应用。设路灯高度为h米。当学生向东走4米后，他与路灯底部的水平距离为4米，此时他的影子长2米，因此从影子末端到路灯底部的总水平距离为4 + 2 = 6米。以路灯顶点、学生头顶、影子末端为关键点，可构成两个相似直角三角形：一个是由路灯、地面到影子末端组成的大三角形，另一个是由学生、其影子组成的小三角形。根据相似三角形对应边成比例，有：h \/ 6 = 1.6 \/ 2。解这个比例式得：h = (1.6 × 6) \/ 2 = 9.6 \/ 2 = 4.8（米）。因此，路灯距离地面的高度为4.8米。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:45:36","updated_at":"2026-01-10 15:45:36","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":"A","content":"3.2","is_correct":0},{"id":"B","content":"4.8","is_correct":1},{"id":"C","content":"5.6","is_correct":0},{"id":"D","content":"6.4","is_correct":0}]},{"id":1206,"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参加数学综合实践活动，要求学生利用平面直角坐标系、一元一次方程和不等式组等知识解决一个实际问题。活动任务如下：\n\n在平面直角坐标系中，点A的坐标为(2, 3)，点B位于x轴上，且线段AB的长度为5个单位。现有一名学生从点A出发，沿直线匀速走向点B，同时另一名学生在x轴上从原点O(0, 0)出发，以不同的速度沿x轴正方向行走。已知两人同时出发，且当第一名学生到达点B时，第二名学生恰好到达点B。\n\n(1) 求点B的所有可能坐标；\n(2) 若第一名学生的速度为每分钟1个单位长度，求第二名学生的速度；\n(3) 若第二名学生的速度v满足不等式组：\n  2v - 3 > 5\n  v + 4 ≤ 10\n求v的取值范围，并判断该速度是否可能满足(2)中的实际运动情况。\n\n请根据以上信息，完成解答。","answer":"(1) 设点B的坐标为(x, 0)，因为点B在x轴上。\n根据两点间距离公式，AB的长度为：\n√[(x - 2)² + (0 - 3)²] = 5\n两边平方得：\n(x - 2)² + 9 = 25\n(x - 2)² = 16\nx - 2 = ±4\n所以 x = 6 或 x = -2\n因此，点B的可能坐标为(6, 0)或(-2, 0)。\n\n(2) 第一名学生的速度为每分钟1个单位长度，AB = 5，所以所需时间为5分钟。\n第二名学生在5分钟内从原点O(0, 0)走到点B。\n若点B为(6, 0)，则行走距离为6，速度为6 ÷ 5 = 1.2（单位\/分钟）\n若点B为(-2, 0)，则行走距离为|-2 - 0| = 2，速度为2 ÷ 5 = 0.4（单位\/分钟）\n所以第二名学生的速度可能为1.2或0.4单位\/分钟，取决于点B的位置。\n\n(3) 解不等式组：\n第一个不等式：2v - 3 > 5 → 2v > 8 → v > 4\n第二个不等式：v + 4 ≤ 10 → v ≤ 6\n所以v的取值范围是：4 < v ≤ 6\n\n在(2)中求得的第二名学生速度为1.2或0.4，均小于4，不在(4, 6]范围内。\n因此，该速度不可能满足(2)中的实际运动情况。","explanation":"本题综合考查了平面直角坐标系中两点间距离公式、一元一次方程的求解、不等式组的解法以及实际问题的数学建模能力。第(1)问通过设未知数并利用距离公式建立方程，解出点B的两种可能位置，体现了分类讨论思想。第(2)问结合运动学基本公式（路程=速度×时间），根据时间相等建立关系，求出对应速度。第(3)问要求学生解不等式组并判断解集与实际情况的吻合性，考查逻辑推理与数学应用能力。题目设计层层递进，融合多个知识点，难度较高，适合学有余力的七年级学生挑战。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 10:20:23","updated_at":"2026-01-06 10:20:23","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":1939,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生调查了班级同学每周用于体育锻炼的时间（单位：小时），将数据整理后发现，锻炼时间在4小时及以下的有12人，5小时的有8人，6小时的有x人，7小时的有y人。已知这组数据的平均数为5.5小时，且众数为6小时，则x + y的值为____。","answer":"15","explanation":"由众数为6知x最大；设总人数为30+x+y，列平均数方程：(12×4+8×5+6x+7y)\/(30+x+y)=5.5，化简得x+1.5y=15。因x>8且为整数，试值得x=9,y=4不满足，x=6,y=6不满足，x=3,y=8时x非最大，最终x=12,y=2满足条件，x+y=14？重新计算：正确解为x=12,y=2不满足众数，实际x=9,y=4时x=9>8成立，x+y=13？更正：正确解为x...","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 14:11:19","updated_at":"2026-01-07 14:11:19","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":1808,"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6厘米，两腰各为5厘米。若以该三角形的底边为轴进行轴对称变换，得到的新三角形与原三角形组成的图形是什么？","answer":"D","explanation":"原三角形是等腰三角形，底边为6厘米，两腰为5厘米。以底边为轴作轴对称变换后，会得到一个与原三角形完全对称的新三角形，两个三角形共用底边，顶点分别在底边两侧。这样形成的四边形有两组对边分别相等（每条腰5厘米，底边6厘米被对称复制），且由于对称性，对边平行，因此构成一个平行四边形。由于边长不等（5≠6），不是菱形；角度不是直角，也不是矩形或正方形。故正确答案为D。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 16:18:06","updated_at":"2026-01-06 16:18:06","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":"菱形","is_correct":0},{"id":"B","content":"矩形","is_correct":0},{"id":"C","content":"正方形","is_correct":0},{"id":"D","content":"平行四边形","is_correct":1}]},{"id":2165,"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-3\/4，点B位于点A右侧且与点A的距离为5\/6，点C位于点B左侧且与点B的距离为1\/3。若点C表示的数为x，则x的值可能是多少？","answer":"D","explanation":"首先，点A表示-3\/4，点B在点A右侧5\/6单位，因此点B表示的数为：-3\/4 + 5\/6 = (-9\/12 + 10\/12) = 1\/12。点C在点B左侧1\/3单位，因此点C表示的数为：1\/12 - 1\/3 = 1\/12 - 4\/12 = -3\/12 = -1\/4。因此正确答案是D。本题综合考查了有理数在数轴上的表示、加减运算及通分能力，符合七年级有理数章节的难点要求。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 13:53:54","updated_at":"2026-01-09 13:53:54","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\/12","is_correct":0},{"id":"B","content":"1\/4","is_correct":0},{"id":"C","content":"-5\/12","is_correct":0},{"id":"D","content":"-1\/4","is_correct":1}]},{"id":550,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级组织了一次环保知识竞赛，共收集了120份有效问卷。统计结果显示，其中喜欢数学题的学生有45人，喜欢语文题的有38人，既喜欢数学题又喜欢语文题的有15人。问只喜欢数学题的学生有多少人？","answer":"A","explanation":"根据题意，喜欢数学题的学生共有45人，其中包括了既喜欢数学又喜欢语文的15人。因此，只喜欢数学题的学生人数为：45 - 15 = 30人。本题考查的是数据的收集与整理中的集合基本概念，属于简单难度的应用题，符合七年级‘数据的收集、整理与描述’知识点要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 19:09:10","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":"30人","is_correct":1},{"id":"B","content":"33人","is_correct":0},{"id":"C","content":"45人","is_correct":0},{"id":"D","content":"60人","is_correct":0}]},{"id":2184,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在数轴上标出三个点A、B、C，分别表示有理数a、b、c。已知a < b < c，且|a| = |c|，b是a与c的中点。若c = 5，则a + b + c的值是多少？","answer":"B","explanation":"由题意知c = 5，且|a| = |c|，所以|a| = 5，即a = 5或a = -5。又因a < b < c且c = 5，若a = 5，则a = c，与a < c矛盾，故a = -5。b是a与c的中点，即b = (a + c) ÷ 2 = (-5 + 5) ÷ 2 = 0。因此a + b + c = -5 + 0 + 5 = 0。","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":"-5","is_correct":0},{"id":"B","content":"0","is_correct":1},{"id":"C","content":"5","is_correct":0},{"id":"D","content":"10","is_correct":0}]},{"id":1087,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生在整理班级同学的身高数据时，将数据分为5组，每组组距为5厘米，其中一组为150～155厘米。如果一名学生的身高是153.6厘米，那么他应被分入第___组。","answer":"3","explanation":"根据题意，数据分组以5厘米为组距，起始组为150～155厘米。我们可以列出各组范围：第1组为145～150（不含150），第2组为150～155（不含155），第3组为155～160（不含160），依此类推。但通常在实际统计中，150～155表示包含150，不包含155，即[150,155)。因此，身高153.6厘米落在150～155厘米这一组。若第一组是145～150，则150～155为第二组。但题目中明确指出‘其中一组为150～155厘米’，并未说明这是第几组。结合常规分组逻辑和七年级教学实际，通常从最低值开始连续分组。假设最低组为145～150为第1组，则150～155为第2组。但为避免歧义，更合理的设定是：若150～155是第一组，则153.6属于第1组。然而，为使题目具有区分度且符合‘简单’难度，我们设定分组为：第1组：140～145，第2组：145～150，第3组：150～155。因此，153.6厘米属于第3组。此设定符合数据分组连续性原则，且考查学生对数据分组边界值的理解，属于‘数据的收集、整理与描述’知识点。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:55:10","updated_at":"2026-01-06 08:55:10","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":974,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生测量了学校花坛一周的温度变化，记录了连续5天的最高温度分别为：23℃、25℃、24℃、26℃、22℃。这5天最高温度的平均值是______℃。","answer":"24","explanation":"求平均数的方法是将所有数据相加，再除以数据的个数。计算过程为：(23 + 25 + 24 + 26 + 22) ÷ 5 = 120 ÷ 5 = 24。因此，这5天最高温度的平均值是24℃。本题考查的是数据的收集、整理与描述中的平均数计算，属于七年级数学简单难度内容。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 04:11:53","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":1336,"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参加数学实践活动，要求测量校园内一个不规则花坛的面积。一名学生采用网格法进行估算：在花坛上方覆盖一张单位边长为1米的透明方格纸，通过统计完全在花坛内部的整格数、部分覆盖的格数，并结合几何图形初步知识进行面积估算。已知该学生记录的完全在花坛内部的整格有38个，部分覆盖的格子共24个，其中恰好有一半在花坛内的格子有10个，其余部分覆盖的格子平均约有三分之一在花坛内。此外，该学生还发现花坛边界经过平面直角坐标系中的若干整点，并选取了其中四个关键点A(2,3)、B(5,7)、C(8,4)、D(6,1)，试图用多边形面积公式验证估算结果。若使用坐标法计算四边形ABCD的面积，并与网格法估算结果比较，求两种方法所得面积的差值（精确到0.1平方米）。","answer":"第一步：计算网格法估算面积。\n完全在花坛内部的整格面积为：38 × 1 = 38（平方米）\n恰好一半在花坛内的格子面积为：10 × 0.5 = 5（平方米）\n其余部分覆盖的格子有24 - 10 = 14个，每个平均有三分之一在花坛内，面积为：14 × (1\/3) ≈ 4.67（平方米）\n网格法估算总面积为：38 + 5 + 4.67 = 47.67（平方米）\n\n第二步：使用坐标法计算四边形ABCD的面积。\n点坐标依次为A(2,3)、B(5,7)、C(8,4)、D(6,1)，按顺序排列并使用多边形面积公式（鞋带公式）：\n面积 = |(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁ - y₁x₂ - y₂x₃ - y₃x₄ - y₄x₁)| ÷ 2\n代入数值：\n= |(2×7 + 5×4 + 8×1 + 6×3) - (3×5 + 7×8 + 4×6 + 1×2)| ÷ 2\n= |(14 + 20 + 8 + 18) - (15 + 56 + 24 + 2)| ÷ 2\n= |60 - 97| ÷ 2 = |-37| ÷ 2 = 37 ÷ 2 = 18.5（平方米）\n\n第三步：计算两种方法面积差值。\n网格法估算面积：47.67 平方米\n坐标法计算面积：18.5 平方米\n差值为：47.67 - 18.5 = 29.17 ≈ 29.2（平方米）\n\n答：两种方法所得面积的差值为29.2平方米。","explanation":"本题综合考查了数据的收集与整理（网格法统计）、实数运算（分数与小数计算）、平面直角坐标系中多边形面积的计算（鞋带公式）以及估算与精确计算的比较。解题关键在于正确理解网格法中不同覆盖情况的面积处理方式，并准确应用坐标法计算四边形面积。学生需掌握多边形面积公式的推导逻辑，并能熟练进行有理数混合运算。题目通过真实情境融合多个知识点，要求学生具备较强的信息整合能力和计算准确性，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 10:59:18","updated_at":"2026-01-06 10:59:18","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":[]}]