习题练习

[{"id":224,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生在计算一个数减去5时，误将减号看成了加号，结果得到20。那么这个数正确的计算结果应该是____。","answer":"10","explanation":"根据题意，某学生把'减去5'误看成'加上5'，得到结果是20。设这个数为x，则有 x + 5 = 20，解得 x = 15。那么正确的计算应是 15 - 5 = 10。因此正确答案是10。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:40:41","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":1300,"subject":"数学","grade":"七年级","stage":"小学","type":"解答题","content":"某城市计划在一条东西走向的主干道旁建设一个矩形公园，公园的边界由四条道路围成。已知公园的东侧边界与主干道平行，且距离主干道120米。公园的北侧边界上有一盏路灯，其位置在平面直角坐标系中表示为点A(3, 8)。公园的南侧边界与北侧边界平行，且南北边界之间的距离为6米。公园的西侧边界是一条直线，经过点B(−2, 5)，且与主干道垂直。现需在公园内部铺设一条从点A正下方地面点C（即点A在x轴上的投影）到点B的步行道，要求步行道为直线段。已知铺设步行道的成本为每米50元，且预算不得超过3000元。请判断该预算是否足够，并说明理由。（注：所有坐标单位均为百米，即1个单位代表100米）","answer":"1. 首先将坐标单位转换为实际距离（米）：点A(3, 8)表示实际位置为(300, 800)米，点B(−2, 5)表示实际位置为(−200, 500)米。\n\n2. 点C是点A在x轴上的投影，因此其坐标为(300, 0)米。\n\n3. 计算步行道长度，即点C(300, 0)到点B(−200, 500)的距离：\n 使用距离公式：\n 距离 = √[(300 − (−200))² + (0 − 500)²]\n = √[(500)² + (−500)²]\n = √[250000 + 250000]\n = √500000\n = 500√2 ≈ 500 × 1.4142 ≈ 707.1米\n\n4. 计算铺设成本：\n 成本 = 707.1 × 50 ≈ 35355元\n\n5. 比较预算：\n 35355元 > 3000元，因此预算不足。\n\n答：该预算不足以铺设步行道，因为所需成本约为35355元，远超3000元的预算。","explanation":"本题综合考查了平面直角坐标系中点的坐标、距离公式、实数运算以及一元一次不等式的实际应用。解题关键在于理解坐标单位的实际意义（1单位=100米），正确确定点C的坐标，并运用勾股定理计算两点间距离。随后通过乘法运算得出总成本，并与预算进行比较，判断是否满足条件。题目融合了坐标几何、实数计算和不等式判断，具有较强的综合性，符合困难难度要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 10:47:48","updated_at":"2026-01-06 10:47:48","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":1695,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市为改善交通状况，计划在一条主干道上设置若干个智能公交站。已知该道路在平面直角坐标系中沿x轴方向延伸，起点坐标为(0, 0)，终点坐标为(12, 0)。规划部门决定在这些站点中设置A、B、C三类站点，其中A类站点每2千米设一个，B类站点每3千米设一个，C类站点每4千米设一个，均从起点开始设置（即起点处同时设有A、B、C三类站点）。若某学生从起点出发，沿道路步行，每经过一个站点就记录一次，问：该学生在到达终点前，共会经过多少个不同的站点？（注：若某位置同时设有多个类型的站点，只算作一个站点）","answer":"1. 确定各类站点的位置：\n - A类站点：每2千米一个，位置为 x = 0, 2, 4, 6, 8, 10, 12\n 共 7 个位置\n - B类站点：每3千米一个，位置为 x = 0, 3, 6, 9, 12\n 共 5 个位置\n - C类站点：每4千米一个，位置为 x = 0, 4, 8, 12\n 共 4 个位置\n\n2. 列出所有站点坐标并去重：\n 合并三类站点的所有x坐标：\n {0, 2, 3, 4, 6, 8, 9, 10, 12}\n 注意：6出现在A和B类，4和12出现在A和C类，0出现在三类中，但每个坐标只算一次\n\n3. 统计不同站点的总数：\n 上述集合中共有 9 个不同的x坐标值\n\n4. 因此，该学生从起点到终点（含起点和终点），共经过 9 个不同的站点\n\n答：该学生共会经过 9 个不同的站点。","explanation":"本题综合考查了平面直角坐标系、有理数（坐标值）、数据的收集与整理（分类统计、去重）以及实际应用建模能力。解题关键在于理解‘不同站点’的含义——即使多个类型站点位于同一位置，也只计为一个物理站点。因此需要分别列出A、B、C三类站点的所有位置，然后合并并去除重复的坐标点。这涉及集合思想的应用，虽然七年级尚未系统学习集合，但通过列表和观察可以实现去重操作。题目背景新颖，结合了城市规划与数学建模，避免了传统行程问题的套路，强调对‘位置唯一性’的理解和数据处理能力，符合困难难度要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 13:39:12","updated_at":"2026-01-06 13:39:12","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":852,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次班级图书整理活动中，某学生统计了同学们捐赠的书籍数量。已知捐赠的数学书比语文书多8本，且两种书共捐赠了36本。设语文书捐赠了x本，则根据题意可列方程为：x + (x + 8) = 36。解这个方程，语文书捐赠了___本。","answer":"14","explanation":"根据题意，语文书为x本，数学书比语文书多8本，即为(x + 8)本。两者总数为36本，因此列出方程：x + (x + 8) = 36。化简得：2x + 8 = 36，移项得：2x = 28，解得：x = 14。所以语文书捐赠了14本。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 01:05:48","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":1062,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"在一次班级环保活动中，某学生收集了废旧纸张和塑料瓶两类物品。若废旧纸张的重量比塑料瓶重量的3倍少2千克，且两类物品总重量为18千克，则塑料瓶的重量是___千克。","answer":"5","explanation":"设塑料瓶的重量为x千克，则废旧纸张的重量为(3x - 2)千克。根据题意，总重量为18千克，可列出一元一次方程：x + (3x - 2) = 18。解这个方程：x + 3x - 2 = 18 → 4x = 20 → x = 5。因此，塑料瓶的重量是5千克。本题考查一元一次方程的实际应用，符合七年级数学课程要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:52:03","updated_at":"2026-01-06 08:52:03","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":1843,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某校八年级开展数学实践活动，测量一座建筑物的高度。一名学生站在距离建筑物底部12米的位置，使用测角仪测得建筑物顶部的仰角为30°。已知该学生的眼睛距离地面1.5米，且测角仪安装在眼睛高度处。若忽略测量误差，则该建筑物的实际高度约为多少米？（结果保留一位小数）","answer":"A","explanation":"本题考查勾股定理与三角函数在实际问题中的应用，属于中等难度。解题思路如下：\n\n1. 建立直角三角形模型：学生眼睛到建筑物底部的水平距离为12米，仰角为30°，建筑物顶部到学生眼睛的视线构成直角三角形的斜边。\n\n2. 设建筑物从学生眼睛高度到顶部的垂直高度为h米，则根据正切函数定义：\n tan(30°) = h \/ 12\n 因为 tan(30°) = √3 \/ 3 ≈ 0.577，\n 所以 h = 12 × (√3 \/ 3) = 4√3 ≈ 4 × 1.732 ≈ 6.928 米。\n\n3. 建筑物的总高度 = h + 学生眼睛离地高度 = 6.928 + 1.5 ≈ 8.428 米。\n\n4. 保留一位小数，得建筑物高度约为 8.4 米。\n\n因此正确答案为 A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-06 16:53:35","updated_at":"2026-01-06 16:53:35","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":"8.4米","is_correct":1},{"id":"B","content":"8.9米","is_correct":0},{"id":"C","content":"9.3米","is_correct":0},{"id":"D","content":"9.8米","is_correct":0}]},{"id":1056,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某班级进行了一次数学测验，成绩整理如下表所示。已知90分及以上为优秀，则该班本次测验的优秀率为___%。（成绩分布：80分以下有6人，80-89分有10人，90-100分有14人）","answer":"46.7","explanation":"首先计算总人数：6 + 10 + 14 = 30人。优秀人数为90-100分的14人。优秀率 = (优秀人数 ÷ 总人数) × 100% = (14 ÷ 30) × 100% ≈ 46.7%。本题考查数据的收集、整理与描述中的百分比计算，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":"2025-12-30 06:42:48","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":759,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级大扫除中，某学生负责统计各小组的垃圾重量。已知第一组收集的垃圾比第二组多3.5千克，两组共收集了12.7千克。设第二组收集的垃圾重量为x千克，则可列出一元一次方程：x + (x + 3.5) = 12.7。解这个方程，第二组收集的垃圾重量为___千克。","answer":"4.6","explanation":"根据题意，设第二组收集的垃圾重量为x千克，则第一组为(x + 3.5)千克。两组共收集12.7千克，因此可列方程：x + (x + 3.5) = 12.7。化简得：2x + 3.5 = 12.7。两边同时减去3.5，得2x = 9.2。再两边同时除以2，得x = 4.6。所以第二组收集的垃圾重量为4.6千克。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 23:29:25","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":397,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次校园植物观察活动中，某学生记录了五种植物一周内每天的生长高度（单位：厘米），并将数据整理如下表。已知这五种植物的平均每日生长高度为1.2厘米，其中四种植物的生长高度分别为0.8、1.0、1.5和1.3厘米，那么第五种植物的每日生长高度是多少？","answer":"C","explanation":"题目考查数据的收集、整理与描述中的平均数计算。已知五种植物的平均每日生长高度为1.2厘米，因此总生长高度为 5 × 1.2 = 6.0 厘米。已知四种植物的生长高度分别为0.8、1.0、1.5和1.3厘米，它们的和为 0.8 + 1.0 + 1.5 + 1.3 = 4.6 厘米。因此第五种植物的生长高度为 6.0 - 4.6 = 1.4 厘米。故正确答案为C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:15:08","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":"1.1厘米","is_correct":0},{"id":"B","content":"1.2厘米","is_correct":0},{"id":"C","content":"1.4厘米","is_correct":1},{"id":"D","content":"1.6厘米","is_correct":0}]},{"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":[]}]