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[{"id":1301,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市计划在一条笔直的主干道旁建设一个矩形公园,公园的一边紧邻道路,因此不需要围栏。其余三边需要用总长为120米的围栏围起来。为了便于管理,公园被划分为两个面积相等的矩形区域,中间用一道与道路垂直的围栏隔开。已知公园的长(平行于道路的一边)比宽(垂直于道路的一边)多20米。现需在该公园内设置若干个边长为2米的正方形花坛,要求花坛之间至少间隔1米,且花坛不能超出公园边界。若每平方米种植成本为50元,且预算为30000元,问:该公园最多可以设置多少个这样的正方形花坛?并验证总种植成本是否在预算范围内。","answer":"设公园的宽为x米(垂直于道路),则长为x + 20米(平行于道路)。\n\n由于公园一边靠路,其余三边加中间一道隔断共需围栏:两条宽和两条长(因为中间隔断与宽同向,增加一条宽的长度)。\n\n围栏总长为:x + x + (x + 20) + x = 4x + 20\n\n根据题意,围栏总长为120米:\n4x + 20 = 120\n4x = 100\nx = 25\n\n所以宽为25米,长为25 + 20 = 45米。\n\n公园总面积为:45 × 25 = 1125 平方米。\n\n每个正方形花坛边长为2米,面积为4平方米。\n\n花坛之间至少间隔1米,且不能靠边(隐含条件:花坛边缘距离公园边界至少0.5米?但题目未明确,故按常规理解:花坛可贴边放置,但彼此之间中心距至少3米,即边缘间距1米)。\n\n更合理的建模是:将每个花坛视为占据一个2×2的区域,并在其四周预留1米间隔。但为避免复杂化,采用网格布局法。\n\n考虑沿长度方向(45米)和宽度方向(25米)布置花坛。\n\n每个花坛占2米,间隔1米,即每个花坛及其右侧\/上侧间隔共占3米,但最后一个花坛后无需间隔。\n\n沿长度方向(45米):设可放n个花坛,则所需长度为:2n + 1×(n - 1) = 3n - 1 ≤ 45\n→ 3n ≤ 46 → n ≤ 15.33 → 最多15个\n验证:3×15 - 1 = 44 ≤ 45,成立。\n\n沿宽度方向(25米):同理,2m + 1×(m - 1) = 3m - 1 ≤ 25\n→ 3m ≤ 26 → m ≤ 8.66 → 最多8个\n验证:3×8 - 1 = 23 ≤ 25,成立。\n\n因此最多可布置:15 × 8 = 120 个花坛。\n\n总种植面积:120 × 4 = 480 平方米。\n\n总种植成本:480 × 50 = 24000 元。\n\n24000 < 30000,在预算范围内。\n\n答案:最多可以设置120个正方形花坛,总种植成本为24000元,在预算范围内。","explanation":"本题综合考查了一元一次方程、几何图形初步、不等式与不等式组以及数据的整理与应用。首先通过建立一元一次方程求出公园的长和宽,利用围栏总长条件解得尺寸。然后结合几何布局思想,分析花坛在矩形区域内的最大排列数量,需考虑间隔约束,转化为不等式问题。最后计算总成本和预算比较,体现数学建模能力。难点在于将实际空间布局问题抽象为数学模型,并正确处理间隔对排列数量的影响。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 10:47:59","updated_at":"2026-01-06 10:47:59","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":1689,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市计划在一条笔直的主干道两侧安装新型节能路灯。道路起点为坐标原点O(0, 0),终点为点A(120, 0),单位为米。路灯必须安装在道路两侧,且每侧路灯的位置关于x轴对称。设计要求如下:\n\n1. 每侧路灯之间的间距必须相等,且为整数米;\n2. 起点和终点都必须安装路灯;\n3. 每侧至少安装6盏路灯(含起点和终点);\n4. 为了美观,两侧路灯在垂直于道路的方向上对齐,即若一侧某盏灯位于(x, y),则另一侧对应灯位于(x, -y),其中y > 0;\n5. 所有路灯的纵坐标y必须满足不等式:2y + 3 ≤ 15;\n6. 若某学生提出安装方案中每侧安装n盏灯,则总灯数为2n,且n必须满足方程:3(n - 4) = 2n - 5。\n\n请根据以上条件,求出:\n(1) 每侧应安装多少盏路灯?\n(2) 相邻两盏路灯之间的间距是多少米?\n(3) 每盏路灯的纵坐标y的最大可能值是多少?\n(4) 若每盏灯的照明范围是以灯为中心、半径为10米的圆,问整条道路是否被完全覆盖?说明理由。","answer":"(1) 设每侧安装n盏路灯。根据条件6,列出方程:\n3(n - 4) = 2n - 5\n展开左边:3n - 12 = 2n - 5\n移项得:3n - 2n = -5 + 12\n解得:n = 7\n所以每侧应安装7盏路灯。\n\n(2) 道路总长为120米,起点和终点都安装灯,共7盏灯,则有6个间隔。\n间距 = 120 ÷ (7 - 1) = 120 ÷ 6 = 20(米)\n所以相邻两盏路灯之间的间距是20米。\n\n(3) 由条件5:2y + 3 ≤ 15\n解不等式:2y ≤ 12 → y ≤ 6\n由于y > 0且为实数,最大可能值为6。\n所以每盏路灯的纵坐标y的最大可能值是6米。\n\n(4) 每盏灯照明半径为10米,即覆盖范围为以灯为中心、直径20米的圆。\n相邻灯间距为20米,恰好等于照明直径,因此在道路方向上,照明范围刚好相接,无重叠也无空隙。\n但由于路灯安装在道路两侧,且关于x轴对称,每盏灯到道路中心线(x轴)的距离为y ≤ 6米。\n灯到道路最远点(如正上方或正下方)的垂直距离为y,而照明半径为10米,因此只要y ≤ 10,道路横向即可被覆盖。\n由于y ≤ 6 < 10,每盏灯在垂直方向上足以覆盖整个道路宽度(假设道路宽度不超过12米,题目隐含道路在x轴附近)。\n又因在道路长度方向上,灯间距等于照明直径,覆盖连续。\n因此,整条道路被完全覆盖。\n答:是,整条道路被完全覆盖。","explanation":"本题综合考查了一元一次方程、不等式、平面直角坐标系和实际问题的建模能力。第(1)问通过建立并求解一元一次方程确定灯的数量;第(2)问利用线段分段模型计算间距;第(3)问解一元一次不等式求最大值;第(4)问结合几何图形初步与实际应用,分析圆的覆盖范围与空间位置关系,要求学生理解对称性、距离与覆盖的逻辑。题目情境新颖,融合多个知识点,强调数学建模与逻辑推理,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 13:35:50","updated_at":"2026-01-06 13:35:50","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":2267,"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之间的距离为7个单位长度,且点B位于点A的右侧。现在将点B向左移动4个单位长度到达点C,再将点C向右移动2个单位长度到达点D。那么点D表示的数是多少?","answer":"B","explanation":"首先,点A表示-3,点B在点A右侧且距离为7,因此点B表示的数是-3 + 7 = 4。将点B向左移动4个单位,到达点C,即4 - 4 = 0。再将点C向右移动2个单位,到达点D,即0 + 2 = 2。因此点D表示的数是2,正确答案是B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-09 16:09:15","updated_at":"2026-01-09 16:09:15","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":"2","is_correct":1},{"id":"C","content":"4","is_correct":0},{"id":"D","content":"6","is_correct":0}]},{"id":242,"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,将原数乘以 -1,得到的结果是 7,那么这个数是____。","answer":"-7","explanation":"根据相反数的定义,一个数的相反数等于这个数乘以 -1。题目中说乘以 -1 后得到 7,说明原数 × (-1) = 7。解这个等式可得:原数 = 7 ÷ (-1) = -7。因此,这个数是 -7。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:42:00","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":2245,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学生在研究温度变化时,记录了连续7天的每日最低气温(单位:℃),这些数据分别为:-3,2,-5,0,-1,4,-2。该学生想计算这7天中,气温低于零度的天数占总天数的几分之几,并进一步求出这些负温度的绝对值的平均数。请完成以下两个任务:(1) 求出气温低于零度的天数占总天数的几分之几(结果用最简分数表示);(2) 求出所有负温度的绝对值的平均数(结果保留一位小数)。","answer":"(1) 4\/7;(2) 2.8","explanation":"本题综合考查了正数、负数的识别,绝对值的概念,以及分数和平均数的计算。七年级学生已掌握负数的意义、绝对值的求法以及基本统计量的计算。题目通过真实情境(气温记录)引导学生分析数据,区分正负数,并进行多步运算,体现了数学在实际生活中的应用,难度较高,符合困难级别要求。","solution_steps":"第一步:确定气温低于零度的天数。观察数据:-3,2,-5,0,-1,4,-2。其中小于0的数有:-3,-5,-1,-2,共4天。总天数为7天,因此所求分数为4\/7,已是最简分数。第二步:找出所有负温度:-3,-5,-1,-2。求它们的绝对值:| -3 | = 3,| -5 | = 5,| -1 | = 1,| -2 | = 2。第三步:计算这些绝对值的和:3 + 5 + 1 + 2 = 11。第四步:求平均数:11 ÷ 4 = 2.75,保留一位小数为2.8。","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":[]},{"id":932,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级环保活动中,某学生记录了5个小组每周回收的废纸重量(单位:千克),分别为:3.5、4.2、3.8、4.0、4.5。为了计算平均每个小组回收的废纸重量,需要先求出总重量,再除以小组数量。那么这5个小组平均每周回收废纸____千克。","answer":"4.0","explanation":"首先将5个小组回收的废纸重量相加:3.5 + 4.2 + 3.8 + 4.0 + 4.5 = 20.0(千克)。然后将总重量除以小组数量5:20.0 ÷ 5 = 4.0(千克)。因此,平均每个小组每周回收废纸4.0千克。本题考查数据的收集与整理中的平均数计算,属于简单难度的基础运算。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 03:03:40","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":1097,"subject":"数学","grade":"七年级","stage":"小学","type":"填空题","content":"某学生在整理班级同学的身高数据时,发现将每个人的身高(单位:厘米)减去150后,得到的新数据中,最小值为-8,最大值为12。那么原始身高数据中,最矮的同学身高是____厘米。","answer":"142","explanation":"题目中给出的是每个身高减去150后的结果,最小值为-8。这意味着最矮的同学的身高比150厘米少8厘米。因此,原始身高为150 - 8 = 142厘米。这是基于有理数中的减法运算和数据的简单变换,符合七年级‘数据的收集、整理与描述’以及‘有理数’的知识点。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:56:53","updated_at":"2026-01-06 08:56:53","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":1344,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某校七年级开展‘校园绿化优化’项目,计划在长方形花坛ABCD中种植花卉。花坛长12米,宽8米,现需在花坛内部修建两条相互垂直的小路:一条平行于长边,一条平行于宽边,且两条小路宽度相同,均为x米。修建后,剩余种植区域的面积为60平方米。已知小路的交叉部分只计算一次面积。若设小路宽度为x米,请根据题意列出方程并求出x的值。此外,若规定小路宽度不得超过花坛较短边长度的1\/4,判断所求得的解是否符合实际要求。","answer":"解:\n\n1. 花坛总面积为:12 × 8 = 96(平方米)\n\n2. 修建两条小路后,剩余种植面积为60平方米,因此两条小路总占地面积为:\n 96 - 60 = 36(平方米)\n\n3. 设小路宽度为x米。\n - 平行于长边(12米)的小路面积为:12x\n - 平行于宽边(8米)的小路面积为:8x\n - 两条小路交叉部分是一个边长为x的正方形,面积为:x²\n - 由于交叉部分被重复计算了一次,因此两条小路的实际总面积为:\n 12x + 8x - x² = 20x - x²\n\n4. 根据题意,小路总面积为36平方米,列方程:\n 20x - x² = 36\n\n5. 整理方程:\n -x² + 20x - 36 = 0\n 两边同乘以-1,得:\n x² - 20x + 36 = 0\n\n6. 解这个一元二次方程(可用因式分解):\n 寻找两个数,乘积为36,和为20:\n 18 和 2 满足条件(18 × 2 = 36,18 + 2 = 20)\n 所以方程可分解为:\n (x - 18)(x - 2) = 0\n\n7. 解得:x = 18 或 x = 2\n\n8. 检验解的合理性:\n - 花坛宽为8米,若x = 18,则小路宽度超过花坛宽度,不符合实际,舍去。\n - 若x = 2,则小路宽度为2米,合理。\n\n9. 检查是否满足‘小路宽度不得超过花坛较短边长度的1\/4’:\n 较短边为8米,其1\/4为:8 ÷ 4 = 2(米)\n x = 2 ≤ 2,满足要求。\n\n答:小路宽度x的值为2米,且符合实际要求。","explanation":"本题综合考查了一元一次方程的建立与求解、整式的加减运算以及实际问题的数学建模能力。题目通过‘校园绿化’这一真实情境,引导学生将几何图形面积计算与代数方程结合。关键在于理解两条垂直小路交叉部分面积不能重复计算,因此总面积应为两条小路面积之和减去重叠的正方形面积。列方程后转化为一元二次方程,但因七年级尚未系统学习一元二次方程求根公式,故设计为可因式分解的形式,符合七年级知识范围。最后结合实际意义和附加约束条件进行解的检验,体现了数学应用的严谨性。题目涉及几何图形初步、整式加减、一元一次方程建模及不等式判断,难度较高,适合学有余力的学生挑战。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:02:45","updated_at":"2026-01-06 11:02:45","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":559,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"18","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 19:22:23","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":18,"subject":"地理","grade":"初一","stage":"初中","type":"选择题","content":"世界上面积最大的洲是?","answer":"A","explanation":"亚洲是世界上面积最大、人口最多的大洲。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-08-29 16:33:04","updated_at":"2025-08-29 16:33:04","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":"亚洲","is_correct":1},{"id":"B","content":"非洲","is_correct":0},{"id":"C","content":"北美洲","is_correct":0},{"id":"D","content":"南美洲","is_correct":0}]}]