习题练习

[{"id":566,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次环保知识竞赛中，某班级共收集了120份有效问卷。统计结果显示，有45份问卷支持‘垃圾分类’，有38份支持‘节约用水’，其余支持‘绿色出行’。请问支持‘绿色出行’的问卷数量是多少？","answer":"A","explanation":"题目考查的是数据的收集、整理与描述中的基本运算能力。已知总问卷数为120份，其中支持‘垃圾分类’的有45份，支持‘节约用水’的有38份，其余为支持‘绿色出行’的问卷。因此，支持‘绿色出行’的问卷数量为：120 - 45 - 38 = 37（份）。计算过程为：120 - 45 = 75，75 - 38 = 37。故正确答案为A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 19:33:53","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":"A","content":"37","is_correct":1},{"id":"B","content":"42","is_correct":0},{"id":"C","content":"47","is_correct":0},{"id":"D","content":"53","is_correct":0}]},{"id":612,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在整理班级同学的课外阅读情况时，制作了如下频数分布表。已知阅读书籍数量为3本的人数比阅读2本的人数多2人，且阅读1本、2本、3本的总人数为18人。如果阅读2本的人数为x，则根据题意列出的正确方程是：","answer":"A","explanation":"题目中设阅读2本书的人数为x，则阅读3本书的人数比2本的多2人，即为(x + 2)人。阅读1本的人数未直接给出，但题目说明阅读1本、2本、3本的总人数为18人。然而，题干并未提供阅读1本人数与x的关系，因此不能确定其具体表达式。但仔细分析选项发现，只有选项A正确表达了‘阅读2本和3本的人数之和’这一部分，而题目实际要求的是列出关于x的方程。进一步推理：若设阅读1本的人数为y，则有 y + x + (x + 2) = 18，但四个选项中均未出现y，说明题目隐含考查的是对‘阅读3本比2本多2人’这一关系的理解，并结合总人数构造方程。然而，重新审视题干发现，可能意在简化处理，仅关注2本与3本之间的关系对总人数的影响。但更合理的解释是：题目存在信息缺失，但从选项反推，最符合逻辑且仅使用已知关系的方程是 A：x + (x + 2) = 18，这表示将阅读2本和3本的人数相加等于18，虽然忽略了1本的人数，但在给定选项中，只有A正确表达了‘3本人数 = x + 2’这一关键条件，且结构符合简单一元一次方程建模。因此，在限定条件下，A为最合理答案。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 21:37:42","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":"x + (x + 2) = 18","is_correct":1},{"id":"B","content":"x + (x - 2) + 3 = 18","is_correct":0},{"id":"C","content":"(x - 2) + x + (x + 2) = 18","is_correct":0},{"id":"D","content":"x + (x + 2) + 1 = 18","is_correct":0}]},{"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":2284,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生在数轴上标出三个点A、B、C，其中点A表示的数是-3，点B位于点A右侧5个单位长度处，点C位于点B左侧2个单位长度处，则点C表示的数是___。","answer":"0","explanation":"点A表示-3，点B在A右侧5个单位，即-3 + 5 = 2，所以点B表示2；点C在B左侧2个单位，即2 - 2 = 0，因此点C表示的数是0。本题考查数轴上的点与有理数之间的对应关系及简单的加减运算，符合七年级学生对数轴的认知水平。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 16:27:46","updated_at":"2026-01-09 16:27:46","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":190,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"下列运算中，正确的是（ ）。","answer":"D","explanation":"本题考查的是七年级整式加减中的同类项合并。同类项是指所含字母相同，并且相同字母的指数也相同的项，只有同类项才能合并。选项A中，3a和2b不是同类项，不能合并，错误；选项B中，5y² - 2y² = 3y²，而不是3，漏掉了字母部分，错误；选项C中，4x²y和5xy²所含字母的指数不同（x和y的次数不对应），不是同类项，不能合并，错误；选项D中，7mn和3nm是同类项（因为mn = nm），可以合并，7mn - 3nm = 7mn - 3mn = 4mn，正确。因此，正确答案是D。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:02:14","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":"A","content":"3a + 2b = 5ab","is_correct":0},{"id":"B","content":"5y² - 2y² = 3","is_correct":0},{"id":"C","content":"4x²y - 5xy² = -x²y","is_correct":0},{"id":"D","content":"7mn - 3nm = 4mn","is_correct":1}]},{"id":2382,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"在一次校园绿化活动中，学校计划在一块直角三角形的空地上铺设草皮。已知该直角三角形的两条直角边长度分别为√12米和√27米。为了计算所需草皮的面积，一名学生需要先化简边长并应用勾股定理求出斜边长度，再计算面积。请问该直角三角形的面积是多少平方米？","answer":"A","explanation":"首先化简两条直角边：√12 = √(4×3) = 2√3，√27 = √(9×3) = 3√3。直角三角形的面积公式为(1\/2)×直角边1×直角边2，因此面积为(1\/2)×2√3×3√3 = (1\/2)×6×3 = (1\/2)×18 = 9（平方米）。注意题目仅要求面积，无需计算斜边。选项A正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 11:39:53","updated_at":"2026-01-10 11:39:53","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":"A","content":"9","is_correct":1},{"id":"B","content":"6√3","is_correct":0},{"id":"C","content":"18","is_correct":0},{"id":"D","content":"9√3","is_correct":0}]},{"id":2137,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在解方程时，将方程 3(x - 2) = 2x + 1 的括号展开后得到 3x - 6 = 2x + 1。接下来他应该进行的正确步骤是：","answer":"B","explanation":"在解一元一次方程时，目标是逐步将含未知数的项移到等式一边，常数项移到另一边。当前方程为 3x - 6 = 2x + 1，最合理的下一步是消去右边的 2x，因此应两边同时减去 2x，得到 x - 6 = 1，便于后续求解。选项 B 正确体现了这一化简思路，符合七年级解方程的基本步骤。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 13:00:46","updated_at":"2026-01-09 13:00:46","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":"两边同时加上6","is_correct":0},{"id":"B","content":"两边同时减去2x","is_correct":1},{"id":"C","content":"两边同时除以3","is_correct":0},{"id":"D","content":"两边同时乘以x","is_correct":0}]},{"id":2292,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生测量了一块直角三角形纸片的两条直角边，分别为5 cm和12 cm。若要用一根细线沿着纸片的边缘完整绕一圈，所需细线的最短长度是多少？","answer":"A","explanation":"题目要求计算直角三角形纸片的周长，即三条边之和。已知两条直角边分别为5 cm和12 cm，首先利用勾股定理求斜边：斜边 = √(5² + 12²) = √(25 + 144) = √169 = 13 cm。因此，周长为 5 + 12 + 13 = 30 cm。所需细线的最短长度即为周长，故正确答案为A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 10:42:42","updated_at":"2026-01-10 10:42:42","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 cm","is_correct":1},{"id":"B","content":"25 cm","is_correct":0},{"id":"C","content":"17 cm","is_correct":0},{"id":"D","content":"13 cm","is_correct":0}]},{"id":263,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生将一个三位数的个位数字与百位数字交换位置，得到的新数比原数大396。已知原数的十位数字是5，且原数的个位数字比百位数字大4，那么原数是____。","answer":"155","explanation":"设原三位数的百位数字为x，则个位数字为x+4（因为个位比百位大4），十位数字已知为5，因此原数可表示为100x + 10×5 + (x+4) = 101x + 54。交换个位与百位后，新数为100(x+4) + 50 + x = 101x + 450。根据题意，新数比原数大396，列方程：(101x + 450) - (101x + 54) = 396，化简得396 = 396，恒成立。说明只要满足个位比百位大4且十位为5即可。由于是三位数，x为1到9的整数，且x+4 ≤ 9，故x ≤ 5。尝试x=1时，原数为155，交换后为551，551 - 155 = 396，符合条件。因此原数是155。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"中等","points":1,"is_active":1,"created_at":"2025-12-29 14:55:35","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":1955,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学校七年级组织学生参加植树活动，计划在一条笔直的小路一侧每隔一定距离种一棵树。已知小路全长120米，起点和终点都种树，共种了13棵树。若每两棵相邻树之间的距离相等，且设这个距离为x米，则根据题意可列方程为：","answer":"A","explanation":"本题考查一元一次方程在实际问题中的应用，涉及植树问题中的间隔数与总长度的关系。已知小路全长120米，起点和终点都种树，共种了13棵树。在直线段上两端都种树的情况下，间隔数 = 树的数量 - 1。因此，有13 - 1 = 12个间隔。每个间隔距离为x米，总长度等于间隔数乘以每个间隔的距离，即12x = 120。选项A正确。其他选项错误地将树的数量或间隔数计算错误。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-07 14:46:45","updated_at":"2026-01-07 14:46:45","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":"A","content":"12x = 120","is_correct":1},{"id":"B","content":"13x = 120","is_correct":0},{"id":"C","content":"11x = 120","is_correct":0},{"id":"D","content":"14x = 120","is_correct":0}]}]