习题练习

[{"id":2547,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"如图，在平面直角坐标系中，抛物线 y = x² - 4x + 3 与反比例函数 y = k\/x 的图像在第一象限内有一个公共点 P，且点 P 到 x 轴的距离为 1。若将该抛物线绕其顶点旋转 180°，得到新的抛物线，则新抛物线与反比例函数图像的交点个数为多少？","answer":"B","explanation":"首先，求原抛物线 y = x² - 4x + 3 的顶点：配方得 y = (x - 2)² - 1，顶点为 (2, -1)。点 P 在第一象限且在抛物线上，且到 x 轴距离为 1，即纵坐标为 1。代入抛物线方程：1 = x² - 4x + 3，解得 x² - 4x + 2 = 0，解得 x = 2 ± √2。因在第一象限，取 x = 2 + √2，故 P(2 + √2, 1)。又 P 在反比例函数 y = k\/x 上，代入得 k = x·y = (2 + √2)·1 = 2 + √2，故反比例函数为 y = (2 + √2)\/x。将原抛物线绕顶点 (2, -1) 旋转 180°，其开口方向反向，形状不变，新抛物线方程为 y = -(x - 2)² - 1 = -x² + 4x - 5。联立新抛物线与反比例函数：-x² + 4x - 5 = (2 + √2)\/x，两边乘以 x（x ≠ 0）得：-x³ + 4x² - 5x = 2 + √2，即 -x³ + 4x² - 5x - (2 + √2) = 0。此三次方程在实数范围内分析图像趋势：当 x → 0⁺ 时，左边 → -∞；当 x → +∞ 时，-x³ 主导，→ -∞；在 x = 2 附近函数值变化分析可知，函数图像仅穿过 x 轴一次，故仅有一个实数解。因此，新抛物线与反比例函数图像有 1 个交点。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 17:02:12","updated_at":"2026-01-10 17:02:12","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":"A","content":"0 个","is_correct":0},{"id":"B","content":"1 个","is_correct":1},{"id":"C","content":"2 个","is_correct":0},{"id":"D","content":"3 个","is_correct":0}]},{"id":2207,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在一条东西走向的直线上做标记，规定向东为正方向。他从原点出发，先向东走了5米，记作+5米，接着又向西走了8米。此时他的位置相对于原点的方向和距离应如何表示？","answer":"B","explanation":"向东走5米记作+5，向西走8米记作-8。总位移为+5 + (-8) = -3，表示最终位于原点西侧3米处，应记作-3米。因此正确答案是B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 14:25:31","updated_at":"2026-01-09 14:25:31","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":"向东3米，记作+3米","is_correct":0},{"id":"B","content":"向西3米，记作-3米","is_correct":1},{"id":"C","content":"向东13米，记作+13米","is_correct":0},{"id":"D","content":"向西13米，记作-13米","is_correct":0}]},{"id":145,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"已知一个三角形的两边长分别为3cm和7cm，第三边的长度可能是以下哪一个？","answer":"B","explanation":"根据三角形三边关系定理：任意两边之和大于第三边，任意两边之差小于第三边。设第三边为x，则需满足：7 - 3 < x < 7 + 3，即4 < x < 10。选项中只有5cm在这个范围内，因此正确答案是B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-24 11:30:06","updated_at":"2025-12-24 11:30:06","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":"3cm","is_correct":0},{"id":"B","content":"5cm","is_correct":1},{"id":"C","content":"10cm","is_correct":0},{"id":"D","content":"11cm","is_correct":0}]},{"id":314,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某学生在平面直角坐标系中描出点 A(3, 4) 和点 B(-2, 1)，他想知道线段 AB 的中点坐标是多少。根据中点坐标公式，正确的结果是：","answer":"A","explanation":"根据平面直角坐标系中两点 A(x₁, y₁) 和 B(x₂, y₂) 的中点坐标公式：中点坐标为 ((x₁ + x₂)\/2, (y₁ + y₂)\/2)。将点 A(3, 4) 和点 B(-2, 1) 代入公式，横坐标为 (3 + (-2))\/2 = 1\/2 = 0.5，纵坐标为 (4 + 1)\/2 = 5\/2 = 2.5。因此，中点坐标为 (0.5, 2.5)，对应选项 A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:36:11","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":"(0.5, 2.5)","is_correct":1},{"id":"B","content":"(1, 5)","is_correct":0},{"id":"C","content":"(2.5, 0.5)","is_correct":0},{"id":"D","content":"(5, 1)","is_correct":0}]},{"id":764,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次班级图书角整理活动中，某学生统计了上周借阅图书的情况：借阅科普类图书的有12人次，借阅文学类图书的有18人次，两类都借阅的有5人次。那么，上周实际参与借阅图书的学生至少有___人。","answer":"25","explanation":"本题考查数据的收集、整理与描述中的集合思想。根据容斥原理，至少参与借阅的学生人数 = 借阅科普类人数 + 借阅文学类人数 - 两类都借阅的人数。即：12 + 18 - 5 = 25（人）。因为‘两类都借阅’的学生被重复计算了一次，所以需要减去一次重复部分，才能得到实际最少参与人数。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 23:39:50","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":634,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"13道","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 21:58:04","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":2150,"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 2x + 5 = 13 的两边同时减去5，得到 2x = 8，然后再将两边同时除以2，得到 x = 4。这名学生使用的解题方法体现了等式的哪一条基本性质？","answer":"D","explanation":"该学生先对等式两边同时减去5，再同时除以2，整个过程体现了对等式两边进行相同运算时，等式依然成立这一基本性质。虽然选项B和C分别描述了其中一步所依据的性质，但整个解题过程综合体现了等式的基本性质：等式两边同时进行相同的运算（加、减、乘、除同一个数，除数不为零），等式仍然成立。因此，最全面且准确的答案是D。","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":"等式两边同时加上同一个数，等式仍然成立","is_correct":0},{"id":"B","content":"等式两边同时减去同一个数，等式仍然成立","is_correct":0},{"id":"C","content":"等式两边同时乘或除以同一个不为零的数，等式仍然成立","is_correct":0},{"id":"D","content":"等式两边同时进行相同的运算，等式仍然成立","is_correct":1}]},{"id":631,"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。在整理数据时，发现有15%的学生选择了‘垃圾分类’作为最关注的环保问题，有40人选择了‘节约用水’，其余学生选择了‘减少塑料使用’。请问选择‘减少塑料使用’的学生人数是多少？","answer":"C","explanation":"首先计算选择‘垃圾分类’的学生人数：120 × 15% = 120 × 0.15 = 18人。已知选择‘节约用水’的有40人。那么选择‘减少塑料使用’的人数为总人数减去前两项：120 - 18 - 40 = 62人。因此正确答案是C。本题考查数据的收集与整理，涉及百分数的基本计算和简单减法运算，符合七年级数学中‘数据的收集、整理与描述’知识点，难度为简单。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 21:55:43","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":"52","is_correct":0},{"id":"B","content":"58","is_correct":0},{"id":"C","content":"62","is_correct":1},{"id":"D","content":"68","is_correct":0}]},{"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":2509,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生设计了一个圆形花坛，花坛中心有一根垂直的灯柱。灯柱顶端投射出的光线在地面上形成一个圆锥形的照明区域。已知灯柱高为3米，光线与地面的夹角为60°，则照明区域在地面上的圆形半径是多少米？","answer":"A","explanation":"本题考查锐角三角函数的应用。灯柱垂直于地面，高度为3米，光线与地面夹角为60°，即光线与灯柱之间的夹角为30°。在由灯柱、地面半径和光线构成的直角三角形中，灯柱为邻边，地面半径为对边，夹角为30°。利用正切函数：tan(30°) = 对边 \/ 邻边 = r \/ 3。因为 tan(30°) = √3 \/ 3，所以 r = 3 × (√3 \/ 3) = √3。因此，照明区域的半径为√3米，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 15:33:23","updated_at":"2026-01-10 15:33:23","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":"A","content":"√3","is_correct":1},{"id":"B","content":"3","is_correct":0},{"id":"C","content":"3√3","is_correct":0},{"id":"D","content":"6","is_correct":0}]}]