1
/ 10
中等
第 1 题
已完成 0 题
练习设置
题目内容
数学
初中
选择题
请选择一道题目开始练习
我的笔记
答案解析
请选择一道题目查看解析
练习完成!
恭喜您完成了本次练习
0
总题数
0
正确
0
错误
答题卡
[{"id":598,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某次数学测验中,某班级共有40名学生参加,其中男生人数是女生人数的1.5倍。设女生人数为x,则根据题意可以列出方程:","answer":"B","explanation":"题目中设女生人数为x,男生人数是女生的1.5倍,因此男生人数为1.5x。全班总人数为男生和女生人数之和,即 x + 1.5x = 40。这个方程正确表达了总人数为40人的条件。选项A错误地将倍数当作具体人数相加;选项C表示的是男女生人数差,不符合题意;选项D将女生人数与倍数关系倒置,也不正确。因此正确答案是B。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 21:00:13","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 + 1.5 = 40","is_correct":0},{"id":"B","content":"x + 1.5x = 40","is_correct":1},{"id":"C","content":"1.5x - x = 40","is_correct":0},{"id":"D","content":"x ÷ 1.5 = 40","is_correct":0}]},{"id":319,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"8人","answer":"答案待完善","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 15:37:32","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":207,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生在计算一个数的相反数时,将原数加上了3,结果得到了0,那么这个数是____。","answer":"-3","explanation":"设这个数为x。根据题意,某学生计算相反数时错误地将原数加上了3,得到结果为0,因此可以列出方程:x + 3 = 0。解这个方程,两边同时减去3,得到x = -3。所以这个数是-3。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:39:39","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":2141,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在解方程 3(x - 2) = 2x + 1 时,第一步去括号得到 3x - 6 = 2x + 1,第二步移项得到 3x - 2x = 1 + 6,第三步合并同类项得到 x = 7。该学生解题过程中哪一步开始出现错误?","answer":"D","explanation":"该学生解题过程完全正确:第一步去括号符合乘法分配律,3(x - 2) = 3x - 6;第二步移项将含x项移到左边,常数项移到右边,符号变换正确;第三步合并同类项得到 x = 7,代入原方程验证成立。因此整个解答过程无误。","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":"第一步","is_correct":0},{"id":"B","content":"第二步","is_correct":0},{"id":"C","content":"第三步","is_correct":0},{"id":"D","content":"没有错误,解答正确","is_correct":1}]},{"id":2763,"subject":"历史","grade":"七年级","stage":"初中","type":"选择题","content":"唐朝时期,长安城是当时世界上最大的城市之一,也是中外文化交流的重要中心。许多外国使节、商人和留学生来到长安,带来了异域的文化和商品。以下哪一项最能体现唐朝长安城作为中外文化交流中心的特点?","answer":"B","explanation":"本题考查唐朝中外交流的特点,重点在于理解长安城作为国际大都市的文化包容性。选项B正确,因为史料记载,唐朝长安城内有大量来自波斯(今伊朗)、大食(阿拉伯帝国)等地的商人,同时存在景教(基督教聂斯脱利派)、祆教(拜火教)等外来宗教的寺庙,这直接体现了中外文化在长安的交融。选项A错误,因为市舶司是宋朝设立的机构,唐朝并未设置;选项C描述的是城市管理制度,虽符合史实,但不直接体现‘中外交流’;选项D强调的是政治功能,与文化交流无关。因此,B项最能体现长安作为中外文化交流中心的特点。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-12 10:40:03","updated_at":"2026-01-12 10:40:03","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":"A","content":"选项A","is_correct":0},{"id":"B","content":"选项B","is_correct":1},{"id":"C","content":"选项C","is_correct":0},{"id":"D","content":"选项D","is_correct":0}]},{"id":511,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"4题","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 18:16:19","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":2176,"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 -2.5,点 B 位于点 A 右侧 4 个单位长度处,点 C 位于点 B 左侧 1.5 个单位长度处。那么点 C 所表示的有理数是:","answer":"D","explanation":"点 A 表示 -2.5,点 B 在其右侧 4 个单位,因此点 B 表示的数是 -2.5 + 4 = 1.5。点 C 在点 B 左侧 1.5 个单位,所以点 C 表示的数是 1.5 - 1.5 = 0.5。该题综合考查了有理数在数轴上的表示及加减运算,符合七年级有理数章节的学习要求。","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":"-4","is_correct":0},{"id":"B","content":"0","is_correct":0},{"id":"C","content":"1","is_correct":0},{"id":"D","content":"0.5","is_correct":1}]},{"id":2483,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"一个圆形花坛被均匀划分为6个扇形区域,分别种植不同颜色的花。若将整个花坛绕其中心顺时针旋转60°,则每个扇形区域会与原来相邻的下一个区域重合。现在随机选择一个点落在花坛上,该点落在红色区域的概率是1\/6。若花坛旋转两次(每次60°),则该点最终落在红色区域的概率是多少?","answer":"A","explanation":"由于花坛被均匀分为6个扇形,每个区域占1\/6的面积,且旋转是绕中心进行的刚体变换,不改变区域的面积和分布。每次顺时针旋转60°,相当于将整个图案向右移动一个扇形位置。旋转两次共120°,即移动两个位置,但整个图案的结构保持不变,每个颜色区域仍然占据1\/6的面积。因此,无论旋转多少次(只要旋转角度是60°的整数倍),每个颜色区域在整体中所占比例不变。所以,随机点落在红色区域的概率始终是1\/6。本题考查的是旋转对称性与概率初步的结合,强调几何变换不改变面积比例这一核心思想。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:10:16","updated_at":"2026-01-10 15:10:16","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\/6","is_correct":1},{"id":"B","content":"1\/3","is_correct":0},{"id":"C","content":"1\/2","is_correct":0},{"id":"D","content":"选项D","is_correct":0}]},{"id":2537,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"一个圆柱形水杯的底面半径为3 cm,高为10 cm。若将杯中的水倒入一个底面为正方形的透明棱柱形容器中,水面高度恰好为6 cm。已知该棱柱形容器的底面边长为5 cm,问原水杯中的水占其总容积的几分之几?","answer":"A","explanation":"首先计算圆柱水杯的总体积:V_圆柱 = π × r² × h = π × 3² × 10 = 90π (cm³)。\n然后计算倒入棱柱形容器中水的体积:V_水 = 底面积 × 高 = 5 × 5 × 6 = 150 (cm³)。\n由于水的体积不变,因此原水杯中水的体积为150 cm³。\n所求比例为:150 \/ (90π) ≈ 150 \/ (90 × 3.14) ≈ 150 \/ 282.6 ≈ 0.53。\n但更精确地,我们保留π符号进行分数化简:150 \/ (90π) = 5 \/ (3π)。然而题目选项为有理数,说明应使用近似值或题目隐含π取3。\n若按π ≈ 3计算,则总体积为90 × 3 = 270 cm³,比例为150 \/ 270 = 5\/9。\n因此正确答案为A。本题考查圆柱与棱柱体积计算及比例关系,属于简单难度,符合九年级‘圆’与‘投影与视图’中立体图形体积的应用。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 16:35:21","updated_at":"2026-01-10 16:35:21","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\/9","is_correct":1},{"id":"B","content":"2\/3","is_correct":0},{"id":"C","content":"5\/6","is_correct":0},{"id":"D","content":"3\/5","is_correct":0}]},{"id":391,"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,随机抽取了50名学生进行调查,发现其中喜欢阅读小说的有28人,喜欢阅读科普书的有15人,两种都不喜欢的有10人。那么既喜欢阅读小说又喜欢阅读科普书的学生至少有多少人?","answer":"A","explanation":"总人数为50人,两种都不喜欢的有10人,因此至少喜欢一种书的学生有50 - 10 = 40人。设既喜欢小说又喜欢科普书的学生人数为x。根据容斥原理,喜欢小说或科普书的人数 = 喜欢小说的人数 + 喜欢科普书的人数 - 两者都喜欢的人数。即:28 + 15 - x = 40。解得:43 - x = 40,所以x = 3。因此,既喜欢阅读小说又喜欢阅读科普书的学生至少有3人。选项A正确。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:13:28","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":"3人","is_correct":1},{"id":"B","content":"5人","is_correct":0},{"id":"C","content":"8人","is_correct":0},{"id":"D","content":"13人","is_correct":0}]}]