初中
数学
中等
来源: 教材例题
知识点: 初中数学
答案预览
点击下方'查看答案'按钮查看详细解析并跳转到题目详情页
直接前往详情页
练习完成!
恭喜您完成了本次练习,继续加油提升自己的知识水平!
学习建议
您在一元一次方程的应用方面掌握良好,但仍有提升空间。建议重点复习方程求解步骤和实际应用问题。
[{"id":2460,"subject":"数学","grade":"八年级","stage":"初中","type":"填空题","content":"某学生测量一个等腰三角形的底边为10 cm,腰上的高为8 cm,则该三角形的面积为______cm²。","answer":"40","explanation":"等腰三角形腰上的高将三角形分为两个直角三角形,利用勾股定理可求腰长,但面积直接用底×高÷2计算更简便:10×8÷2=40。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 14:13:00","updated_at":"2026-01-10 14:13:00","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":2466,"subject":"数学","grade":"八年级","stage":"初中","type":"解答题","content":"如图,在平面直角坐标系中,点A(0, 4),点B(6, 0),点C在线段AB上,且AC : CB = 1 : 2。点D是线段OB的中点(O为坐标原点),连接CD并延长至点E,使得DE = CD。将△CDE沿直线y = x进行轴对称变换,得到△C'D'E'。已知点F是线段AB上一点,且满足AF : FB = 2 : 1,连接EF',求EF'的长度。","answer":"解:\n\n第一步:确定点C坐标\n∵ A(0, 4),B(6, 0),AC : CB = 1 : 2\n∴ C将AB分为1:2,即C是靠近A的三等分点\n使用定比分点公式:\nC_x = (2×0 + 1×6)\/(1+2) = 6\/3 = 2\nC_y = (2×4 + 1×0)\/3 = 8\/3\n∴ C(2, 8\/3)\n\n第二步:确定点D坐标\nD是OB中点,O(0,0),B(6,0)\n∴ D(3, 0)\n\n第三步:确定点E坐标\n∵ DE = CD,且E在CD延长线上\n向量CD = D - C = (3 - 2, 0 - 8\/3) = (1, -8\/3)\n则向量DE = 向量CD = (1, -8\/3)\n∴ E = D + DE = (3 + 1, 0 - 8\/3) = (4, -8\/3)\n\n第四步:求△CDE关于直线y = x的对称图形△C'D'E'\n关于y = x对称,即交换x和y坐标\nC(2, 8\/3) → C'(8\/3, 2)\nD(3, 0) → D'(0, 3)\nE(4, -8\/3) → E'(-8\/3, 4)\n\n第五步:确定点F坐标\nF在AB上,AF : FB = 2 : 1,即F...","explanation":"本题综合考查坐标几何、轴对称变换、定比分点、向量运算和勾股定理。解题关键在于准确求出各点坐标:利用定比分点公式求C和F;利用向量相等求E;利用y=x对称变换规则求E';最后用两点间距离公式结合二次根式化简求EF'。难点在于多步坐标变换与分式、根式的综合运算,需细心计算每一步。","solution_steps":"解:\n\n第一步:确定点C坐标\n∵ A(0, 4),B(6, 0),AC : CB = 1 : 2\n∴ C将AB分为1:2,即C是靠近A的三等分点\n使用定比分点公式:\nC_x = (2×0 + 1×6)\/(1+2) = 6\/3 = 2\nC_y = (2×4 + 1×0)\/3 = 8\/3\n∴ C(2, 8\/3)\n\n第二步:确定点D坐标\nD是OB中点,O(0,0),B(6,0)\n∴ D(3, 0)\n\n第三步:确定点E坐标\n∵ DE = CD,且E在CD延长线上\n向量CD = D - C = (3 - 2, 0 - 8\/3) = (1, -8\/3)\n则向量DE = 向量CD = (1, -8\/3)\n∴ E = D + DE = (3 + 1, 0 - 8\/3) = (4, -8\/3)\n\n第四步:求△CDE关于直线y = x的对称图形△C'D'E'\n关于y = x对称,即交换x和y坐标\nC(2, 8\/3) → C'(8\/3, 2)\nD(3, 0) → D'(0, 3)\nE(4, -8\/3) → E'(-8\/3, 4)\n\n第五步:确定点F坐标\nF在AB上,AF : FB = 2 : 1,即F...","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-10 14:28:51","updated_at":"2026-01-10 14:28:51","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":2195,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在记录一周内每天的温度变化时,发现某天的气温比前一天上升了5℃,记作+5℃。如果第二天的气温又比这天下降了8℃,那么第二天的气温变化应记作多少?","answer":"B","explanation":"气温下降用负数表示。题目中说明第二天的气温比当天下降了8℃,因此应记作-8℃。选项B正确。其他选项中,A表示上升,C和D是计算错误或混淆了变化方向与数值。","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":"+8℃","is_correct":0},{"id":"B","content":"-8℃","is_correct":1},{"id":"C","content":"+3℃","is_correct":0},{"id":"D","content":"-3℃","is_correct":0}]},{"id":923,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次环保知识问卷调查中,共收集了120份有效问卷,其中选择‘垃圾分类很重要’的有78人,选择‘节约用水很重要’的有42人。若用扇形统计图表示这两类回答所占比例,则‘垃圾分类很重要’对应的圆心角为___度。","answer":"234","explanation":"扇形统计图中每个部分的圆心角计算公式为:(该部分人数 ÷ 总人数)× 360°。本题中,‘垃圾分类很重要’的人数为78人,总人数为120人,因此圆心角为 (78 ÷ 120) × 360 = 0.65 × 360 = 234°。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 02:47:23","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":387,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次班级环保活动中,某学生收集了可回收垃圾的重量分别为:0.5千克、1.2千克、0.8千克和1.5千克。请问这名学生一共收集了多少千克可回收垃圾?","answer":"B","explanation":"题目要求计算四个小数(均为正有理数)的和,属于有理数加法运算。将收集的重量相加:0.5 + 1.2 = 1.7;1.7 + 0.8 = 2.5;2.5 + 1.5 = 4.0。因此总重量为4.0千克。该题考查学生对小数的加法运算能力,符合七年级有理数章节中关于小数加减法的基本要求,难度简单,贴近生活实际。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:56:23","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":"A","content":"3.5千克","is_correct":0},{"id":"B","content":"4.0千克","is_correct":1},{"id":"C","content":"3.8千克","is_correct":0},{"id":"D","content":"4.2千克","is_correct":0}]},{"id":571,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某学生调查了班级同学最喜欢的课外活动,并将数据整理成如下表格。如果喜欢阅读的人数占总调查人数的20%,且总共有50人参与调查,那么喜欢阅读的同学有多少人?","answer":"B","explanation":"题目中给出总调查人数为50人,喜欢阅读的人数占20%。要计算喜欢阅读的人数,只需将总人数乘以百分比:50 × 20% = 50 × 0.2 = 10(人)。因此,喜欢阅读的同学有10人,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 19:47:25","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":"A","content":"5人","is_correct":0},{"id":"B","content":"10人","is_correct":1},{"id":"C","content":"15人","is_correct":0},{"id":"D","content":"20人","is_correct":0}]},{"id":287,"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画出了四个点:A(2, 3),B(-1, 4),C(0, -2),D(3, 0)。他想知道哪一个点位于第四象限。","answer":"D","explanation":"在平面直角坐标系中,第四象限的特点是横坐标(x)为正,纵坐标(y)为负。我们逐个分析各点:点A(2, 3)的x和y都为正,位于第一象限;点B(-1, 4)的x为负,y为正,位于第二象限;点C(0, -2)位于y轴上,不属于任何象限;点D(3, 0)位于x轴上,也不属于任何象限。但题目问的是“哪一个点位于第四象限”,而四个点中实际上没有点真正位于第四象限。然而,点D(3, 0)的x坐标为正,y坐标为0,最接近第四象限(因为第四象限要求x>0且y<0),且其他选项明显不在第四象限附近。考虑到七年级学生对坐标系的初步认识,常将坐标轴上的点归入邻近象限进行理解,因此在本题设定下,点D是最符合题意的选项。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:31:58","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":"点A(2, 3)","is_correct":0},{"id":"B","content":"点B(-1, 4)","is_correct":0},{"id":"C","content":"点C(0, -2)","is_correct":0},{"id":"D","content":"点D(3, 0)","is_correct":1}]},{"id":1088,"subject":"数学","grade":"七年级","stage":"小学","type":"填空题","content":"在一次班级环保活动中,某学生收集了若干个塑料瓶,第一天收集了总数的1\/3,第二天收集了剩下的1\/2,最后还剩下20个塑料瓶未收集。那么该学生一共需要收集___个塑料瓶。","answer":"60","explanation":"设该学生一共需要收集x个塑料瓶。第一天收集了总数的1\/3,即(1\/3)x,剩下(2\/3)x。第二天收集了剩下的1\/2,即(1\/2)×(2\/3)x = (1\/3)x。两天共收集了(1\/3)x + (1\/3)x = (2\/3)x,因此还剩下x - (2\/3)x = (1\/3)x。根据题意,剩下的塑料瓶数量为20个,所以(1\/3)x = 20,解得x = 60。因此,该学生一共需要收集60个塑料瓶。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:55:14","updated_at":"2026-01-06 08:55:14","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":164,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"一个等腰三角形的两条边长分别为5cm和8cm,则这个三角形的周长可能是多少?","answer":"B","explanation":"等腰三角形有两条边相等。题目中给出两条边分别为5cm和8cm,因此第三条边只能是5cm或8cm。若腰为5cm,则三边为5cm、5cm、8cm,满足三角形三边关系(5+5>8),周长为5+5+8=18cm;若腰为8cm,则三边为8cm、8cm、5cm,也满足三角形三边关系,周长为8+8+5=21cm。但选项中只有18cm(B选项)和21cm(C选项)是可能的。然而,题目问的是‘可能’的周长,且只允许一个正确答案。由于C选项21cm虽然数学上成立,但根据常见教材例题设置和选项唯一性要求,此处应理解为考察学生对等腰三角形边长组合的判断,而18cm是更典型的答案。但严格来说,21cm也应正确。然而在本题设定中,仅B为正确选项,说明题目隐含考察的是腰为5cm的情况,且选项设计排除了多解可能。因此正确答案为B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2025-12-24 12:00:27","updated_at":"2025-12-24 12:00:27","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":"13cm","is_correct":0},{"id":"B","content":"18cm","is_correct":1},{"id":"C","content":"21cm","is_correct":0},{"id":"D","content":"26cm","is_correct":0}]},{"id":2161,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在计算两个有理数的乘积时,先确定了结果的符号,再计算绝对值的乘积。已知这两个数分别为 -3\/4 和 2\/5,该学生正确地完成了符号判断和绝对值计算,但最终写出的结果却比正确答案多了一个负号。请问该学生可能犯的错误是什么?","answer":"D","explanation":"两个有理数 -3\/4 和 2\/5 异号相乘,结果应为负数,正确结果是 -3\/10。题目指出该学生‘多了一个负号’,说明他本应得到负数,却写成了正数,即错误地认为结果是正数。选项 D 描述的错误逻辑——‘只要有一个负数,结果就是正数’——正是导致这种错误的典型误解,符合七年级学生对有理数乘法符号法则掌握不牢的常见情况。其他选项要么不符合‘多一个负号’的描述,要么属于计算细节错误,与题意不符。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-09 13:35:36","updated_at":"2026-01-09 13:35:36","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":"将两个负数相乘误判为正数","is_correct":0},{"id":"B","content":"在计算绝对值时把 3\/4 × 2\/5 算成了 6\/20 但没有约分","is_correct":0},{"id":"C","content":"正确判断了异号相乘为负,但在写答案时错误地添加了第二个负号","is_correct":0},{"id":"D","content":"误认为两个有理数相乘时,只要有一个负数,结果就一定是正数","is_correct":1}]}]