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":2238,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生在数轴上从原点出发,先向右移动5个单位长度,再向左移动8个单位长度,接着向右移动3个单位长度,最后向左移动6个单位长度。该学生最终所在位置的数与其起始位置(原点)的距离是___。","answer":"6","explanation":"该学生从原点0出发,按照顺序移动:+5 → -8 → +3 → -6。计算总位移:5 - 8 + 3 - 6 = -6。最终位置是-6,与原点0的距离是|−6| = 6。题目考查正负数在数轴上的实际应用及绝对值的理解,要求学生掌握连续正负数运算和距离的非负性,属于综合应用型难题。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 14:39:22","updated_at":"2026-01-09 14:39:22","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":2758,"subject":"历史","grade":"七年级","stage":"初中","type":"选择题","content":"考古学家在河南安阳发现了一处大型商代遗址,出土了大量刻有文字的龟甲和兽骨。这些文字主要用于记录商王占卜的内容,对研究商朝历史具有重要价值。这种文字被称为:","answer":"A","explanation":"题干中提到‘刻有文字的龟甲和兽骨’以及‘用于记录商王占卜的内容’,这是甲骨文的典型特征。甲骨文是商朝时期刻在龟甲和兽骨上的文字,主要用于占卜记事,是中国已发现的古代文字中时代最早、体系较为完整的文字。金文主要铸刻在青铜器上,盛行于西周;小篆是秦朝统一后的标准字体;隶书则流行于汉代。因此,根据出土文物的材质和用途,正确答案为A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-12 10:39:39","updated_at":"2026-01-12 10:39:39","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":"甲骨文","is_correct":1},{"id":"B","content":"金文","is_correct":0},{"id":"C","content":"小篆","is_correct":0},{"id":"D","content":"隶书","is_correct":0}]},{"id":2433,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某公园计划修建一个等腰三角形花坛ABC,其中AB = AC,且底边BC长为12米。为了美观,设计师在底边BC上取一点D,使得AD将花坛分成两个面积相等的部分。已知AD垂直于BC,且花坛的高为8米。若一名学生想计算线段BD的长度,他应如何求解?以下选项中正确的是:","answer":"A","explanation":"由于花坛ABC是等腰三角形(AB = AC),且AD垂直于底边BC,根据等腰三角形的性质,底边上的高、中线、角平分线三线合一。因此,AD不仅是高,还是中线,即D是BC的中点。已知BC = 12米,所以BD = 12 ÷ 2 = 6米。同时,AD将三角形分成两个面积相等的部分,也符合中线的性质。选项A正确。其他选项错误:B误认为面积相等意味着三等分;C错误应用勾股定理而未正确分析几何关系;D虽提到列方程,但未体现等腰三角形的核心性质,且结果不符。本题综合考查等腰三角形性质、轴对称、面积与几何推理,符合八年级学生认知水平。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 13:00:16","updated_at":"2026-01-10 13:00:16","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":"BD = 6米,因为AD是底边上的高,也是中线,所以D是BC的中点","is_correct":1},{"id":"B","content":"BD = 4米,因为面积相等意味着BD是BC的三分之一","is_correct":0},{"id":"C","content":"BD = 8米,根据勾股定理在△ABD中计算得出","is_correct":0},{"id":"D","content":"BD = 5米,通过设BD = x,利用面积公式列出方程求解","is_correct":0}]},{"id":2249,"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5个单位长度,且点B在原点右侧。那么点B表示的数是多少?","answer":"B","explanation":"点A在数轴上表示-3,点B与点A的距离为5个单位长度。由于点B在原点右侧,说明点B表示的数大于0。从-3向右移动5个单位,即-3 + 5 = 2,因此点B表示的数是2。选项B正确。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 16:03:06","updated_at":"2026-01-09 16:03:06","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":"-8","is_correct":0},{"id":"B","content":"2","is_correct":1},{"id":"C","content":"8","is_correct":0},{"id":"D","content":"5","is_correct":0}]},{"id":135,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"已知一个长方形的长比宽多3厘米,且它的周长是26厘米,那么这个长方形的宽是____厘米。","answer":"5","explanation":"本题考查初一学生对方程的应用能力,结合几何图形(长方形)的周长公式进行列方程求解。题目涉及有理数运算和一元一次方程的建立与求解,符合初一数学课程中‘代数初步’与‘简单几何应用’的学习内容。通过设未知数、列方程、解方程的过程,帮助学生理解实际问题向数学模型的转化。","solution_steps":"设长方形的宽为x厘米,则长为(x + 3)厘米。根据长方形周长公式:周长 = 2 × (长 + 宽),代入已知条件得:2 × (x + x + 3) = 26。化简得:2 × (2x + 3) = 26 → 4x + 6 = 26 → 4x = 20 → x = 5。因此,宽是5厘米。","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-24 09:40:59","updated_at":"2025-12-24 09:40:59","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":1249,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学生在研究平面直角坐标系中的几何问题时,发现一个有趣的规律:若将一个点P(x, y)先向右平移3个单位,再向上平移2个单位,得到点P';然后将点P'绕原点逆时针旋转90°,得到点P''。已知点P''的坐标为(-5, 4),求原点P的坐标(x, y)。此外,若该点P满足不等式组:2x - y > 1 且 x + 3y ≤ 10,请验证所求得的点P是否满足该不等式组。","answer":"解:\n\n第一步:设原点P的坐标为(x, y)。\n\n根据题意,点P先向右平移3个单位,再向上平移2个单位,得到点P'。\n平移变换规则:向右平移a个单位,横坐标加a;向上平移b个单位,纵坐标加b。\n因此,P'的坐标为:\n P' = (x + 3, y + 2)\n\n第二步:将点P'绕原点逆时针旋转90°,得到点P''。\n旋转90°逆时针的坐标变换公式为:\n 若点A(a, b)绕原点逆时针旋转90°,则新坐标为(-b, a)\n\n对P'(x + 3, y + 2)应用该公式:\nP'' = (-(y + 2), x + 3) = (-y - 2, x + 3)\n\n题目已知P''的坐标为(-5, 4),因此列出方程组:\n -y - 2 = -5\n x + 3 = 4\n\n解第一个方程:\n -y - 2 = -5\n → -y = -3\n → y = 3\n\n解第二个方程:\n x + 3 = 4\n → x = 1\n\n所以,原点P的坐标为(1, 3)。\n\n第三步:验证点P(1, 3)是否满足不等式组:\n 2x - y > 1\n x + 3y ≤ 10\n\n代入x = 1,y = 3:\n\n第一式:2(1) - 3 = 2 - 3 = -1\n -1 > 1? 不成立。\n\n第二式:1 + 3×3 = 1 + 9 = 10\n 10 ≤ 10? 成立。\n\n由于第一式不满足,因此点P(1, 3)不满足整个不等式组。\n\n最终答案:\n点P的坐标为(1, 3),但该点不满足给定的不等式组。","explanation":"本题综合考查了平面直角坐标系中的平移变换、旋转变换、二元一次方程组的建立与求解,以及不等式组的验证。解题关键在于掌握坐标变换的代数表示:平移是坐标的加减,旋转90°逆时针的公式为(a, b) → (-b, a)。通过逆向推理,从P''的坐标反推出P',再反推出P。最后将所得坐标代入不等式组进行验证,体现了数形结合与逻辑推理能力。题目设计新颖,融合了多个知识点,要求学生具备较强的综合运用能力,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 10:31:09","updated_at":"2026-01-06 10:31:09","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":427,"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20名学生进行调查,记录他们每周课外阅读的小时数如下:3, 5, 4, 6, 3, 7, 5, 4, 6, 5, 3, 4, 6, 7, 5, 4, 3, 5, 6, 4。如果该学生想用一个统计量来代表这组数据的集中趋势,并且希望这个值能反映大多数学生的阅读时间,那么他应该选择以下哪个统计量?","answer":"C","explanation":"题目要求选择一个能反映‘大多数学生’阅读时间的统计量,即关注数据中出现次数最多的值。观察数据:3出现4次,4出现5次,5出现5次,6出现4次,7出现2次。其中4和5都出现了5次,是出现频率最高的数值,因此众数为4和5(双众数)。众数正是用来描述数据集中出现最频繁的数值,最能体现‘大多数’的情况。而平均数受所有数据影响,中位数是中间值,极差反映的是数据波动范围,均不能直接体现‘大多数’的集中趋势。因此正确答案是C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:34:19","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":"A","content":"平均数","is_correct":0},{"id":"B","content":"中位数","is_correct":0},{"id":"C","content":"众数","is_correct":1},{"id":"D","content":"极差","is_correct":0}]},{"id":787,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次班级数学测验成绩整理中,某学生将10名同学的成绩按从小到大的顺序排列,得到的数据为:72,75,78,80,82,85,88,90,93,96。这组数据的中位数是____。","answer":"83.5","explanation":"中位数是指将一组数据按大小顺序排列后,处于中间位置的数。当数据个数为偶数时,中位数是中间两个数的平均值。本题中有10个数据(偶数个),因此中位数是第5个和第6个数据的平均数。第5个数是82,第6个数是85,所以中位数为 (82 + 85) ÷ 2 = 167 ÷ 2 = 83.5。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 00:06:37","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":1475,"subject":"数学","grade":"七年级","stage":"小学","type":"解答题","content":"某学生在研究平面直角坐标系中的点与图形关系时,设计了如下实验:在坐标系中,点A的坐标为(2, 3),点B位于x轴上,且线段AB的长度为5。点C是线段AB的中点,点D在y轴上,且满足CD的长度等于AB长度的一半。已知点D位于y轴正半轴,求点D的坐标。","answer":"解题步骤如下:\n\n1. 设点B的坐标为(x, 0),因为点B在x轴上。\n\n2. 根据两点间距离公式,AB的长度为:\n AB = √[(x - 2)² + (0 - 3)²] = 5\n 即:(x - 2)² + 9 = 25\n (x - 2)² = 16\n x - 2 = ±4\n 所以x = 6 或 x = -2\n 因此点B有两个可能位置:(6, 0) 或 (-2, 0)\n\n3. 分别求两种情况下点C的坐标(AB中点):\n - 若B为(6, 0),则C = ((2+6)\/2, (3+0)\/2) = (4, 1.5)\n - 若B为(-2, 0),则C = ((2-2)\/2, (3+0)\/2) = (0, 1.5)\n\n4. 点D在y轴上,设其坐标为(0, y),且y > 0(因在正半轴)\n 已知CD = AB \/ 2 = 5 \/ 2 = 2.5\n\n5. 分情况讨论CD的距离:\n\n 情况一:C为(4, 1.5)\n CD = √[(0 - 4)² + (y - 1.5)²] = 2.5\n 16 + (y - 1.5)² = 6.25\n (y - 1.5)² = -9.75 → 无实数解(舍去)\n\n 情况二:C为(0, 1.5)\n CD = √[(0 - 0)² + (y - 1.5)²] = |y - 1.5| = 2.5\n 所以 y - 1.5 = 2.5 或 y - 1.5 = -2.5\n 解得 y = 4 或 y = -1\n 但y > 0,故y = 4\n\n6. 因此点D的坐标为(0, 4)\n\n答案:点D的坐标是(0, 4)","explanation":"本题综合考查了平面直角坐标系、两点间距离公式、中点坐标公式以及实数运算。解题关键在于分类讨论点B的两种可能位置,并通过距离条件排除不符合的情况。特别需要注意的是,当点C在y轴上时,CD的距离计算简化为纵坐标差的绝对值,这是解题的突破口。同时,题目设置了无解情况以检验学生对方程解的合理性判断能力,体现了对数学严谨性的考查。整个过程涉及代数运算、几何直观和逻辑推理,属于较高难度的综合题。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:53:23","updated_at":"2026-01-06 11:53:23","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":251,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生在解方程 3(x - 2) + 5 = 2x + 7 时,第一步将等式左边展开得到 3x - 6 + 5,合并同类项后为 3x - 1;第二步将方程写成 3x - 1 = 2x + 7;第三步将 2x 移到左边,-1 移到右边,得到 3x - 2x = 7 + 1;第四步解得 x = ___。","answer":"8","explanation":"根据题目描述的解方程步骤:第一步展开括号正确,3(x - 2) = 3x - 6,再加5得 3x - 1;第二步方程为 3x - 1 = 2x + 7;第三步移项,将含x的项移到左边,常数项移到右边,即 3x - 2x = 7 + 1;第四步计算得 x = 8。此过程符合解一元一次方程的基本步骤,移项变号规则应用正确,最终结果为8。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"中等","points":1,"is_active":1,"created_at":"2025-12-29 14:54:13","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":[]}]