初中
数学
中等
来源: 教材例题
知识点: 初中数学
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[{"id":131,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"一个长方形的长比宽多5厘米,若其周长为30厘米,则这个长方形的宽是______厘米。","answer":"5","explanation":"设长方形的宽为x厘米,则长为(x + 5)厘米。根据长方形周长公式:周长 = 2 × (长 + 宽),代入得:2 × (x + x + 5) = 30,即2 × (2x + 5) = 30,化简得4x + 10 = 30,解得4x = 20,x = 5。因此,宽为5厘米。本题结合代数设未知数与一元一次方程求解,符合初一学生对方程和几何基础的学习要求。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-24 08:59:12","updated_at":"2025-12-24 08:59:12","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":130,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"小明在解一个一元一次方程时,将方程 3(x - 2) = 2x + 1 的括号展开后,写成了 3x - 6 = 2x + 1。接着他正确地移项并合并同类项,最终得到的解是 x = a。请问 a 的值是多少?","answer":"7","explanation":"本题考查初一学生对方程变形和求解的掌握情况,涉及去括号、移项、合并同类项等基本代数操作。题目通过描述解题过程,引导学生关注方程求解的逻辑步骤,而非直接给出方程求解,具有一定的思维引导性。学生需要理解每一步变形的合理性,并正确执行计算。","solution_steps":"1. 原方程为:3(x - 2) = 2x + 1\n2. 去括号得:3x - 6 = 2x + 1\n3. 移项(将含x的项移到左边,常数项移到右边):3x - 2x = 1 + 6\n4. 合并同类项:x = 7\n5. 因此,a = 7","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-24 08:59:12","updated_at":"2025-12-24 08:59:12","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":"7","is_correct":1},{"id":"B","content":"8","is_correct":0},{"id":"C","content":"6","is_correct":0},{"id":"D","content":"3.5","is_correct":0}]},{"id":497,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"5","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 18:08:49","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":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":2184,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在数轴上标出三个点A、B、C,分别表示有理数a、b、c。已知a < b < c,且|a| = |c|,b是a与c的中点。若c = 5,则a + b + c的值是多少?","answer":"B","explanation":"由题意知c = 5,且|a| = |c|,所以|a| = 5,即a = 5或a = -5。又因a < b < c且c = 5,若a = 5,则a = c,与a < c矛盾,故a = -5。b是a与c的中点,即b = (a + c) ÷ 2 = (-5 + 5) ÷ 2 = 0。因此a + b + c = -5 + 0 + 5 = 0。","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":"-5","is_correct":0},{"id":"B","content":"0","is_correct":1},{"id":"C","content":"5","is_correct":0},{"id":"D","content":"10","is_correct":0}]},{"id":2164,"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和时,先将两个数的绝对值相加,再根据两数符号确定结果的符号。若他计算的是 -7 与 3 的和,按照他的方法会得到什么结果?实际正确答案又是什么?以下哪一项正确描述了他的错误?","answer":"A","explanation":"该学生错误地将两个有理数的绝对值相加(7 + 3 = 10),然后因两数异号而误判符号为负,得出 -10。但正确方法应为异号相加时用大绝对值减小绝对值(7 - 3 = 4),符号取绝对值较大数的符号(-7 的绝对值大),因此正确答案是 -4。他的错误本质是未掌握异号有理数相加的运算法则,应相减而非相加绝对值。","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":"他得到的结果是 -10,正确答案是 -4,错误在于没有考虑两数异号时应相减","is_correct":1},{"id":"B","content":"他得到的结果是 10,正确答案是 4,错误在于符号判断错误","is_correct":0},{"id":"C","content":"他得到的结果是 -4,正确答案是 -10,错误在于绝对值相加不正确","is_correct":0},{"id":"D","content":"他得到的结果是 4,正确答案是 -4,错误在于没有取绝对值","is_correct":0}]},{"id":768,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次环保知识竞赛中,某班级共收集了120份有效问卷,其中支持垃圾分类的有78人,支持节约用水的有65人,两项都支持的有40人。那么,只支持垃圾分类而不支持节约用水的有___人。","answer":"38","explanation":"根据题意,支持垃圾分类的人数为78人,其中40人同时支持节约用水,因此只支持垃圾分类的人数为78减去40,即78 - 40 = 38人。此题考查的是数据的收集与整理中的集合思想,利用集合的交集与差集进行简单计算,符合七年级数学中‘数据的收集、整理与描述’的知识点。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 23:45:54","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":2148,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在解方程 2x + 3 = 9 时,第一步将等式两边同时减去3,得到 2x = 6。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,目标是求出未知数 x 的值。某学生已经通过移项得到 2x = 6,说明 2 是 x 的系数。为了求出 x,需要将等式两边同时除以 2,从而得到 x = 3。这是解方程的基本步骤,符合七年级学生对方程求解的学习要求。","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":"将等式两边同时加上2","is_correct":0},{"id":"B","content":"将等式两边同时除以2","is_correct":1},{"id":"C","content":"将等式两边同时乘以2","is_correct":0},{"id":"D","content":"将等式两边同时减去2","is_correct":0}]},{"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":2168,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在数轴上标出三个有理数 a、b、c,已知 a < b < c,且 |a| = |c|,b 是 a 与 c 的算术平均数。若 a + c = -8,则下列说法正确的是:","answer":"B","explanation":"由已知 a + c = -8,且 b 是 a 与 c 的算术平均数,得 b = (a + c) \/ 2 = -8 \/ 2 = -4,因此选项 B 正确。又因为 |a| = |c|,说明 a 和 c 到原点的距离相等,但 a + c = -8 ≠ 0,所以 a 和 c 不互为相反数(相反数之和为 0),排除 A。由于 |a| = |c|,C 错误。a 与 c 不相等(因 a < b < c),距离不可能为 0,D 错误。本题综合考查有理数在数轴上的表示、绝对值、相反数及平均数概念,需多步推理,符合七年级困难题要求。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 13:53:54","updated_at":"2026-01-09 13:53:54","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":"a 和 c 互为相反数","is_correct":0},{"id":"B","content":"b 的值为 -4","is_correct":1},{"id":"C","content":"c 的绝对值小于 a 的绝对值","is_correct":0},{"id":"D","content":"a 与 c 之间的距离为 0","is_correct":0}]}]