初中
数学
中等
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知识点: 初中数学
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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":[]},{"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":889,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在某次班级大扫除中,一组学生负责擦窗户。如果每扇窗户需要2分钟擦干净,他们一共用了24分钟完成任务,那么这组学生一共擦了____扇窗户。","answer":"12","explanation":"题目中给出每扇窗户需要2分钟,总用时24分钟。要求擦了多少扇窗户,可以用总时间除以每扇窗户所需时间:24 ÷ 2 = 12。这是一道简单的一元一次方程应用题,设擦了x扇窗户,则2x = 24,解得x = 12。考查学生将实际问题转化为方程并求解的能力,属于一元一次方程知识点,难度简单。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 02:01:51","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":1788,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在平面直角坐标系中绘制了一个四边形ABCD,其顶点坐标分别为A(2, 3)、B(5, 7)、C(8, 4)、D(6, 1)。该学生想验证这个四边形是否为平行四边形,于是计算了四条边的长度和对角线AC与BD的长度。已知两点间距离公式为√[(x₂−x₁)² + (y₂−y₁)²],若该四边形是平行四边形,则必须满足对边相等且对角线互相平分。根据这些条件,以下哪一项是该四边形为平行四边形的充分必要条件?","answer":"D","explanation":"判断一个四边形是否为平行四边形,有多种方法。选项A只说明对边长度相等,但在平面直角坐标系中,仅边长相等不能保证是平行四边形(可能是空间扭曲的四边形)。选项B中AC和BD是对角线,它们的长度相等是矩形的特征之一,不是平行四边形的必要条件。选项C提到对边平行,虽然正确,但题目中并未提供斜率信息,且‘平行’需要通过斜率计算验证,不如中点法直接。而选项D指出‘对角线AC与BD的中点重合’,这是平行四边形的一个核心判定定理:若四边形的两条对角线互相平分,则该四边形必为平行四边形。计算AC中点:((2+8)\/2, (3+4)\/2) = (5, 3.5);BD中点:((5+6)\/2, (7+1)\/2) = (5.5, 4),实际不相等,说明本题中四边形不是平行四边形,但题目问的是‘充分必要条件’,即理论上正确的判定方法,因此D是正确答案。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 15:58:52","updated_at":"2026-01-06 15:58:52","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":"AB = CD 且 BC = DA","is_correct":0},{"id":"B","content":"AB = CD 且 AC = BD","is_correct":0},{"id":"C","content":"AB ∥ CD 且 BC ∥ DA","is_correct":0},{"id":"D","content":"对角线AC与BD的中点重合","is_correct":1}]},{"id":1094,"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,某学生收集废旧纸张的重量比另一名学生的3倍还多2千克。如果两人一共收集了26千克,那么这名学生自己收集了___千克。","answer":"20","explanation":"设这名学生收集的废旧纸张重量为x千克,则另一名学生收集的为(3x + 2)千克。根据题意,两人共收集26千克,可列方程:x + (3x + 2) = 26。化简得4x + 2 = 26,解得4x = 24,x = 6。但注意:题目中描述的是“某学生收集的重量比另一名学生的3倍还多2千克”,因此应设另一名学生为x千克,则该学生为(3x + 2)千克。于是方程为x + (3x + 2) = 26,解得4x = 24,x = 6,那么该学生收集了3×6 + 2 = 20千克。因此答案是20。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:56:06","updated_at":"2026-01-06 08:56:06","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":2450,"subject":"数学","grade":"八年级","stage":"初中","type":"填空题","content":"在一次函数 y = kx + b 的图像中,已知该函数与 x 轴交于点 (4, 0),与 y 轴交于点 (0, -6),则 k 的值为___。","answer":"3\/2","explanation":"由 y 轴交点得 b = -6,代入 x 轴交点 (4, 0) 得 0 = 4k - 6,解得 k = 3\/2。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 13:54:56","updated_at":"2026-01-10 13:54:56","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":[]}]