题目 #24059
- UUID
1c851475-b4eb-4f83-913d-64635f14d0df- 来源
- 人教版/九上/2026江苏九年级期中数学试卷+答案115.pdf 第 1-2 页，bbox
[30.22, 731.29, 200.26, 196.94] - 题型 / 学科
- subjective / math / JUNIOR
- 知识树
KT_JUNIOR_MATH_STANDARD- 难度
- 难度 1 / 5 最简单 · 识记/直接套公式
- QA 状态
- PUBLISHED 重试 0 次，得分 0.988
- 最终题 ID
11121
最终题干渲染（f_body 预览）
解方程：\(\left(x+1\right)^{2}-3\left(x+1\right)+2=0\)填空题，每题6分
答案 / 解析
答案（f_answer）
\(x_{1}\)= 0,\(x_{2}\)= 1
解析（f_ways）
由题可得: (x + 1)^{2}−3(x + 1) + 2 = 0, 设x + 1 = t, 则原式为\(t^{2}\)−3t + 2 = 0,\(\frac{十字}{x}\)相乘法得(t −1)(t −2) = 0, t −1 = 0或t −2 = 0, 解得t = 1或t = 2, 当t = 1时, +1 = 1; 当t = 2时, x + 1 = 2; 解得\(x_{1}\)= 0,\(x_{2}\)= 1, 故原方程的解为\(x_{1}\)= 0,\(x_{2}\)= 1 3 + √17 3 −√17\(x_{1}\)=,\(x_{2}\)=
原图与资产（0）
该题没有裁出图片资产
审计记录
| 阶段 | 审计器 | 结果 | 得分 | 依据（简短） |
|---|---|---|---|---|
| STRUCTURE | structure-auditor | PASS | 1.000 | |
| FORMULA | formula-auditor | PASS | 1.000 | |
| KNOWLEDGE | knowledge-auditor | WARN | - | |
| SOLVER | blind-solver | PASS | 1.000 | |
| CRITIC | critic | PASS | 1.000 | |
| VISION | vision-auditor | PASS | 0.950 | |
| SYMBOLIC | symbolic-auditor | WARN | 0.550 | |
| COHERENCE | coherence-auditor | PASS | 1.000 | |
| QUALITY_GATE | quality-gate | PASS | 0.988 |
知识点（0）
无
正式题库记录（t_question.f_id=11121）
- f_qkey
ee4a1239b489f22587ff7e9de46f7176- f_style / 难度
- 5 / 2
- 入库题干
- 解方程：\(\left(x+1\right)^{2}-3\left(x+1\right)+2=0\)填空题，每题6分
- 入库答案
- \(x_{1}\)= 0,\(x_{2}\)= 1
- 入库解析
- 由题可得: (x + 1)^{2}−3(x + 1) + 2 = 0, 设x + 1 = t, 则原式为\(t^{2}\)−3t + 2 = 0,\(\frac{十字}{x}\)相乘法得(t −1)(t −2) = 0, t −1 = 0或t −2 = 0, 解得t = 1或t = 2, 当t = 1时, +1 = 1; 当t = 2时, x + 1 = 2; 解得\(x_{1}\)= 0,\(x_{2}\)= 1, 故原方程的解为\(x_{1}\)= 0,\(x_{2}\)= 1 3 + √17 3 −√17\(x_{1}\)=,\(x_{2}\)=
事件
| 时间 | 阶段 | 级别 | 消息 |
|---|---|---|---|
| 2026-09-25 21:39:11 | PUBLISH | INFO | 已发布 t_question.f_id=11121 qkey=ee4a1239b489f22587ff7e9de46f7176 |
| 2026-09-25 21:39:11 | QA | INFO | QA: PASS (score=0.988) |
| 2026-09-25 21:37:13 | FORMULA | WARN | 公式转换未确认，保留原文本: solution |