{"ok":true,"question":{"id":21391,"question_uuid":"182d7c76-8c08-4b54-bf0a-1f92031f309b","job_id":4,"source_file":"A中考/2026江苏中考期末数学试卷+答案95.pdf","source_page_start":1,"source_page_end":1,"source_bbox":"[45.0, 94.2, 323.04, 175.53]","subject_code":"math","grade_level":"SENIOR","tree_code":"KT_SENIOR_MATH_STANDARD","question_type":"subjective","canonical_json":"{\"year\": null, \"blocks\": [{\"text\": \"选择题，每题6 分 ★ 1.先化简,再求值: (1 + _{a−1}) ÷ _{a−1}, 其中a = 4 + 3． 3 a2−4\", \"type\": \"text\"}, {\"type\": \"answer\", \"blocks\": [{\"text\": \"a + b =−1, ab =−\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"6\\\\frac{1}{x(x−1)}\"}, {\"text\": \"，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"x_{x−2}\"}, {\"text\": \"−^\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"{a}_{2a}_{2}\"}, {\"text\": \"2 x x2−x(x−2) _{x−2}−^\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"{1}_{2}\"}, {\"text\": \"ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"1 \\\\frac{1}{x^{2}−4x+4}\"}, {\"text\": \"（x−2） _{x−2}= 3时\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{11.−10}{13.2}\"}, {\"text\": \"原式=\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"2 \\\\times ( −3)\"}, {\"text\": \"−4 =−6 −4 =−10. 原式( 原式= (4a^{\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"2\\\\frac{_{x−2}−x) ÷}{_{a+1}−}\"}, {\"text\": \"}−12a + 9) −(\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"a^{2}\"}, {\"text\": \"−25) =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"4a^{2}\"}, {\"text\": \"−12a + 9 −\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"a^{2}\"}, {\"text\": \"+ 25=\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"3a^{2}\"}, {\"text\": \"−12a + 34. _{x2}_{−4x+4}= x−2 ⋅ x =x2−x2+2x ⋅ x2−4x+4 = _{x−2}⋅2x （x−2） = 2(x −2) = 2x −4,当 x =−\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"2 \\\\frac{x−2}{2_{2}}\"}, {\"text\": \"x x 12.\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"3a^{2}\"}, {\"text\": \"−12a + 34 原式= _{(a−1)(a+1)}⋅ = _{a+1}− _{a+1}= _{a+1}，∵a + 1 ≠0, a −1 ≠0, ∴a ≠−1, a≠1, ∴当a = 0时\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{2(a−2)}{x−y} \\\\frac{2a−2}{x+y}\"}, {\"text\": \"（a−1） 2a ,原式 = _{0+1}= 2. a−\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"2_{2}\"}, {\"text\": \"14.2 −1 原式= + − + − = x − y + x + （x）^{2}−2xy+（y） （x）^{2}+2xy+（y） 2x−y （x−y） （x+ y） 2x−y 2\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"2 _{x}\"}, {\"text\": \"= x−y x+ y x y − + _{x}=2x −2x + _{x}= _{x}. 当x = 2 + 1, y = 2 −1 时,原式 = _{2+1}=\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\sqrt{（）} 2\"}, {\"text\": \"−\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"1 ^{2}\"}, {\"text\": \"= 2 2（x） y 2−1 y y x 2 −1. 15.−4 原式=\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"a^{2}\"}, {\"text\": \"+ 6ab +\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"9b^{2}\"}, {\"text\": \"+\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"a^{2}\"}, {\"text\": \"−\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"9b^{2}\"}, {\"text\": \"=\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"2a^{2}\"}, {\"text\": \"+ 6ab, 当a = 2, b =−1时,原式=\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"2 \\\\times 2^{2}+ 6 \\\\times 2 \\\\times ( −1) =−4\"}, {\"text\": \".\", \"type\": \"text\"}]}], \"source\": {\"bbox\": [45.0, 94.2, 323.04, 175.53], \"file\": \"A中考/2026江苏中考期末数学试卷+答案95.pdf\", \"regions\": [{\"bbox\": [45.0, 94.2, 323.04, 175.53], \"page\": 1}], \"section\": \"generic\", \"page_end\": 1, \"page_start\": 1, \"section_label\": \"\", \"answer_regions\": [{\"bbox\": [30.0, 85.77799987792969, 570.3590087890625, 764.845947265625], \"page\": 6}, {\"bbox\": [30.0, 54.097999572753906, 565.6439819335938, 186.4459991455078], \"page\": 7}]}, \"version\": 1, \"grade_level\": \"SENIOR\", \"number_label\": \"1\", \"subject_code\": \"math\", \"question_type\": \"subjective\", \"content_sha256\": \"5acaf32fa1c5b8d5e9b0d5cb2a3bdb848294c23f6b2a90cf39bf86119b77e8f8\"}","body_html":"<div>选择题，每题6 分 ★ 1.先化简,再求值: (1 + _{a−1}) ÷ _{a−1}, 其中a = 4 + 3． 3 a2−4</div>","answer_html":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\">\\(2frac_x\\)−2−x) ÷_a+1−</span>}−12a + 9) −(\\(a^{2}\\)−25) =\\(4a^{2}\\)−12a + 9 −\\(a^{2}\\)+ 25=\\(3a^{2}\\)−12a + 34. _{x2}_{−4x+4}= x−2 ⋅ x =x2−x2+2x ⋅ x2−4x+4 = _{x−2}⋅2x （x−2） = 2(x −2) = 2x −4,当 x =−\\(2 \\frac{x−2}{2_{2}}\\)x x 12.\\(3a^{2}\\)−12a + 34 原式= _{(a−1)(a+1)}⋅ = _{a+1}− _{a+1}= _{a+1}，∵a + 1 ≠0, a −1 ≠0, ∴a ≠−1, a≠1, ∴当a = 0时\\(\\frac{2(a−2)}{x−y} \\frac{2a−2}{x+y}\\)（a−1） 2a ,原式 = _{0+1}= 2. a−\\(2_{2}\\)14.2 −1 原式= + − + − = x − y + x + （x）^{2}−2xy+（y） （x）^{2}+2xy+（y） 2x−y （x−y） （x+ y） 2x−y 2\\(2 _{x}\\)= x−y x+ y x y − + _{x}=2x −2x + _{x}= _{x}. 当x = 2 + 1, y = 2 −1 时,原式 = _{2+1}=\\(\\sqrt{（）} 2\\)−\\(1 ^{2}\\)= 2 2（x） y 2−1 y y x 2 −1. 15.−4 原式=\\(a^{2}\\)+ 6ab +\\(9b^{2}\\)+\\(a^{2}\\)−\\(9b^{2}\\)=\\(2a^{2}\\)+ 6ab, 当a = 2, b =−1时,原式=\\(2 \\times 2^{2}+ 6 \\times 2 \\times ( −1) =−4\\).","ways_html":"","explain_html":null,"difficulty":2,"score":null,"year":null,"parse_status":"PARSED","qa_status":"QUARANTINED","retry_count":0,"final_score":null,"error_message":"处理异常: LLM call failed after retries: HTTPConnectionPool(host='127.0.0.1', port=8091): Read timed out. (read timeout=120) [endpoints: http://127.0.0.1:8092/v1 -> http://127.0.0.1:8091/v1]","published_question_id":null,"content_sha256":"5acaf32fa1c5b8d5e9b0d5cb2a3bdb848294c23f6b2a90cf39bf86119b77e8f8","created_at":"2026-09-25T00:44:08","updated_at":"2026-09-25T00:53:47","answer_missing":false,"formula_fallback":true,"_number":"1","_section":"generic","_section_label":""},"assets":[],"knowledge":[],"audits":[{"id":421384,"import_question_id":21391,"audit_type":"STRUCTURE","auditor_name":"structure-auditor","status":"PASS","score":0.9,"input_snapshot":"选择题，每题6 分 ★ 1.先化简,再求值: (1 + _{a−1}) ÷ _{a−1}, 其中a = 4 + 3． 3 a2−4","result_json":"{\"ok\": true, \"issues\": [], \"confidence\": 0.9}","created_at":"2026-09-25T00:49:47"},{"id":421385,"import_question_id":21391,"audit_type":"FORMULA","auditor_name":"formula-auditor","status":"FAIL","score":1.0,"input_snapshot":"6\\frac{1}{x(x−1)}; x_{x−2}; {a}_{2a}_{2}; {1}_{2}; 1 \\frac{1}{x^{2}−4x+4}","result_json":"{\"risky\": 25, \"issues\": [\"非法 LaTeX #8: unbalanced delimiters\"], \"checked\": 25, \"invalid\": 1, \"visual_check\": \"delegated_to_question_vision\", \"invalid_details\": [{\"index\": 7, \"latex\": \"2\\\\frac{_{x−2}−x) ÷}{_{a+1}−}\", \"issues\": [\"unbalanced delimiters\"]}], \"visually_verified\": 0}","created_at":"2026-09-25T00:49:47"},{"id":421386,"import_question_id":21391,"audit_type":"KNOWLEDGE","auditor_name":"knowledge-auditor","status":"WARN","score":0.0,"input_snapshot":"选择题，每题6 分 ★ 1.先化简,再求值: (1 + _{a−1}) ÷ _{a−1}, 其中a = 4 + 3． 3 a2−4","result_json":"{\"picks\": [], \"tree_code\": \"KT_SENIOR_MATH_STANDARD\", \"candidates\": 0, \"classifier_version\": null}","created_at":"2026-09-25T00:49:47"},{"id":421393,"import_question_id":21391,"audit_type":"VISION","auditor_name":"vision-auditor","status":"PASS","score":0.9,"input_snapshot":"题干：\n选择题，每题6 分 ★ 1.先化简,再求值: (1 + _{a−1}) ÷ _{a−1}, 其中a = 4 + 3． 3 a2−4\n答案：\na + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{ 2frac_x−2−x) ÷_a+1− }−12a + 9) −(\\(a^{2}\\)−25) =\\(4a^{2}\\)−12a + 9 −\\(a^{2}\\)+ 25=\\(3a^{2}\\)−12a + 34. _{x2}_{−4x+4}= x−2 ⋅ x =x2−x2+2x ⋅ x2−4x+4 = _{x−2}⋅2x （x−2） = 2(x −2) = 2x −4,当 x =−\\(2 \\frac{x−2}{2_{2}}\\)x x 12.\\(3a^{2}\\)−12a + 34 原式= _{(a−1)(a+1)}⋅ = _{a+1}− _{a+1}= _{a+1}，∵a + 1 ≠0, a −1 ≠0, ∴a ≠−1, a≠1, ∴当a = 0时\\(\\frac{2(a−2)}{x−y} \\frac{2a−2}{x+y}\\)（a−1） 2a ,原式 = _{0+1}= 2. a−\\(2_{2}\\)14.2 −1 原式= + − + − = x − y + x + （x）^{2}−2xy+（y） （x）^{2}+2xy+（y） 2x−y （x−y） （x+ y） 2x−y 2\\(2 _{x}\\)= x−y x+ y x y − + _{x}=2x −2x + _{x}= _{x}. 当x = 2 + 1, y = 2 −1 时,原式 = _{2+1}=\\(\\sqrt{（）} 2\\)−\\(1 ^{2}\\)= 2 2（x） y 2−1 y y x 2 −1. 15.−4 原式=\\(a^{2}\\)+ 6ab +\\(9b^{2}\\)+\\(a^{2}\\)−\\(9b^{2}\\)=\\(2a^{2}\\)+ 6ab, 当a = 2, b =−1时,原式=\\(2 \\times 2^{2}+ 6 \\times 2 \\times ( −1) =−4\\).\n解析：\n","result_json":"{\"ok\": true, \"hard\": false, \"issues\": [], \"confidence\": 0.9, \"suggestions\": [\"题干中的 a = 4 + 3 应为 a = 4 + √3，建议核对原题确认是否漏印根号\"]}","created_at":"2026-09-25T00:50:42"},{"id":421751,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"PENDING\", \"status_from\": \"PENDING\"}","created_at":"2026-09-25T00:51:47"},{"id":422517,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T01:06:52"},{"id":423394,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T01:21:56"},{"id":424243,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T01:37:00"},{"id":424940,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T01:52:05"},{"id":425830,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T02:07:09"},{"id":426562,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T02:22:13"},{"id":427222,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T02:37:17"},{"id":427899,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T02:52:22"},{"id":428425,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T03:07:26"},{"id":429207,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T03:22:30"},{"id":430101,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T03:37:35"},{"id":430851,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T03:52:39"},{"id":431807,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T04:07:44"},{"id":433003,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T04:22:48"},{"id":434017,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T04:37:52"},{"id":434963,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T04:52:56"},{"id":435839,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T05:08:01"},{"id":436790,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T05:23:05"},{"id":437699,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T05:38:10"},{"id":438531,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T05:53:14"},{"id":439425,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T06:08:18"},{"id":440393,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T06:23:23"},{"id":441218,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T06:38:27"},{"id":442162,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T06:53:32"},{"id":442959,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T07:08:36"},{"id":444093,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T07:23:40"},{"id":444736,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T07:38:45"},{"id":445492,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T07:53:49"},{"id":446440,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T08:08:54"},{"id":447276,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T08:23:58"},{"id":448057,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T08:39:02"},{"id":448572,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T08:54:07"},{"id":449325,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T09:09:11"},{"id":450176,"import_question_id":21391,"audit_type":"RETRO_REPAIR","auditor_name":"retro-deterministic","status":"FAIL","score":0.3,"input_snapshot":"a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= _{a+b}= _{−1}= 6. _{(a+b)(a−b)}−_{x} ab −6\\(x_{x−2}\\)−^\\({a}_{2a}_{2}\\)2 x x2−x(x−2) _{x−2}−^\\({1}_{2}\\)ab 原式= _{x}⋅ _{2}= _{x−2}, 当x = 1时, 原式= _{1−2}=−1. x−2 x x−1\\(1 \\frac{1}{x^{2}−4x+4}\\)（x−2） _{x−2}= 3时\\(\\frac{11.−10}{13.2}\\)原式=\\(2 \\times ( −3)\\)−4 =−6 −4 =−10. 原式( 原式= (4a^{<span class=\"formula-fallback\" title=\"公式识别结果以文本显示\">\\(2frac_x\\)−2−x) ÷_a+1−</s","result_json":"{\"hard\": {\"answer_integrity\": \"答案包含设问或连续选项，疑似题干串入答案栏\"}, \"fixes\": {}, \"status_to\": \"QUARANTINED\", \"status_from\": \"QUARANTINED\"}","created_at":"2026-09-25T09:24:15"}]}