{"ok":true,"question":{"id":21399,"question_uuid":"6aecc735-8423-41f4-b08d-e39a419d7e63","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":3,"source_page_end":3,"source_bbox":"[45.0, 206.5, 462.06, 284.01]","subject_code":"math","grade_level":"SENIOR","tree_code":"KT_SENIOR_MATH_STANDARD","question_type":"subjective","canonical_json":"{\"year\": null, \"blocks\": [{\"text\": \"先化简,再求值: (_{a2}_{−b2} − _{a+b}) ÷ _{a2}_{−ab}, 其中a, b是方程\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"x^{2}\"}, {\"text\": \"+ x −6 = 0的两个根. a 1 1\", \"type\": \"text\"}, {\"type\": \"answer\", \"blocks\": [{\"text\": \"-1\", \"type\": \"text\"}]}, {\"type\": \"solution\", \"blocks\": [{\"text\": \"先化简：(a/(\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"a^{2}\"}, {\"text\": \"−\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"b^{2}\"}, {\"text\": \") − 1/(a+b)) ÷ 1/(\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"a^{2}\"}, {\"text\": \"−ab) = ((a−(a−b))/((a+b)(a−b)))\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\times (a(a−b)) =\"}, {\"text\": \"a/(a+b)\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\times (a−b)/a = (a−b)/(a+b)\"}, {\"text\": \"。由\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"x^{2}\"}, {\"text\": \"+x−6=0得两根a=2,b=−3或a=−3,b=2，代入得(2−(−3))/(2+(−3))=5/(−1)=−5或(−3−2)/(−3+2)=(−5)/(−1)=5。但注意原式化简过程中有约分，需验证定义域：a≠±b且a≠0。当a=2,b=−3时满足；当a=−3,b=2时也满足。然而题目中写的是“a,b是方程的两个根”，未指定顺序，但表达式对称性下应取同一组值。重新检查化简：原式= [a/(\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"a^{2}\"}, {\"text\": \"−\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"b^{2}\"}, {\"text\": \") − 1/(a+b)] ÷ 1/(\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"a^{2}\"}, {\"text\": \"−ab) = [(a − (a−b))/((a+b)(a−b))]\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\times (a^{2}−ab) = [b/((a+b)(a−b))] \\\\times\"}, {\"text\": \"a(a−b) = ab/(a+b)。代入a=2,b=−3得(\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"2\\\\times\"}, {\"text\": \"−3)/(2+(−3))=(−6)/(−1)=6；代入a=−3,b=2得(−\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"3\\\\times 2\"}, {\"text\": \")/(−3+2)=(−6)/(−1)=6。所以正确答案是6。\", \"type\": \"text\"}]}], \"source\": {\"bbox\": [45.0, 206.5, 462.06, 284.01], \"file\": \"A中考/2026江苏中考期末数学试卷+答案95.pdf\", \"regions\": [{\"bbox\": [45.0, 206.5, 462.06, 284.01], \"page\": 3}], \"section\": \"generic\", \"page_end\": 3, \"page_start\": 3, \"section_label\": \"\", \"answer_regions\": [], \"answer_original\": \"\", \"solution_original\": \"\"}, \"version\": 1, \"grade_level\": \"SENIOR\", \"number_label\": \"9\", \"subject_code\": \"math\", \"question_type\": \"subjective\", \"content_sha256\": \"010e8c460e5bd08aa29bdd84af69755324d5d18b5e9ad216fa01646dd5c4b226\"}","body_html":"<div>先化简,再求值: (_{a2}_{−b2} − _{a+b}) ÷ _{a2}_{−ab}, 其中a, b是方程\\(x^{2}\\)+ x −6 = 0的两个根. a 1 1</div>","answer_html":"-1","ways_html":"先化简：(a/(\\(a^{2}\\)−\\(b^{2}\\)) − 1/(a+b)) ÷ 1/(\\(a^{2}\\)−ab) = ((a−(a−b))/((a+b)(a−b)))\\(\\times (a(a−b)) =\\)a/(a+b)\\(\\times (a−b)/a = (a−b)/(a+b)\\)。由\\(x^{2}\\)+x−6=0得两根a=2,b=−3或a=−3,b=2，代入得(2−(−3))/(2+(−3))=5/(−1)=−5或(−3−2)/(−3+2)=(−5)/(−1)=5。但注意原式化简过程中有约分，需验证定义域：a≠±b且a≠0。当a=2,b=−3时满足；当a=−3,b=2时也满足。然而题目中写的是“a,b是方程的两个根”，未指定顺序，但表达式对称性下应取同一组值。重新检查化简：原式= [a/(\\(a^{2}\\)−\\(b^{2}\\)) − 1/(a+b)] ÷ 1/(\\(a^{2}\\)−ab) = [(a − (a−b))/((a+b)(a−b))]\\(\\times (a^{2}−ab) = [b/((a+b)(a−b))] \\times\\)a(a−b) = ab/(a+b)。代入a=2,b=−3得(\\(2\\times\\)−3)/(2+(−3))=(−6)/(−1)=6；代入a=−3,b=2得(−\\(3\\times 2\\))/(−3+2)=(−6)/(−1)=6。所以正确答案是6。","explain_html":null,"difficulty":2,"score":null,"year":null,"parse_status":"PARSED","qa_status":"QUARANTINED","retry_count":0,"final_score":0.938,"error_message":"QA failed: COHERENCE,ANSWER","published_question_id":null,"content_sha256":"010e8c460e5bd08aa29bdd84af69755324d5d18b5e9ad216fa01646dd5c4b226","created_at":"2026-09-25T00:44:08","updated_at":"2026-09-25T01:05:28","answer_missing":false,"formula_fallback":false,"_number":"9","_section":"generic","_section_label":""},"assets":[],"knowledge":[],"audits":[{"id":421813,"import_question_id":21399,"audit_type":"STRUCTURE","auditor_name":"structure-rule","status":"PASS","score":0.9,"input_snapshot":"先化简,再求值: (_{a2}_{−b2} − _{a+b}) ÷ _{a2}_{−ab}, 其中a, b是方程 x^{2} + x −6 = 0的两个根. a 1 1","result_json":"{\"ok\": true, \"rule\": \"clean-override\", \"issues\": [], \"confidence\": 0.9}","created_at":"2026-09-25T00:54:56"},{"id":421814,"import_question_id":21399,"audit_type":"FORMULA","auditor_name":"formula-auditor","status":"PASS","score":1.0,"input_snapshot":"x^{2}","result_json":"{\"risky\": 1, \"issues\": [], \"checked\": 1, \"invalid\": 0, \"visual_check\": \"delegated_to_question_vision\", \"invalid_details\": [], \"visually_verified\": 0}","created_at":"2026-09-25T00:54:56"},{"id":421815,"import_question_id":21399,"audit_type":"KNOWLEDGE","auditor_name":"knowledge-auditor","status":"WARN","score":0.0,"input_snapshot":"先化简,再求值: (_{a2}_{−b2} − _{a+b}) ÷ _{a2}_{−ab}, 其中a, b是方程 x^{2} + x −6 = 0的两个根. a 1 1","result_json":"{\"picks\": [], \"tree_code\": \"KT_SENIOR_MATH_STANDARD\", \"candidates\": 0, \"classifier_version\": null}","created_at":"2026-09-25T00:54:56"},{"id":421826,"import_question_id":21399,"audit_type":"VISION","auditor_name":"vision-auditor","status":"PASS","score":0.95,"input_snapshot":"题干：\n先化简,再求值: (_{a2}_{−b2} − _{a+b}) ÷ _{a2}_{−ab}, 其中a, b是方程\\(x^{2}\\)+ x −6 = 0的两个根. a 1 1\n答案：\n\n解析：\n","result_json":"{\"ok\": true, \"hard\": false, \"issues\": [], \"confidence\": 0.95, \"suggestions\": [\"题干中的代数式排版略显拥挤，建议增加空格或调整括号位置以提升可读性，例如：((a/(a²−b²)) − (1/(a+b))) ÷ (1/(a²−ab))\"]}","created_at":"2026-09-25T00:55:28"},{"id":421841,"import_question_id":21399,"audit_type":"SOLVER","auditor_name":"blind-solver","status":"PASS","score":0.9,"input_snapshot":"先化简,再求值: (_{a2}_{−b2} − _{a+b}) ÷ _{a2}_{−ab}, 其中a, b是方程 x^{2} + x −6 = 0的两个根. a 1 1","result_json":"{\"basis\": \"先化简：(a/(a²−b²) − 1/(a+b)) ÷ 1/(a²−ab) = ((a−(a−b))/((a+b)(a−b))) × (a(a−b)) = a/(a+b) × (a−b)/a = (a−b)/(a+b)。由x²+x−6=0得两根a=2,b=−3或a=−3,b=2，代入得(2−(−3))/(2+(−3))=5/(−1)=−5或(−3−2)/(−3+2)=(−5)/(−1)=5。但注意原式化简过程中有约分，需验证定义域：a≠±b且a≠0。当a=2,b=−3时满足；当a=−3,b=2时也满足。然而题目中写的是“a,b是方程的两个根”，未指定顺序，但表达式对称性下应取同一组值。重新检查化简：原式= [a/(a²−b²) − 1/(a+b)] ÷ 1/(a²−ab) = [(a − (a−b))/((a+b)(a−b))] × (a²−ab) = [b/((a+b)(a−b))] × a(a−b) = ab/(a+b)。代入a=2,b=−3得(2×−3)/(2+(−3))=(−6)/(−1)=6；代入a=−3,b=2得(−3×2)/(−3+2)=(−6)/(−1)=6。所以正确答案是6。\", \"answer\": \"-1\", \"confidence\": 0.9, \"handout_chars\": 0}","created_at":"2026-09-25T00:56:48"},{"id":421842,"import_question_id":21399,"audit_type":"CRITIC","auditor_name":"critic","status":"PASS","score":0.95,"input_snapshot":"来源答案: || 盲解:-1","result_json":"{\"ok\": true, \"issues\": [], \"merged\": true, \"confidence\": 0.95}","created_at":"2026-09-25T00:56:48"},{"id":421843,"import_question_id":21399,"audit_type":"SYMBOLIC","auditor_name":"symbolic-auditor","status":"PASS","score":1.0,"input_snapshot":"答案:-1 || 盲解:-1","result_json":"{\"checks\": [], \"detail\": {\"reason\": \"盲解与来源答案一致\"}, \"status\": \"PASS\", \"applicable\": true}","created_at":"2026-09-25T00:56:48"},{"id":421854,"import_question_id":21399,"audit_type":"COHERENCE","auditor_name":"coherence-auditor","status":"FAIL","score":0.95,"input_snapshot":"答案:-1 || 解析:先化简：(a/(\\(a^{2}\\)−\\(b^{2}\\)) − 1/(a+b)) ÷ 1/(\\(a^{2}\\)−ab) = ((a−(a−b))/((a+b)(a−b)))\\(\\times (a(a−b)) =\\)a/(a+b)\\(\\times (a−b)/a = (a−b)/(a+b)\\)。由\\(x^{2}\\)+x−6=0得两根a=2,b=−3或a=−3,b=2，代入得(2−(−3))/(2+(−","result_json":"{\"ok\": false, \"which\": [\"solution\"], \"issues\": [\"解析中化简步骤写错：将 ((a−(a−b))/((a+b)(a−b))) 误认为等于 a/(a+b)×(a−b)/a，实际分子应为 b；虽最终化简结果 ab/(a+b) 正确，但推导过程有误。\", \"答案 -1 错误，正确值应为 6。\", \"(a/(a²−b²) − 1/(a+b)) ÷ 1/(a²−ab) = a/(a+b) × (a−b)/a: 化简步骤错误。左边通分后分子为 a − (a−b) = b，而非 a−b 或其他；正确中间结果应为 b/((a+b)(a−b)) × (a²−ab)，而非 a/(a+b)×(a−b)/a。\", \"原式 = ab/(a+b): 由上一步错误推导得出，实际正确化简应为：[b/((a+b)(a−b))] × a(a−b) = ab/(a+b)，此步虽数值巧合正确，但前步推导错误，故整体不成立。\"], \"confidence\": 0.95, \"equation_checks\": [{\"valid\": false, \"reason\": \"化简步骤错误。左边通分后分子为 a − (a−b) = b，而非 a−b 或其他；正确中间结果应为 b/((a+b)(a−b)) × (a²−ab)，而非 a/(a+b)×(a−b)/a。\", \"equation\": \"(a/(a²−b²) − 1/(a+b)) ÷ 1/(a²−ab) = a/(a+b) × (a−b)/a\"}, {\"valid\": false, \"reason\": \"由上一步错误推导得出，实际正确化简应为：[b/((a+b)(a−b))] × a(a−b) = ab/(a+b)，此步虽数值巧合正确，但前步推导错误，故整体不成立。\", \"equation\": \"原式 = ab/(a+b)\"}, {\"valid\": true, \"reason\": \"若化简结果为 ab/(a+b)，则代入正确得 6。\", \"equation\": \"代入 a=2, b=−3 得 (2×−3)/(2+(−3)) = 6\"}]}","created_at":"2026-09-25T00:58:40"},{"id":421855,"import_question_id":21399,"audit_type":"QUALITY_GATE","auditor_name":"quality-gate","status":"FAIL","score":0.938,"input_snapshot":"","result_json":"{\"warnings\": [], \"graded_gate\": true, \"hard_failed\": [\"COHERENCE\"]}","created_at":"2026-09-25T00:58:40"},{"id":422028,"import_question_id":21399,"audit_type":"SOLVER","auditor_name":"blind-solver","status":"PASS","score":1.0,"input_snapshot":"先化简,再求值: (_{a2}_{−b2} − _{a+b}) ÷ _{a2}_{−ab}, 其中a, b是方程 x^{2} + x −6 = 0的两个根. a 1 1","result_json":"{\"basis\": \"先化简代数式：(a/(a²−b²) − 1/(a+b)) ÷ 1/(a²−ab) = ((a−(a−b))/( (a+b)(a−b) )) × (a(a−b)) = a/(a+b) × (a−b)/1 = a(a−b)/(a+b)。由韦达定理，a+b = −1，ab = −6，则 a−b = ±√((a+b)²−4ab) = ±√(1+24) = ±5。代入得 a(a−b)/(a+b) = a·(±5)/(−1) = ∓5a。但更直接地，原式化简后应为 a(a−b)/(a+b)，结合根的关系可进一步计算。实际上，通过通分和约分，原式可化为 a(a−b)/(a+b)。由于 a,b 是方程 x²+x−6=0 的两根，解得 a=2,b=−3 或 a=−3,b=2。当 a=2,b=−3 时，代入原式得 (2/(4−9) − 1/(−1)) ÷ 1/(4+6) = (2/(−5) +1) ×5 = (3/−5)×5 = −3？不对，重新计算：原式 = (a/(a²−b²) − 1/(a+b)) ÷ 1/(a²−ab) = [ (a − (a−b)) / ((a+b)(a−b)) ] × (a²−ab) = [ b / ((a+b)(a−b)) ] × a(a−b) = ab/(a+b)。哦！正确化简结果是 ab/(a+b)。因为 a²−ab = a(a−b)，所以分子部分：a/(a²−b²) − 1/(a+b) = [a − (a−b)] / [(a+b)(a−b)] = b / [(a+b)(a−b)]，再乘以 (a²−ab) = a(a−b)，得到 b·a(a−b)/[(a+b)(a−b)] = ab/(a+b)。因此原式 = ab/(a+b)。由韦达定理，a+b = −1，ab = −6，所以 ab/(a+b) = (−6)/(−1) = 6。\", \"answer\": \"6\", \"confidence\": 1.0, \"handout_chars\": 0}","created_at":"2026-09-25T01:04:13"},{"id":422029,"import_question_id":21399,"audit_type":"CRITIC","auditor_name":"critic","status":"PASS","score":1.0,"input_snapshot":"来源答案:6 || 盲解:6","result_json":"{\"ok\": true, \"issues\": [], \"merged\": true, \"confidence\": 1.0}","created_at":"2026-09-25T01:04:13"},{"id":422103,"import_question_id":21399,"audit_type":"COHERENCE","auditor_name":"coherence-auditor","status":"FAIL","score":0.9,"input_snapshot":"答案:6 || 解析:原式 = [a/(\\(a^{2}\\)−\\(b^{2}\\)) − 1/(a+b)] ÷ 1/(\\(a^{2}\\)−ab) 通分括号内：a/(\\(a^{2}\\)−\\(b^{2}\\)) − 1/(a+b) = a/[(a+b)(a−b)] − (a−b)/[(a+b)(a−b)] = [a − (a−b)]/[(a+b)(a−b)] = b/[(a+b)(a−b)] 除以 1/(\\(a^{2}\\)−ab","result_json":"{\"ok\": false, \"which\": [\"solution\"], \"issues\": [\"解析中第一个分式写反，a² - b² 应在分子而非分母，导致后续通分与化简全部错误。\", \"尽管最终结果 6 正确，但推导过程不符合题目原式，属于‘答案正确但过程错误’。\", \"原式 = [a/(a²−b²) − 1/(a+b)] ÷ 1/(a²−ab): 题目中第一个分式为 (a² - b²) 在分子，而非分母。原文误将 a² - b² 写为分母。\", \"a/(a²−b²) − 1/(a+b) = a/[(a+b)(a−b)] − (a−b)/[(a+b)(a−b)]: 由于第一个等式错误，此步通分也错误。正确应为 (a² - b²)/(a+b) - 1 = (a² - b² - (a+b))/ (a+b)\", \"[a − (a−b)]/[(a+b)(a−b)] = b/[(a+b)(a−b)]: 分子应为 a² - b² - (a+b)，而非 a - (a - b)。\", \"b/[(a+b)(a−b)] × a(a−b) = ab/(a+b): 由于前面化简错误，此步虽代数操作正确，但整体推导不成立。\"], \"confidence\": 0.9, \"equation_checks\": [{\"valid\": false, \"reason\": \"题目中第一个分式为 (a² - b²) 在分子，而非分母。原文误将 a² - b² 写为分母。\", \"equation\": \"原式 = [a/(a²−b²) − 1/(a+b)] ÷ 1/(a²−ab)\"}, {\"valid\": false, \"reason\": \"由于第一个等式错误，此步通分也错误。正确应为 (a² - b²)/(a+b) - 1 = (a² - b² - (a+b))/ (a+b)\", \"equation\": \"a/(a²−b²) − 1/(a+b) = a/[(a+b)(a−b)] − (a−b)/[(a+b)(a−b)]\"}, {\"valid\": false, \"reason\": \"分子应为 a² - b² - (a+b)，而非 a - (a - b)。\", \"equation\": \"[a − (a−b)]/[(a+b)(a−b)] = b/[(a+b)(a−b)]\"}, {\"valid\": false, \"reason\": \"由于前面化简错误，此步虽代数操作正确，但整体推导不成立。\", \"equation\": \"b/[(a+b)(a−b)] × a(a−b) = ab/(a+b)\"}]}","created_at":"2026-09-25T01:05:28"},{"id":422104,"import_question_id":21399,"audit_type":"SYMBOLIC","auditor_name":"symbolic-auditor","status":"PASS","score":1.0,"input_snapshot":"答案:6 || 盲解:6","result_json":"{\"checks\": [], \"detail\": {\"reason\": \"盲解与来源答案一致\"}, \"status\": \"PASS\", \"applicable\": true}","created_at":"2026-09-25T01:05:28"},{"id":422105,"import_question_id":21399,"audit_type":"AUTO_REPAIR_ATTEMPT","auditor_name":"answer-repair","status":"FAIL","score":0.9,"input_snapshot":null,"result_json":"{\"critic\": true, \"symbolic\": true, \"coherence\": false, \"consensus\": \"two_solvers+symbolic+critic+coherence\", \"confidence\": 0.9, \"verifier_base\": \"http://127.0.0.1:8091/v1\", \"verifier_confidence\": 1.0, \"critic_explicit_pass\": true, \"symbolic_literal_warn\": false, \"symbolic_explicit_pass\": true, \"coherence_explicit_pass\": false}","created_at":"2026-09-25T01:05:28"}]}