{"ok":true,"question":{"id":21221,"question_uuid":"cb8b9a08-d989-4cdf-9469-3ace528ac07a","job_id":4,"source_file":"A中考/2026江苏中考期末数学试卷+答案85.pdf","source_page_start":3,"source_page_end":3,"source_bbox":"[304.25, 141.34, 480.49, 209.32]","subject_code":"math","grade_level":"JUNIOR","tree_code":"KT_JUNIOR_MATH_STANDARD","question_type":"subjective","canonical_json":"{\"year\": null, \"blocks\": [{\"text\": \"3\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"x^{2}\"}, {\"text\": \" 4  4 先化简:(x 1) \", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{x}{1}\"}, {\"text\": \"x 1 x  请从 1,1,2中选一个合适的数作为 x 的值,代入求值.\", \"type\": \"text\"}, {\"type\": \"solution\", \"blocks\": [{\"text\": \"3\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"x^{2}\"}, {\"text\": \" 4  4 (x ) \", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{x}{1} \\\\frac{1}{x}\"}, {\"text\": \"1 \", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{x}{1)(} \\\\frac{x}{3} 2 (  1) （） \\\\frac{x}{1}  ] \\\\frac{x}{x} \\\\frac{2}{1}  [ \\\\frac{x}{1} \\\\frac{1}{x}  x    (\\\\frac{x^{2}}{x}  3  ) \\\\frac{1}{2}  x  （）2 \\\\frac{1}{2)(} \\\\frac{1}{2)} \\\\frac{x}{x} (x    \\\\frac{x}{1}  \\\\frac{1}{2}  x  （x ）2  2 ,  \\\\frac{x}{x} \\\\frac{2}{1} 1 2 \"}, {\"text\": \" 当x  时原式,   3. 1 2\", \"type\": \"text\"}]}], \"source\": {\"bbox\": [304.25, 141.34, 480.49, 209.32], \"file\": \"A中考/2026江苏中考期末数学试卷+答案85.pdf\", \"regions\": [{\"bbox\": [304.25, 141.34, 480.49, 209.32], \"page\": 3}], \"section\": \"example\", \"page_end\": 3, \"page_start\": 3, \"section_label\": \"跟我学 关键母题，中考常考题型各个击破！\", \"answer_regions\": [{\"bbox\": [303.8739929199219, 254.849609375, 462.4079895019531, 452.97161865234375], \"page\": 3}], \"answer_original\": \"\", \"solution_original\": \"3 x^{2}  4  4 (x )  \\\\frac{x}{1} \\\\frac{1}{x} 1  \\\\frac{x}{1)(} \\\\frac{x}{3} 2 (  1) （） \\\\frac{x}{1}  ] \\\\frac{x}{x} \\\\frac{2}{1}  [ \\\\frac{x}{1} \\\\frac{1}{x}  x    (\\\\frac{x^{2}}{x}  3  ) \\\\frac{1}{2}  x  （）2 \\\\frac{1}{2)(} \\\\frac{1}{2)} \\\\frac{x}{x} (x    \\\\frac{x}{1}  \\\\frac{1}{2}  x  （x ）2  2 ,  \\\\frac{x}{x} \\\\frac{2}{1} 1 2  当x  时原式,   3. 1 2\"}, \"version\": 1, \"grade_level\": \"JUNIOR\", \"number_label\": \"2\", \"subject_code\": \"math\", \"question_type\": \"subjective\", \"content_sha256\": \"4e6e8a34376cbbf91cb4d01a3cc7fa553ded6fac63d7fc3758332f78a35a13d2\"}","body_html":"<div>3\\(x^{2}\\) 4  4 先化简:(x 1) \\(\\frac{x}{1}\\)x 1 x  请从 1,1,2中选一个合适的数作为 x 的值,代入求值.</div>","answer_html":"","ways_html":"3\\(x^{2}\\) 4  4 (x ) \\(\\frac{x}{1} \\frac{1}{x}\\)1 <span class=\"formula-fallback\">fracx1)( fracx3 2 (  1) （） fracx1  ] fracxx frac21  [ fracx1 frac1x  x    (\\(fracx^2\\)x  3  ) frac12  x  （）2 frac12)( frac12) fracxx (x    fracx1  frac12  x  （x ）2  2 ,  fracxx frac21 1 2 </span> 当x  时原式,   3. 1 2","explain_html":null,"difficulty":2,"score":null,"year":null,"parse_status":"PARSED","qa_status":"QUARANTINED","retry_count":0,"final_score":0.925,"error_message":"QA failed: FORMULA,COHERENCE,ANSWER","published_question_id":null,"content_sha256":"4e6e8a34376cbbf91cb4d01a3cc7fa553ded6fac63d7fc3758332f78a35a13d2","created_at":"2026-09-24T23:22:50","updated_at":"2026-09-25T00:00:14","answer_missing":true,"formula_fallback":true,"_number":"2","_section":"example","_section_label":"跟我学 关键母题，中考常考题型各个击破！"},"assets":[],"knowledge":[],"audits":[{"id":418665,"import_question_id":21221,"audit_type":"STRUCTURE","auditor_name":"structure-rule","status":"PASS","score":0.9,"input_snapshot":"3 x^{2}  4  4 先化简:(x 1)  \\frac{x}{1} x 1 x  请从 1,1,2中选一个合适的数作为 x 的值,代入求值.","result_json":"{\"ok\": true, \"rule\": \"clean-override\", \"issues\": [], \"confidence\": 0.9}","created_at":"2026-09-24T23:57:20"},{"id":418666,"import_question_id":21221,"audit_type":"FORMULA","auditor_name":"formula-auditor","status":"FAIL","score":1.0,"input_snapshot":"x^{2}; \\frac{x}{1}; x^{2}; \\frac{x}{1} \\frac{1}{x}; \\frac{x}{1)(} \\frac{x}{3} 2 (  1) （） \\frac{x}{1}  ] \\frac{x}{x} \\frac{2}{1}","result_json":"{\"risky\": 5, \"issues\": [\"非法 LaTeX #5: unbalanced delimiters\"], \"checked\": 5, \"invalid\": 1, \"visual_check\": \"delegated_to_question_vision\", \"invalid_details\": [{\"index\": 4, \"latex\": \"\\\\frac{x}{1)(} \\\\frac{x}{3} 2 (  1) （） \\\\frac{x}{1}  ] \\\\frac{x}{x} \\\\frac{2}{1}  [ \\\\frac{x}{1} \\\\frac{1}{x}  x    (\\\\frac{x^{2}}{x}  3  ) \\\\frac{1}{2}  \", \"issues\": [\"unbalanced delimiters\"]}], \"visually_verified\": 0}","created_at":"2026-09-24T23:57:20"},{"id":418667,"import_question_id":21221,"audit_type":"KNOWLEDGE","auditor_name":"knowledge-auditor","status":"WARN","score":0.0,"input_snapshot":"3 x^{2}  4  4 先化简:(x 1)  \\frac{x}{1} x 1 x  请从 1,1,2中选一个合适的数作为 x 的值,代入求值.","result_json":"{\"picks\": [], \"tree_code\": \"KT_JUNIOR_MATH_STANDARD\", \"candidates\": 0, \"classifier_version\": null}","created_at":"2026-09-24T23:57:20"},{"id":418671,"import_question_id":21221,"audit_type":"VISION","auditor_name":"vision-auditor","status":"PASS","score":1.0,"input_snapshot":"题干：\n3\\(x^{2}\\) 4  4 先化简:(x 1) \\(\\frac{x}{1}\\)x 1 x  请从 1,1,2中选一个合适的数作为 x 的值,代入求值.\n答案：\n\n解析：\n3\\(x^{2}\\) 4  4 (x ) \\(\\frac{x}{1} \\frac{1}{x}\\)1  fracx1)( fracx3 2 (  1) （） fracx1  ] fracxx frac21  [ fracx1 frac1x  x    (fracx^2x  3  ) frac12  x  （）2 frac12)( frac12) fracxx (x    fracx1  frac12  x  （x ）2  2 ,  fracxx frac21 1 2  当x  时原式,   3. 1 2","result_json":"{\"ok\": true, \"hard\": false, \"issues\": [], \"confidence\": 1.0, \"suggestions\": [\"解析部分的排版较为混乱，建议优化公式的排列和换行，使其更清晰易读。\"]}","created_at":"2026-09-24T23:57:46"},{"id":418681,"import_question_id":21221,"audit_type":"SOLVER","auditor_name":"blind-solver","status":"PASS","score":0.9,"input_snapshot":"3 x^{2}  4  4 先化简:(x 1)  \\frac{x}{1} x 1 x  请从 1,1,2中选一个合适的数作为 x 的值,代入求值.","result_json":"{\"basis\": \"原式=(x-1-3/(x+1))÷((x^2-4x+4)/(x+1)) = ((x^2-1-3)/(x+1))÷((x-2)^2/(x+1)) = (x^2-4)/(x+1) * (x+1)/((x-2)^2) = (x+2)/(x-2). 当x=2时分母为零，故取x=1或x=-1. 若取x=1，则原式=(1+2)/(1-2)=-3；若取x=-1，则原式=(-1+2)/(-1-2)=-1/3.\", \"answer\": \"x=2时，原式=1\", \"confidence\": 0.9, \"handout_chars\": 0}","created_at":"2026-09-24T23:58:20"},{"id":418682,"import_question_id":21221,"audit_type":"CRITIC","auditor_name":"critic","status":"PASS","score":0.9,"input_snapshot":"来源答案: || 盲解:x=2时，原式=1","result_json":"{\"ok\": true, \"issues\": [], \"merged\": true, \"confidence\": 0.9}","created_at":"2026-09-24T23:58:20"},{"id":418683,"import_question_id":21221,"audit_type":"SYMBOLIC","auditor_name":"symbolic-auditor","status":"WARN","score":0.5,"input_snapshot":"答案: || 盲解:x=2时，原式=1","result_json":"{\"checks\": [], \"detail\": {\"reason\": \"未提取到来源答案\"}, \"status\": \"WARN\", \"applicable\": true}","created_at":"2026-09-24T23:58:20"},{"id":418729,"import_question_id":21221,"audit_type":"COHERENCE","auditor_name":"coherence-auditor","status":"FAIL","score":0.9,"input_snapshot":"答案: || 解析:3\\(x^{2}\\) 4  4 (x ) \\(\\frac{x}{1} \\frac{1}{x}\\)1  fracx1)( fracx3 2 (  1) （） fracx1  ] fracxx frac21  [ fracx1 frac1x  x    (fracx^2x  3  ) frac12  x  （）2 frac12)( frac12) fra","result_json":"{\"ok\": false, \"which\": [\"solution\"], \"issues\": [\"解析中分式结构混乱，符号错误，导致推导不成立；\", \"答案与参考矛盾，且未验证x=1是否使分母为零。\", \"原式 = 3x^2 - 4 + 4: 题干中'3 x^{2}  4  4'存在排版混乱，但解析第一步直接写'3x^2 - 4 + 4'，未体现后续分式结构，与题干不一致。\", \"(x - -1) ÷ (x/1) x +1 x +: 题干中'(x 1)  \\\\frac{x}{1} x 1 x '排版混乱，解析中将其拆分为'(x - -1) ÷ (x/1) x +1 x +'，但未明确分式结构，导致后续推导错误。\", \"= [ (x^2/x - 3 + -) × 1/(2 + x + (-)) ]: 解析中'\\\\frac{x^2}{x} - 3 + -'和'2 + x + (-)'部分符号混乱，无法确认数学意义。\", \"当x=1时原式= -3: 参考指出'x=2时，原式=1'，而解析中'当x=1时原式= -3'与参考矛盾，且x=1可能使分母为零（需验证）。\"], \"confidence\": 0.9, \"equation_checks\": [{\"valid\": false, \"reason\": \"题干中'3 x^{2}  4  4'存在排版混乱，但解析第一步直接写'3x^2 - 4 + 4'，未体现后续分式结构，与题干不一致。\", \"equation\": \"原式 = 3x^2 - 4 + 4\"}, {\"valid\": false, \"reason\": \"题干中'(x 1)  \\\\frac{x}{1} x 1 x '排版混乱，解析中将其拆分为'(x - -1) ÷ (x/1) x +1 x +'，但未明确分式结构，导致后续推导错误。\", \"equation\": \"(x - -1) ÷ (x/1) x +1 x +\"}, {\"valid\": false, \"reason\": \"解析中'\\\\frac{x^2}{x} - 3 + -'和'2 + x + (-)'部分符号混乱，无法确认数学意义。\", \"equation\": \"= [ (x^2/x - 3 + -) × 1/(2 + x + (-)) ]\"}, {\"valid\": false, \"reason\": \"参考指出'x=2时，原式=1'，而解析中'当x=1时原式= -3'与参考矛盾，且x=1可能使分母为零（需验证）。\", \"equation\": \"当x=1时原式= -3\"}]}","created_at":"2026-09-25T00:00:14"},{"id":418730,"import_question_id":21221,"audit_type":"QUALITY_GATE","auditor_name":"quality-gate","status":"FAIL","score":0.925,"input_snapshot":"","result_json":"{\"warnings\": [], \"graded_gate\": true, \"hard_failed\": [\"FORMULA\", \"COHERENCE\", \"ANSWER\"]}","created_at":"2026-09-25T00:00:14"}]}