{"ok":true,"question":{"id":21083,"question_uuid":"159aa0e2-9c92-4259-9b5b-eb5a982aead2","job_id":3,"source_file":"人教版/七下/2026江苏七年级期末数学试卷+答案390.pdf","source_page_start":2,"source_page_end":2,"source_bbox":"[298.11, 377.72, 491.08, 426.14]","subject_code":"math","grade_level":"SENIOR","tree_code":"KT_SENIOR_MATH_STANDARD","question_type":"fill_blank","canonical_json":"{\"year\": null, \"blocks\": [{\"text\": \"求下列各式中的值：x （）\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{1}{2}1 (x + 2)^{2} = 7（）23(x −1)^{3} + 81 = 0\"}, {\"type\": \"solution\", \"blocks\": [{\"type\": \"math_inline\", \"latex\": \"\\\\frac{1}{14}\"}, {\"text\": \"由题意,（）解：原方程可变形为： (x + =，所以 + 2 = ，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{2)^{2}}{x} \\\\frac{x}{x} \\\\frac{\\\\sqrt{14}}{\\\\sqrt{14}} =\"}, {\"text\": \"−\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"+2 \\\\sqrt{14}\"}, {\"text\": \"或 = −−；\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{所以}{2} \\\\frac{2}{x}\"}, {\"text\": \"（）解：原方程即为： −1)^{3} = −81，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{3(}{x}\"}, {\"text\": \" 即(x −1)^{3} = −27，所以 −= −13，所以x = −2．\", \"type\": \"text\"}]}], \"source\": {\"bbox\": [298.11, 377.72, 491.08, 426.14], \"file\": \"人教版/七下/2026江苏七年级期末数学试卷+答案390.pdf\", \"regions\": [{\"bbox\": [298.11, 377.72, 491.08, 426.14], \"page\": 2}], \"section\": \"example\", \"page_end\": 2, \"page_start\": 2, \"section_label\": \"跟我学        关键母题，常考题型各个击破！\", \"answer_regions\": [{\"bbox\": [298.1283874511719, 475.8661804199219, 533.8668212890625, 569.4227905273438], \"page\": 2}], \"answer_original\": \"\", \"solution_original\": \"\\\\frac{1}{14} 由题意,（）解：原方程可变形为： (x + =，所以 + 2 = ， \\\\frac{2)^{2}}{x} \\\\frac{x}{x} \\\\frac{\\\\sqrt{14}}{\\\\sqrt{14}} = −+2 \\\\sqrt{14}或 = −−； \\\\frac{所以}{2} \\\\frac{2}{x} （）解：原方程即为： −1)^{3} = −81， \\\\frac{3(}{x} 即(x −1)^{3} = −27，所以 −= −13，所以x = −2．\"}, \"version\": 1, \"grade_level\": \"SENIOR\", \"number_label\": \"2\", \"subject_code\": \"math\", \"question_type\": \"fill_blank\", \"content_sha256\": \"a47ed597a270588a0e3e87f16a59f82a3ce5c93c622d7757b44294702ed25fdd\"}","body_html":"<div>求下列各式中的值：x （）\\(\\frac{1}{2}1 (x + 2)^{2} = 7（）23(x −1)^{3} + 81 = 0\\)</div>","answer_html":"","ways_html":"\\(\\frac{1}{14}\\)由题意,（）解：原方程可变形为： (x + =，所以 + 2 = ，<span class=\"formula-fallback\">frac2)^2x fracxx fracsqrt14sqrt14 =</span>−\\(+2 \\sqrt{14}\\)或 = −−；\\(\\frac{所以}{2} \\frac{2}{x}\\)（）解：原方程即为： −1)^{3} = −81，<span class=\"formula-fallback\">frac3(x</span> 即(x −1)^{3} = −27，所以 −= −13，所以x = −2．","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: FORMULA,COHERENCE,ANSWER","published_question_id":null,"content_sha256":"a47ed597a270588a0e3e87f16a59f82a3ce5c93c622d7757b44294702ed25fdd","created_at":"2026-09-24T22:18:26","updated_at":"2026-09-24T22:35:40","answer_missing":true,"formula_fallback":true,"_number":"2","_section":"example","_section_label":"跟我学        关键母题，常考题型各个击破！"},"assets":[],"knowledge":[],"audits":[{"id":414646,"import_question_id":21083,"audit_type":"STRUCTURE","auditor_name":"structure-rule","status":"PASS","score":0.9,"input_snapshot":"求下列各式中的值：x （） \\frac{1}{2}1 (x + 2)^{2} = 7（）23(x −1)^{3} + 81 = 0","result_json":"{\"ok\": true, \"rule\": \"clean-override\", \"issues\": [], \"confidence\": 0.9}","created_at":"2026-09-24T22:33:48"},{"id":414647,"import_question_id":21083,"audit_type":"FORMULA","auditor_name":"formula-auditor","status":"FAIL","score":1.0,"input_snapshot":"\\frac{1}{2}1 (x + 2)^{2} = 7（）23(x −1)^{3} + 81 = 0; \\frac{1}{14}; \\frac{2)^{2}}{x} \\frac{x}{x} \\frac{\\sqrt{14}}{\\sqrt{14}} =; +2 \\sqrt{14}; \\frac{所以}{2} \\frac{2}{x}","result_json":"{\"risky\": 6, \"issues\": [\"非法 LaTeX #3: unbalanced delimiters\", \"非法 LaTeX #6: unbalanced delimiters\"], \"checked\": 6, \"invalid\": 2, \"visual_check\": \"delegated_to_question_vision\", \"invalid_details\": [{\"index\": 2, \"latex\": \"\\\\frac{2)^{2}}{x} \\\\frac{x}{x} \\\\frac{\\\\sqrt{14}}{\\\\sqrt{14}} =\", \"issues\": [\"unbalanced delimiters\"]}, {\"index\": 5, \"latex\": \"\\\\frac{3(}{x}\", \"issues\": [\"unbalanced delimiters\"]}], \"visually_verified\": 0}","created_at":"2026-09-24T22:33:48"},{"id":414648,"import_question_id":21083,"audit_type":"KNOWLEDGE","auditor_name":"knowledge-auditor","status":"WARN","score":0.0,"input_snapshot":"求下列各式中的值：x （） \\frac{1}{2}1 (x + 2)^{2} = 7（）23(x −1)^{3} + 81 = 0","result_json":"{\"picks\": [], \"tree_code\": \"KT_SENIOR_MATH_STANDARD\", \"candidates\": 0, \"classifier_version\": null}","created_at":"2026-09-24T22:33:48"},{"id":414658,"import_question_id":21083,"audit_type":"SOLVER","auditor_name":"blind-solver","status":"PASS","score":0.8,"input_snapshot":"求下列各式中的值：x （） \\frac{1}{2}1 (x + 2)^{2} = 7（）23(x −1)^{3} + 81 = 0","result_json":"{\"basis\": \"(1) 由 \\\\frac{1}{2}(x+2)^2 = 7 得 (x+2)^2 = 14，开方得 x+2 = \\\\pm\\\\sqrt{14}，故 x = -2 \\\\pm\\\\sqrt{14}。\\\\n(2) 由 3(x-1)^3 + 81 = 0 得 (x-1)^3 = -27，开立方得 x-1 = -3，故 x = -2。\", \"answer\": \"(1) x = 2 或 x = -6；(2) x = 4\", \"confidence\": 0.8, \"handout_chars\": 0}","created_at":"2026-09-24T22:34:23"},{"id":414659,"import_question_id":21083,"audit_type":"CRITIC","auditor_name":"critic","status":"PASS","score":1.0,"input_snapshot":"来源答案: || 盲解:(1) x = 2 或 x = -6；(2) x = 4","result_json":"{\"ok\": true, \"issues\": [], \"merged\": true, \"confidence\": 1.0}","created_at":"2026-09-24T22:34:23"},{"id":414667,"import_question_id":21083,"audit_type":"VISION","auditor_name":"vision-auditor","status":"PASS","score":0.95,"input_snapshot":"题干：\n求下列各式中的值：x （）\\(\\frac{1}{2}1 (x + 2)^{2} = 7（）23(x −1)^{3} + 81 = 0\\)\n答案：\n\n解析：\n\\(\\frac{1}{14}\\)由题意,（）解：原方程可变形为： (x + =，所以 + 2 = ， frac2)^2x fracxx fracsqrt14sqrt14 = −\\(+2 \\sqrt{14}\\)或 = −−；\\(\\frac{所以}{2} \\frac{2}{x}\\)（）解：原方程即为： −1)^{3} = −81， frac3(x 即(x −1)^{3} = −27，所以 −= −13，所以x = −2．","result_json":"{\"ok\": true, \"hard\": false, \"issues\": [], \"confidence\": 0.95, \"suggestions\": [\"题干中的数学表达式排版较为混乱，建议优化为更清晰的格式，例如：\\n(1) \\\\frac{1}{2}(x+2)^2 = 7\\n(2) 3(x-1)^3 + 81 = 0\", \"答案部分为空，建议补充最终答案：\\n(1) x = -2 + \\\\sqrt{14} 或 x = -2 - \\\\sqrt{14}\\n(2) x = -2\"]}","created_at":"2026-09-24T22:34:44"},{"id":414668,"import_question_id":21083,"audit_type":"SYMBOLIC","auditor_name":"symbolic-auditor","status":"WARN","score":0.5,"input_snapshot":"答案: || 盲解:(1) x = 2 或 x = -6；(2) x = 4","result_json":"{\"checks\": [], \"detail\": {\"reason\": \"未提取到来源答案\"}, \"status\": \"WARN\", \"applicable\": true}","created_at":"2026-09-24T22:34:44"},{"id":414687,"import_question_id":21083,"audit_type":"COHERENCE","auditor_name":"coherence-auditor","status":"FAIL","score":0.9,"input_snapshot":"答案: || 解析:\\(\\frac{1}{14}\\)由题意,（）解：原方程可变形为： (x + =，所以 + 2 = ， frac2)^2x fracxx fracsqrt14sqrt14 = −\\(+2 \\sqrt{14}\\)或 = −−；\\(\\frac{所以}{2} \\frac{2}{x}\\)（）解：原方程即为： −1)^{3} = −81， frac3(x 即(x −1)^{3} = −27，所以 −= −1","result_json":"{\"ok\": false, \"which\": [\"solution\"], \"issues\": [\"第一个方程解析错误地将 7 写成 14\", \"第二个方程解析错误地将 -81 写成 -27\", \"(x + 2)^2 = 7: 解析中写为 (x+2)^2 = 14，与题干 7 不符\", \"(x-1)^3 = -81: 解析中写为 (x-1)^3 = -27，与题干 -81 不符\"], \"confidence\": 0.9, \"equation_checks\": [{\"valid\": false, \"reason\": \"解析中写为 (x+2)^2 = 14，与题干 7 不符\", \"equation\": \"(x + 2)^2 = 7\"}, {\"valid\": false, \"reason\": \"解析中写为 (x-1)^3 = -27，与题干 -81 不符\", \"equation\": \"(x-1)^3 = -81\"}]}","created_at":"2026-09-24T22:35:40"},{"id":414688,"import_question_id":21083,"audit_type":"QUALITY_GATE","auditor_name":"quality-gate","status":"FAIL","score":0.938,"input_snapshot":"","result_json":"{\"warnings\": [], \"graded_gate\": true, \"hard_failed\": [\"FORMULA\", \"COHERENCE\", \"ANSWER\"]}","created_at":"2026-09-24T22:35:40"}]}