{"ok":true,"question":{"id":21393,"question_uuid":"d2cc7818-9be8-4c68-b1ed-26dd0889f6b9","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, 454.9, 351.3, 527.96]","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: (2x + 1)(2x −1) + x(3 −4x),其中 x = _{3}. 1\", \"type\": \"text\"}, {\"type\": \"answer\", \"blocks\": [{\"type\": \"math_inline\", \"latex\": \"\\\\frac{8.3}{10.−1}\"}]}, {\"type\": \"solution\", \"blocks\": [{\"text\": \"原式= _{a−2}⋅ _{2}= _{a+2}, 当a = _{3−2+2}= a+2 3(a−2) 3 （a+2） 原式= [ _{a+b}] ÷ _{a2}_{−ab}= _{(a+b)(a−b)}⋅a(a −b) = _{a+b}.因为a, b是方程\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"x^{2}\"}, {\"text\": \"+ x −6 = 0的两个根, ∴ 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, 454.9, 351.3, 527.96], \"file\": \"A中考/2026江苏中考期末数学试卷+答案95.pdf\", \"regions\": [{\"bbox\": [45.0, 454.9, 351.3, 527.96], \"page\": 1}], \"section\": \"generic\", \"page_end\": 1, \"page_start\": 1, \"section_label\": \"\", \"answer_regions\": [{\"bbox\": [30.0, 375.3380126953125, 570.3590087890625, 764.845947265625], \"page\": 6}, {\"bbox\": [30.0, 54.097999572753906, 565.6439819335938, 186.4459991455078], \"page\": 7}], \"answer_original\": \"\\\\frac{8.3}{10.−1}\", \"solution_original\": \"原式= _{a−2}⋅ _{2}= _{a+2}, 当a = _{3−2+2}= a+2 3(a−2) 3 （a+2） 原式= [ _{a+b}] ÷ _{a2}_{−ab}= _{(a+b)(a−b)}⋅a(a −b) = _{a+b}.因为a, b是方程x^{2} + x −6 = 0的两个根, ∴ 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 × ( −3) −4 =−6 −4 =−10. 原式( 原式= (4a^{2\\\\frac{_{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 × 2^{2}+ 6 × 2 × ( −1) =−4.\"}, \"version\": 1, \"grade_level\": \"SENIOR\", \"number_label\": \"3\", \"subject_code\": \"math\", \"question_type\": \"subjective\", \"content_sha256\": \"2d1491296cf404e4b2dcb8b28959ee994ad0a3c5af5976ac9559a570b83d6e0a\"}","body_html":"<div>先化简,再求值: (2x + 1)(2x −1) + x(3 −4x),其中 x = _{3}. 1</div>","answer_html":"\\(\\frac{8.3}{10.−1}\\)","ways_html":"原式= _{a−2}⋅ _{2}= _{a+2}, 当a = _{3−2+2}= a+2 3(a−2) 3 （a+2） 原式= [ _{a+b}] ÷ _{a2}_{−ab}= _{(a+b)(a−b)}⋅a(a −b) = _{a+b}.因为a, b是方程\\(x^{2}\\)+ x −6 = 0的两个根, ∴ 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\\).","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","published_question_id":null,"content_sha256":"2d1491296cf404e4b2dcb8b28959ee994ad0a3c5af5976ac9559a570b83d6e0a","created_at":"2026-09-25T00:44:08","updated_at":"2026-09-25T00:53:00","answer_missing":false,"formula_fallback":true,"_number":"3","_section":"generic","_section_label":""},"assets":[],"knowledge":[],"audits":[{"id":421395,"import_question_id":21393,"audit_type":"STRUCTURE","auditor_name":"structure-auditor","status":"PASS","score":0.95,"input_snapshot":"先化简,再求值: (2x + 1)(2x −1) + x(3 −4x),其中 x = _{3}. 1","result_json":"{\"ok\": true, \"issues\": [], \"confidence\": 0.95}","created_at":"2026-09-25T00:51:02"},{"id":421396,"import_question_id":21393,"audit_type":"FORMULA","auditor_name":"formula-auditor","status":"FAIL","score":1.0,"input_snapshot":"\\frac{8.3}{10.−1}; x^{2}; 6\\frac{1}{x(x−1)}; x_{x−2}; {a}_{2a}_{2}","result_json":"{\"risky\": 27, \"issues\": [\"非法 LaTeX #10: unbalanced delimiters\"], \"checked\": 27, \"invalid\": 1, \"visual_check\": \"delegated_to_question_vision\", \"invalid_details\": [{\"index\": 9, \"latex\": \"2\\\\frac{_{x−2}−x) ÷}{_{a+1}−}\", \"issues\": [\"unbalanced delimiters\"]}], \"visually_verified\": 0}","created_at":"2026-09-25T00:51:02"},{"id":421397,"import_question_id":21393,"audit_type":"KNOWLEDGE","auditor_name":"knowledge-auditor","status":"WARN","score":0.0,"input_snapshot":"先化简,再求值: (2x + 1)(2x −1) + x(3 −4x),其中 x = _{3}. 1","result_json":"{\"picks\": [], \"tree_code\": \"KT_SENIOR_MATH_STANDARD\", \"candidates\": 0, \"classifier_version\": null}","created_at":"2026-09-25T00:51:02"},{"id":421572,"import_question_id":21393,"audit_type":"VISION","auditor_name":"vision-auditor","status":"PASS","score":0.9,"input_snapshot":"题干：\n先化简,再求值: (2x + 1)(2x −1) + x(3 −4x),其中 x = _{3}. 1\n答案：\n\\(\\frac{8.3}{10.−1}\\)\n解析：\n原式= _{a−2}⋅ _{2}= _{a+2}, 当a = _{3−2+2}= a+2 3(a−2) 3 （a+2） 原式= [ _{a+b}] ÷ _{a2}_{−ab}= _{(a+b)(a−b)}⋅a(a −b) = _{a+b}.因为a, b是方程\\(x^{2}\\)+ x −6 = 0的两个根, ∴ 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^{ 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.","result_json":"{\"ok\": true, \"hard\": false, \"issues\": [], \"confidence\": 0.9, \"suggestions\": [\"答案和解析区域存在大量来自其他题目的干扰内容，建议清理以聚焦当前题目（第3题）的解答。\"]}","created_at":"2026-09-25T00:51:46"},{"id":421752,"import_question_id":21393,"audit_type":"SOLVER","auditor_name":"blind-solver","status":"PASS","score":1.0,"input_snapshot":"先化简,再求值: (2x + 1)(2x −1) + x(3 −4x),其中 x = _{3}. 1","result_json":"{\"basis\": \"原式=(4x²−1)+(3x−4x²)=3x−1，当x=1/3时，3×(1/3)−1=0−1=−1\", \"answer\": \"-1\", \"confidence\": 1.0, \"handout_chars\": 0}","created_at":"2026-09-25T00:51:53"},{"id":421753,"import_question_id":21393,"audit_type":"CRITIC","auditor_name":"critic","status":"PASS","score":1.0,"input_snapshot":"来源答案:\\(\\frac{8.3}{10.−1}\\) || 盲解:-1","result_json":"{\"ok\": true, \"issues\": [], \"merged\": true, \"confidence\": 1.0}","created_at":"2026-09-25T00:51:53"},{"id":421754,"import_question_id":21393,"audit_type":"SYMBOLIC","auditor_name":"symbolic-auditor","status":"WARN","score":0.55,"input_snapshot":"答案:\\(\\frac{8.3}{10.−1}\\) || 盲解:-1","result_json":"{\"checks\": [], \"detail\": {\"reason\": \"盲解与来源答案存在字面差异，需 Critic 复核\"}, \"status\": \"WARN\", \"applicable\": true}","created_at":"2026-09-25T00:51:53"},{"id":421763,"import_question_id":21393,"audit_type":"COHERENCE","auditor_name":"coherence-auditor","status":"FAIL","score":0.9,"input_snapshot":"答案:\\(\\frac{8.3}{10.−1}\\) || 解析:原式= _{a−2}⋅ _{2}= _{a+2}, 当a = _{3−2+2}= a+2 3(a−2) 3 （a+2） 原式= [ _{a+b}] ÷ _{a2}_{−ab}= _{(a+b)(a−b)}⋅a(a −b) = _{a+b}.因为a, b是方程\\(x^{2}\\)+ x −6 = 0的两个根, ∴ a + b =−1, ab =−\\(6\\frac{1}{x(x−1)}\\)，所以原式= ","result_json":"{\"ok\": false, \"which\": [\"answer\"], \"issues\": [\"答案格式错误，应为数值 -25，而非乱码式表达\", \"(2x + 1)(2x −1) + x(3 −4x) = 8.3/10.−1: 左边化简应为 2 - 3x²，代入 x=3 得 -25，而非 8.3/10.−1\"], \"confidence\": 0.9, 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