{"ok":true,"question":{"id":21392,"question_uuid":"45f5cbc5-8b71-4f21-83b4-086be1f150dd","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, 278.74, 247.26, 351.81]","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: _{x−3}+ _{3−x}, 其中x = 5. 2x 3\", \"type\": \"text\"}, {\"type\": \"answer\", \"blocks\": [{\"text\": \"_{3}时, 原式=−1 +\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"3 \\\\times _{3}= 0\"}, {\"text\": \".\", \"type\": \"text\"}]}, {\"type\": \"solution\", \"blocks\": [{\"text\": \"_{x−1},由x =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"2024^{0}\"}, {\"text\": \"+\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"2^{−1}\"}, {\"text\": \"= 1 + _{2}= _{2}得, 原式= _{2−1}=3 1 = 3. 原式= _{x+1}= _{(x−1)(x+1)}\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\times _{x−2}= 3 3\"}, {\"text\": \"x2−2x x−2 x+1\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"x \\\\frac{x(x−2)}{x−3} 1 3 2 2 2\"}, {\"text\": \"原式=\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"x^{2}\"}, {\"text\": \"−6−3(x−2)_{2} ⋅ (x+2)(x−2) = x2−\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"3x _{2}\"}, {\"text\": \"⋅ (x+2)(x−2)=_{(x−2)2} ⋅ (x+2)(x−2) = x(x+2)_{x−2}=\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"x_{x−2}\"}, {\"text\": \", ∵要使原分式有意义, ∴2+2x x的值不能取2,3, 所以x可取的值为1, 当x = 1时,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{x(x−3)}{1}\"}, {\"text\": \"原式= 1+2 （x−2） （x−2） x−3 x−\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"3 _{1−2}\"}, {\"text\": \"=−3. 原\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{6.6_{2}_{7}}{9.6}\"}, {\"text\": \"式= 1 − _{a}⋅ _{(a−3)(a+3)}= 1 − _{a+3}= _{a+3}, 当a = 4时,原式= _{4+3}= _{7}. 原式= 4 −\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"a^{2}\"}, {\"text\": \"−\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"2a^{2}\"}, {\"text\": \"−6a +\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"3a^{2}\"}, {\"text\": \"= 4 −6a. 当a =− _{3}时, 原式= 6. a−3 a(a+1) a+1 3 −\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{2}{b}2\"}, {\"text\": \"时,原式= 3 2 2\", \"type\": \"text\"}]}], \"source\": {\"bbox\": [45.0, 278.74, 247.26, 351.81], \"file\": \"A中考/2026江苏中考期末数学试卷+答案95.pdf\", \"regions\": [{\"bbox\": [45.0, 278.74, 247.26, 351.81], \"page\": 1}], \"section\": \"generic\", \"page_end\": 1, \"page_start\": 1, \"section_label\": \"\", \"answer_regions\": [{\"bbox\": [30.0, 164.85800170898438, 570.2760009765625, 486.8255004882813], \"page\": 6}], \"answer_original\": \"_{3}时, 原式=−1 + 3 × _{3}= 0.\", \"solution_original\": \"_{x−1},由x = 2024^{0} + 2^{−1}= 1 + _{2}= _{2}得, 原式= _{2−1}=3 1 = 3. 原式= _{x+1}= _{(x−1)(x+1)}× _{x−2}= 3 3 x2−2x x−2 x+1 x \\\\frac{x(x−2)}{x−3} 1 3 2 2 2 原式= x^{2}−6−3(x−2)_{2} ⋅ (x+2)(x−2) = x2−3x _{2}⋅ (x+2)(x−2)=_{(x−2)2} ⋅ (x+2)(x−2) = x(x+2)_{x−2}= x_{x−2}, ∵要使原分式有意义, ∴2+2x x的值不能取2,3, 所以x可取的值为1, 当x = 1时,\\\\frac{x(x−3)}{1}原式= 1+2 （x−2） （x−2） x−3 x−3 _{1−2}=−3. 原\\\\frac{6.6_{2}_{7}}{9.6}式= 1 − _{a}⋅ _{(a−3)(a+3)}= 1 − _{a+3}= _{a+3}, 当a = 4时,原式= _{4+3}= _{7}. 原式= 4 −a^{2}−2a^{2}−6a + 3a^{2}= 4 −6a. 当a =− _{3}时, 原式= 6. a−3 a(a+1) a+1 3 −\\\\frac{2}{b}2时,原式= 3 2 2\"}, \"version\": 1, \"grade_level\": \"SENIOR\", \"number_label\": \"2\", \"subject_code\": \"math\", \"question_type\": \"subjective\", \"content_sha256\": \"987c8ab5ed4f28734fb6d7a1870eff8b02e3457b46569a9a5020cc622f4df1d5\"}","body_html":"<div>先化简,再求值: _{x−3}+ _{3−x}, 其中x = 5. 2x 3</div>","answer_html":"_{3}时, 原式=−1 +\\(3 \\times _{3}= 0\\).","ways_html":"_{x−1},由x =\\(2024^{0}\\)+\\(2^{−1}\\)= 1 + _{2}= _{2}得, 原式= _{2−1}=3 1 = 3. 原式= _{x+1}= _{(x−1)(x+1)}\\(\\times _{x−2}= 3 3\\)x2−2x x−2 x+1\\(x \\frac{x(x−2)}{x−3} 1 3 2 2 2\\)原式=\\(x^{2}\\)−6−3(x−2)_{2} ⋅ (x+2)(x−2) = x2−\\(3x _{2}\\)⋅ (x+2)(x−2)=_{(x−2)2} ⋅ (x+2)(x−2) = x(x+2)_{x−2}=\\(x_{x−2}\\), ∵要使原分式有意义, ∴2+2x x的值不能取2,3, 所以x可取的值为1, 当x = 1时,\\(\\frac{x(x−3)}{1}\\)原式= 1+2 （x−2） （x−2） x−3 x−\\(3 _{1−2}\\)=−3. 原\\(\\frac{6.6_{2}_{7}}{9.6}\\)式= 1 − _{a}⋅ _{(a−3)(a+3)}= 1 − _{a+3}= _{a+3}, 当a = 4时,原式= _{4+3}= _{7}. 原式= 4 −\\(a^{2}\\)−\\(2a^{2}\\)−6a +\\(3a^{2}\\)= 4 −6a. 当a =− _{3}时, 原式= 6. a−3 a(a+1) a+1 3 −\\(\\frac{2}{b}2\\)时,原式= 3 2 2","explain_html":null,"difficulty":2,"score":null,"year":null,"parse_status":"PARSED","qa_status":"QUARANTINED","retry_count":0,"final_score":0.75,"error_message":"QA failed: VISION,CRITIC,COHERENCE","published_question_id":null,"content_sha256":"987c8ab5ed4f28734fb6d7a1870eff8b02e3457b46569a9a5020cc622f4df1d5","created_at":"2026-09-25T00:44:08","updated_at":"2026-09-25T00:54:41","answer_missing":false,"formula_fallback":false,"_number":"2","_section":"generic","_section_label":""},"assets":[],"knowledge":[],"audits":[{"id":421755,"import_question_id":21392,"audit_type":"STRUCTURE","auditor_name":"structure-auditor","status":"PASS","score":0.9,"input_snapshot":"先化简,再求值: _{x−3}+ _{3−x}, 其中x = 5. 2x 3","result_json":"{\"ok\": true, \"issues\": [], \"confidence\": 0.9}","created_at":"2026-09-25T00:52:36"},{"id":421756,"import_question_id":21392,"audit_type":"FORMULA","auditor_name":"formula-auditor","status":"PASS","score":1.0,"input_snapshot":"3 \\times _{3}= 0; 2024^{0}; 2^{−1}; \\times _{x−2}= 3 3; x \\frac{x(x−2)}{x−3} 1 3 2 2 2","result_json":"{\"risky\": 15, \"issues\": [], \"checked\": 15, \"invalid\": 0, \"visual_check\": \"delegated_to_question_vision\", \"invalid_details\": [], \"visually_verified\": 0}","created_at":"2026-09-25T00:52:36"},{"id":421757,"import_question_id":21392,"audit_type":"KNOWLEDGE","auditor_name":"knowledge-auditor","status":"WARN","score":0.0,"input_snapshot":"先化简,再求值: _{x−3}+ _{3−x}, 其中x = 5. 2x 3","result_json":"{\"picks\": [], \"tree_code\": \"KT_SENIOR_MATH_STANDARD\", \"candidates\": 0, \"classifier_version\": null}","created_at":"2026-09-25T00:52:36"},{"id":421767,"import_question_id":21392,"audit_type":"SOLVER","auditor_name":"blind-solver","status":"PASS","score":1.0,"input_snapshot":"先化简,再求值: _{x−3}+ _{3−x}, 其中x = 5. 2x 3","result_json":"{\"basis\": \"原式=\\\\frac{2x}{x-3}+\\\\frac{3}{3-x}=\\\\frac{2x}{x-3}-\\\\frac{3}{x-3}=\\\\frac{2x-3}{x-3}，当x=5时，原式=\\\\frac{2\\\\times5-3}{5-3}=\\\\frac{7}{2}=3.5\", \"answer\": \"-1\", \"confidence\": 1.0, \"handout_chars\": 0}","created_at":"2026-09-25T00:53:15"},{"id":421768,"import_question_id":21392,"audit_type":"CRITIC","auditor_name":"critic","status":"FAIL","score":1.0,"input_snapshot":"来源答案:_{3}时, 原式=−1 +\\(3 \\times _{3}= 0\\). || 盲解:-1","result_json":"{\"ok\": false, \"issues\": [\"来源答案计算错误：代入x=5后得到-1+9=8，但正确结果应为3.5\"], \"merged\": true, \"confidence\": 1.0}","created_at":"2026-09-25T00:53:15"},{"id":421781,"import_question_id":21392,"audit_type":"VISION","auditor_name":"vision-auditor","status":"WARN","score":0.2,"input_snapshot":"题干：\n先化简,再求值: _{x−3}+ _{3−x}, 其中x = 5. 2x 3\n答案：\n_{3}时, 原式=−1 +\\(3 \\times _{3}= 0\\).\n解析：\n_{x−1},由x =\\(2024^{0}\\)+\\(2^{−1}\\)= 1 + _{2}= _{2}得, 原式= _{2−1}=3 1 = 3. 原式= _{x+1}= _{(x−1)(x+1)}\\(\\times _{x−2}= 3 3\\)x2−2x x−2 x+1\\(x \\frac{x(x−2)}{x−3} 1 3 2 2 2\\)原式=\\(x^{2}\\)−6−3(x−2)_{2} ⋅ (x+2)(x−2) = x2−\\(3x _{2}\\)⋅ (x+2)(x−2)=_{(x−2)2} ⋅ (x+2)(x−2) = x(x+2)_{x−2}=\\(x_{x−2}\\), ∵要使原分式有意义, ∴2+2x x的值不能取2,3, 所以x可取的值为1, 当x = 1时,\\(\\frac{x(x−3)}{1}\\)原式= 1+2 （x−2） （x−2） x−3 x−\\(3 _{1−2}\\)=−3. 原\\(\\frac{6.6_{2}_{7}}{9.6}\\)式= 1 − _{a}⋅ _{(a−3)(a+3)}= 1 − _{a+3}= _{a+3}, 当a = 4时,原式= _{4+3}= _{7}. 原式= 4 −\\(a^{2}\\)−\\(2a^{2}\\)−6a +\\(3a^{2}\\)= 4 −6a. 当a =− _{3}时, 原式= 6. a−3 a(a+1) a+1 3 −\\(\\frac{2}{b}2\\)时,原式= 3 2 2","result_json":"{\"note\": \"视觉比对不可用\", \"error\": \"invalid_json\", \"unavailable\": true}","created_at":"2026-09-25T00:54:03"},{"id":421782,"import_question_id":21392,"audit_type":"SYMBOLIC","auditor_name":"symbolic-auditor","status":"WARN","score":0.55,"input_snapshot":"答案:_{3}时, 原式=−1 +\\(3 \\times _{3}= 0\\). || 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":"2026-09-25T00:54:03"},{"id":421801,"import_question_id":21392,"audit_type":"COHERENCE","auditor_name":"coherence-auditor","status":"FAIL","score":0.9,"input_snapshot":"答案:_{3}时, 原式=−1 +\\(3 \\times _{3}= 0\\). || 解析:_{x−1},由x =\\(2024^{0}\\)+\\(2^{−1}\\)= 1 + _{2}= _{2}得, 原式= _{2−1}=3 1 = 3. 原式= _{x+1}= _{(x−1)(x+1)}\\(\\times _{x−2}= 3 3\\)x2−2x x−2 x+1\\(x \\frac{x(x−2)}{x−3} 1 3 2 2 2\\)原式=\\(x^{2}\\)−6−3(x−2)_{2} ⋅ (x+2)","result_json":"{\"ok\": false, \"which\": [\"answer\"], \"issues\": [\"答案计算结果错误，−1 + 9 应等于 8，而非 0。\", \"原式=−1 + (3 × 3) = 0: 计算错误：−1 + 9 = 8，而非 0。\"], \"confidence\": 0.9, \"equation_checks\": [{\"valid\": false, \"reason\": \"计算错误：−1 + 9 = 8，而非 0。\", \"equation\": \"原式=−1 + (3 × 3) = 0\"}]}","created_at":"2026-09-25T00:54:41"},{"id":421802,"import_question_id":21392,"audit_type":"QUALITY_GATE","auditor_name":"quality-gate","status":"FAIL","score":0.75,"input_snapshot":"","result_json":"{\"warnings\": [], \"graded_gate\": true, \"hard_failed\": [\"VISION\", \"CRITIC\", \"COHERENCE\"]}","created_at":"2026-09-25T00:54:41"}]}