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[讨论] Mathematica简化测试!Why?

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发表于 2020-12-10 15:00:58 | 显示全部楼层 |阅读模式

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  1. Simplify[{Abs[(-Subscript[m, 2] Subscript[n, 1]+Subscript[m, 1] Subscript[n, 2])],Abs[(-Subscript[m, 3] Subscript[n, 2]+Subscript[m, 2] Subscript[n, 3])],Abs[(-Subscript[m, 3] Subscript[n, 1]+Subscript[m, 1] Subscript[n, 3])]},Subscript[m, 1]>0&&Subscript[n, 1]>0&&Subscript[m, 2]>0&&Subscript[n, 2]>0&&Subscript[m, 3]>0&&Subscript[n, 3]>0&&Subscript[m, 3]/Subscript[n, 3]>Subscript[m, 2]/Subscript[n, 2]>Subscript[m, 1]/Subscript[n, 1]]

  2. FullSimplify[{Abs[(-Subscript[m, 2] Subscript[n, 1]+Subscript[m, 1] Subscript[n, 2])],Abs[(-Subscript[m, 3] Subscript[n, 2]+Subscript[m, 2] Subscript[n, 3])],Abs[(-Subscript[m, 3] Subscript[n, 1]+Subscript[m, 1] Subscript[n, 3])]},Subscript[m, 1]>0&&Subscript[n, 1]>0&&Subscript[m, 2]>0&&Subscript[n, 2]>0&&Subscript[m, 3]>0&&Subscript[n, 3]>0&&Subscript[m, 3]/Subscript[n, 3]>Subscript[m, 2]/Subscript[n, 2]&&Subscript[m, 2]/Subscript[n, 2]>Subscript[m, 1]/Subscript[n, 1]&&Subscript[m, 3]/Subscript[n, 3]>Subscript[m, 1]/Subscript[n, 1]]
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毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2020-12-10 15:56:23 | 显示全部楼层
How
  1. assum=m1>0&&n1>0&&m2>0&&n2>0&&m3>0&&n3>0&&m3/n3>m2/n2>m1/n1;

  2. expr={Abs[-m2 n1+m1 n2],Abs[-m3 n2+m2 n3],Abs[-m3 n1+m1 n3]};

  3. PiecewiseExpand[expr,Reals,Method->{"ConditionSimplifier" -> (Reduce[#&&assum]&)}]

  4. FullSimplify[expr,TransformationFunctions->{Automatic,PiecewiseExpand[#,Reals,Method->{"ConditionSimplifier" -> (Reduce[#&&assum]&)}]&}]
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毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2020-12-10 17:57:47 | 显示全部楼层
  1. PiecewiseExpand[{Abs[-m2 n1 + m1 n2], Abs[-m3 n2 + m2 n3],
  2.   Abs[-m3 n1 + m1 n3]}, Reals,
  3. Method -> {"ConditionSimplifier" -> (Reduce[# && n1 > 0 && n2 > 0 &&
  4.         n3 > 0 && m3/n3 > m2/n2 > m1/n1 > 0] &)}]
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毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
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