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How to remove "Abs" from "Simplify" output?

Posted 1 month ago
 Hello, In Mathemtiaca Simplify[Sqrt[(a b c + d e g)^2], a b c + d e g > 0]  will produce Abs[a b c + d e g]  It is a little wired, as I think the answer should be a b d+ d e g 
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Posted 1 month ago
 Sqrt[(a b c + d e g)^2] // PowerExpand 
Posted 1 month ago
Posted 1 month ago
 Try this: Simplify[Sqrt[(a b c+d e f)^2],(a b c+d e f)\[Element]NonNegativeReals] It gives: a b c + d e f 
Posted 1 month ago
 Your way is very good. Do you feel a lilttle weird as \[Element] NonNegativeReals works but >0 doesn't?
Posted 1 month ago
 Well, looking at this further I see that this would also work: Simplify[Sqrt[(a b c + d e f)^2], (a b c + d e f) >= 0] So zero needs to be included in the domain of this case.But when the expression is shorter there is no such need: Simplify[Sqrt[(a b + d e)^2], (a b + d e) > 0] This is puzzling
Posted 1 month ago
 If in this situation Simplify[Sqrt[1/(a b c + d e f)^2], (a b c + d e f) > 0] we can't use >=0 here.
Posted 1 month ago
 Simplify[Sqrt[(a b c + d e g)^2], a b c>0&&d e g>0]returns a b c+d e g You might be able to loosen that up a little bit by indicating the result is still positive if one of those triples is negative
Posted 1 month ago
Posted 1 month ago
 Your example shows the square of the root. In that case no domain specification is needed. The OP asked for the root of the square,
Posted 1 month ago
 You are of course correct. Apologies.
Posted 1 month ago
 Does this help?  Assuming[a b c + d e g > 0, Simplify[Sqrt[(a b c + d e g)^2]]] Out[22]= Abs[a b c + d e g] Never mind; that's what you started with!
Posted 1 month ago
 The easiest would just be removing the absolutes: yourexpression /. Abs -> Identity 
Posted 1 month ago
 May you tell why does this work? I don't know the reason. For example, this doesn't work List[{a, b, c, d}] /. List -> Identity 
Posted 1 month ago
 Just my guess. Simplify transforms the condition a*b*c + d*e*f>0 into the complicated result of Reduce[a*b*c + d*e*f>0] and then gets bogged down. It does not recognize that it fits the pattern Simplify[Sqrt[x^2], x > 0]. Funny that the following does work: With[{x = Sin[a*b*c + d*e*f ]}, Simplify[Sqrt[x^2], x > 0]] 
Posted 1 month ago
 Yes. It is fuuny. Is there a way to Simplify[Sqrt[(a b c + d e g)^2], a b c + d e g > 0]