Karl Posted June 1, 2012 Report Share Posted June 1, 2012 Stie cineva cum se rezolva dupa ce inlocuiesc cu 3 la x ? Ma chinui si nu-i dau de cap. Quote Link to comment Share on other sites More sharing options...
informatician Posted June 1, 2012 Report Share Posted June 1, 2012 Notezi 3 la x=t rezulta t la a3a+t la a2a-17 * t+15=0.Aceasta ecuatie se rezolva cu schema lui Horner si are ca solutii numerele:t1=1,t2=3 si t3=-5 care nu convine.Inlocuind rezulta x=0 si x=1. Quote Link to comment Share on other sites More sharing options...
cifratorul Posted June 1, 2012 Report Share Posted June 1, 2012 parca e ceva de genu :y^3+y^2-17y+15=0; y^3+3*1/3*y^2+3*1/9*y+ 1/27 -(17+1/3y) +14+26/27=0; (y+1/3)^3 -1/3(y+1/3) ....z^3-1/3z+C=0.... Quote Link to comment Share on other sites More sharing options...
Karl Posted June 1, 2012 Author Report Share Posted June 1, 2012 @informatician cum o rezolv cu schema lui horner, am incercat dar imi prind urechile.Te rog imi poti arata cum, sau altcineva daca stie. Quote Link to comment Share on other sites More sharing options...
staticwater Posted June 1, 2012 Report Share Posted June 1, 2012 Quote Link to comment Share on other sites More sharing options...