Karl Posted June 1, 2012 Report 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
informatician Posted June 1, 2012 Report 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
cifratorul Posted June 1, 2012 Report 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
Karl Posted June 1, 2012 Author Report 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