(x^4)+4(x^2)-5=0^2=au carré^4 au cubeMerci :).... Pergunta de ideia deBreizh

April 2019 1 58 Report

(x^4)+4(x^2)-5=0

^2=au carré

^4 au cube

Merci :)

Lista de comentários

Commentaires

équation "bicarrée" à résoudre:

$x^4+4 x^2-5=0$

on effectue le changement de variable

$X=x^2$

donc on obtient : $X^2+4X-5=0$

donc $(X-1)(X+5)=0$

donc $X=1 \ ou \ X=-5$

or $X=x^2>0$

donc $X=x^2=1$

donc $x=-1 \ ou \ x=1$

0 votes Thanks 0

More Questions From This User See All

Breizh April 2019 | 0 Respostas

500(1+t/100)*(1+((t-1)/100))= 546

Breizh April 2019 | 0 Respostas

Breizh April 2019 | 0 Respostas

Helpful Links

Política de Privacidade

Termos e Condições

direito autoral

Helpful Social

Copyright © 2024 ELIBRARY.TIPS - All rights reserved.