it's more simple if you observe that a-b=1 then you know (a-b)^2=1 and you have a^3-b^3=1 . Finally you find ab=0. The next is simple

You told that solition very well.Thank you.

@user-ee7nw2rx9s

Ай бұрын

Если умножить на знаменатель, то не надо возиться с дробями

Если присмотреться то если из первого числителя отнять второй равно знаменателю, это не зря

Eksi....😂

How about the same equation but make it functional?

Un truc que je voudrais voir sur tes vidéos, c’est le domaine de définition. Ici par exemple : X^2-3≠0 ===> Df = R - { -√3 ; √3}

It just need patience

Nice problem

please more equation like this ! I love this so much

asnwer=x1 isit

Another method: The original equation is: ((x^2 + x) / (x^2 + 3))^3 - ((x - 3) / (x^2 + 3))^3 = 1 (1) This equation may be written as: a^3 + b^3 + c^3 = 0 , denoting a, b, c as: a = (x^2 + x) / (x^2 + 3) , b = - (x - 3) / (x^2 + 3) , c = -1 Is easy to show that: a + b + c = 0 and in this case: a^3 + b^3 + c^3 = 3abc = 0 . Therefore, eq. (1) reduces to: ((x^2 + x) / (x^2 + 3)) ((x - 3) / (x^2 + 3)) = 0 (2) However, x^2 + 3 ≠ 0 and therefore eq. (2) reduces to: x (x + 1) (x - 3) = 0 which gives 3 roots: x=0 , x = - 1 , x = 3

Kwerage tight ywello wy werji x aydelem xsquaru yishalegnal middre chikko azawoch ekhkhkhkheeey yihe dirty ymiyasetella bahriyachun bmin y9gelets y9hon ainstien rasu binor qala5 yattalet nbere kirfatamoch