Japanese | Can you solve this ? | A Nice Olympiad Algebra Problem
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- čas přidán 4. 01. 2024
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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.
Если умножить на знаменатель, то не надо возиться с дробями
Eksi....😂
let y=(x-3)/(x^2+3) and note that (x^2+x)/(x^2+3) = 1+y (use synthetic division if this is not obvious). Problem maybe written (1+y)^3-y^3-1=0
expand to obtain 3*y^2+3*y= 0 , y=0 or y=-1; y=0=> x=3, & y=-1 => x^2+x=0 or x = 0, or x=-1,
It just need patience
Если присмотреться то если из первого числителя отнять второй равно знаменателю, это не зря
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}
How about the same equation but make it functional?
Nice problem
asnwer=x1 isit
please more equation like this ! I love this so much
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
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