China | Can you solve this ? | A Nice Math Olympiad Radical Simplification

This is an interesting question that tests a lot of concepts!
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Questão de Matemática Olímpica || Isso é da China? || Matemática
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[@ichdu6710]

an amazing solution.
also an amazing discovery is; you can solve this directly by squaring the given fraction, square the result two times and multiply with the fraction at last

[@ГалинаПархомчик]

Как всегда интересною. Спасибо! Удачи!

[@АндрейПергаев-з4н]

Можно всего лишь два раза возвести в куб
Первый раз
Числитель 5sqrt5-3*5*1+3*1* sqrt 5-1=5sqrt 5+sqrt 5-15-1=8 sqrt 5-16=8(sqrt 5-2)
Знаменатель 2^3=8
Делим в итоге получается sqrt 5-2
Повторно возводим в куб
5sqrt 5-3*5*2+3*4*sqrt 5-8=5sqrt 5+12sqrt 5-30-8=17 sqrt 5-38

[@StephenWong14]

More straightforward to just simply power 4 (binomial theorem), square it and multiply once more. 😢

[@ТатьянаДергачева-л9ъ]

А вообще то в жизни кому нужны такие вычисления? Мне не пригодились от слова совсем 😢

[@anjalidilipogale4208]

You have given accurate sol

[@learncommunolizer]

Thank you very much

[@deckexcellentfoundationaca2390]

In fact this solution is nonsense. The answer is simply 1. What were you solving

[@michaelblankenau6598]

In fact the answer is NOT 1 … what were YOU solving ?

[@bradycall1889]

Ru trolling? lol

[@nkverma1577]

No sence

[@Rosomaxa-wu2ps]

Wunderbar

[@learncommunolizer]

Thank you very much

[@sirikumaradayananda368]

😜

[@gabrielebariletti6045]

more easy, and logic too:
((5^(1/2)-1)/2)^9 ;
((5^(1/2)-1))^2 * ((5^(1/2)-1))^2 * ((5^(1/2)-1))^2 * ((5^(1/2)-1))^2 * (5^(1/2)-1) ;
((5^(1/2)-1))^2 = 2(3-5^(1/2)) => (2(3-5^(1/2)))^2 * (2(3-5^(1/2)))^2 * (5^(1/2)-1) ;
(2(3-5^(1/2)))^2 = 8(7-3*5^(1/2)) => (8(7-3*5^(1/2)))^2 = 128(47-21*5^(1/2)) ;
2^(7)*(47-21*5^(1/2)) * (5^(1/2)-1)/2) = 2^(7)*4*(17*5^(1/2)-38) =>
((5^(1/2)-1)/2)^9 = (2^(9)*(17*5^(1/2)-38)) / 2^(9) =>
((5^(1/2)-1)/2)^9 = 17*5^(1/2)-38 = circa 0,013156
C.V.D. as we wanted to show !