วีดีโอ

Too complicated you must use isqare = - 1 So 32 square / (1-i) square= a Then develop ( 1-i )square = -2i So à = 32x32×i/-2 Thats all

L' equazione del titolo è diversa ed ha come soluzione 1/4 .

This is needlessly complicated.

Math Olympiade? This is basic stuff of the 8th grade.

Il existe une démonstration bien plus simple, pour arriver au résultat.😮

Congs 👏👏👏The secret is proceeding step by step, needing no hurry👍

The Lord Jesus give you life.

Это точно олимпиадная задача? Решается в уме

More easily, just squaring both sides, 32^2 = (√a + √-a)^2 = a +- 2√(-a^2) + (-a) = +- 2√(-a^2). Squaring again, 32^4 = 4(-a^2). So, a = +- √( 32^4 / (-4) ) = +- 512i. Be aware. √a √-a not= √(-a^2), √a √-a = +- √(-a^2). √(-a) not= (√a)i, √(-a) = +- (√a)i.

The statement is missing that "a" and "b" are integers

Correct result, long way. It is simple with complex numbers from the beginning.

La réponse est longue😮 alors que le résultat est tout simple à condition que a#0 donc a =3 *2=4,24

A short route can be √a(1 + i) = 32 √a = 32/(1 + i) a =[32/(1+i) ]^2 = 1024/2i = 512/i = - 512 i

Я решила в уме

Nice Explanation ❤

The equation in the title is not the equation you solve. For Olympiad mathematics I find it extremely simple.

Better if we square as it is, both sides at the first step: a - a + 2ai = 1024 a = - 512i

Вот что я сразу увидел, так это каллиграфический почерк!!! Браво!

(a×a×a)÷3a=6 (a×a)=18 a=√{9×2) =3√2

What a tediously tasking work at problem one

t=3^x t^3 + t^2 = 150 the key is to notice that 150 =125 + 25 = 5^3 + 5^2 => t =5 => x = ln(5)/ln(3) that gives one root, the other twoo rots calculated regular way from quadratic equation ( t^3+t^2-150 )/( t - 5 ) =0

( sqrt( y )+ sqtr( -y ) )^2 = y + 2iy - y = 2iy =4^2 =16 y ={+/-) i8

a=3racinede 2

You made it more complicated than needed. Assume y>0 √y+√-y=4 √y+i√y=4 (1+i)√y=4 √y=4/(1+i) √y=4(1-i)/[(1+i)(1-i)] √y=(4-4i)/2 √y=2-2i y=(2-2i)^2 y=4+2*2*2i+4i*i y=4+8i-4 y=8i Assume y<0. By symetry y=-8i

Esta música na velocidade 2x fica muito divertida hahaha

√a+√-a=2^5 √a+i√a=2^5 (√a+i√a)^2 = 2^10 a + 2 * √a * i√a - a = 2^10 2 * i * a = 2^10 i * a = 2^9 a = 2^9/i a^2 = 2^18/-1 = -(2^18) a = 512i and because i = -i: a = +- 512i

la resolví mentalmente en 10 segundos

Much easier ... divide both sides by y. then x/y +1 = 6*sqrt(x/y) then x/y - 6*sqrt(x/y) + 1 = 0 then solve Sqrt(x/y) = 3 +/- 2*sqrt(2) then square each solution then x/y = 17 +/- 12*sqrt(2)

Juste AFFLIGEANT...🥱🥱🥵

👍👍👍

This music is annoying.

4^x(4 - 1/4) = 25 4^x * 15 = 100 4^x = 20/3 4^(x-1) = 5/3 x - 1 = ln(5/3) / ln(4) x = 1 + ln(5/3) / ln(4)

I wanna say thanks for verifying. I hate when they don't verify. On one video, I watched them get three answers and not verify, but then I looked in the comments section, and one commenter pointed out that two of the three answers did not work when plugged in.

√a+√(-a)=32 → a = ?? √a + √(-a) = 32 √a + i √a = 32 √a (1+i) = 32 → √a = 32/(1+i) = z a = z^2 z = ρ(cos ⁡θ+i sin ⁡θ ) z^n = ρ^n (cos⁡ nθ+i sin ⁡nθ ) de Moivre z = 32/(1+i) = 32/(1+i) ∙ (1-i)/(1-i) = 32/2 (1-i) = 16√2∙√2/2 (1-i) z = 16√2 (√2/2 - i √2/2) = 16√2 [cos⁡(-π/4) + i sin⁡(-π/4) ] ρ =2^4 √2 θ = -π/4 a = z^2 = ρ^2 (cos ⁡2θ + i sin⁡2θ) = 2^9 (cos⁡(-π/2) +i sin⁡(-π/2) ) a = -2^9 i= -512 i

2,715

(a)^(1/2)+(-a)^(1/2)=32 ((a)^(1/2)+(-a)^(1/2))^2=a+2*(a)^(1/2)*(-a)^(1/2)-a =2ai=1024 ai=512 (a)^2=-(512)^2 a=512i or a=-512i

P=given formula divide by rt5 P=(rt7+1)/(rt7-1) =(8+2rt7)/6 =(4+rt7)/3

I did it in less lines. However this is the correct solution - tested by substitution.

If Brazil accepts this level of maths challenge as Olympiad level 🤦- I sincerely regret the future of this country...

log3/log2

受益匪淺

So. Easy

И что в итоге мы получили? Множество неявных корней и 9 минут потраченных в пустую. 😂😂😂 Вот что происходит, когда занимаешься херней😂😂

Особенно умилительна строчка 4 про Both side. ,Это задание для олимпиадников 😅😅😅

X=16 Sa demande pas tous ses tapes

Насколько нас хорошо учили в СССР! Хватило половины действий для ответа... Зачем так всё разжёвывать?

Congratulations

Pourquoi ne pas écrire directement 2a/2 = (512-32✓-a) sans diviser les 2 membres par 2. On gagne une ligne

Very noisy

Зачем надо переписывать то, что уже написано? И всё это можно показать за пару минут.