By simplify, I think the kind teacher meant the expression should be written with a rational denominator and in terms of n-th roots of positive integers. So, (1/9)¹ᐟ⁹ = (1/3²)¹ᐟ⁹ = (1/3²ᐟ⁹) = 3⁷ᐟ⁹ ⁻¹ = 3⁷ᐟ⁹/3¹ = {⁹√(3⁷)}/3 = {⁹√(2187)}/3. .

How is that result simpler than the original? Simplest form technically would be ²(1/9) but without using tetration, surely the simplest form would be 3^(-2/9).

Simply you are wasting the time by repenting the trivial steps.

Unnecessary lengthy

This is a call nay an appeal to all maths teachers. Please make your presentations shorter , avoid repetition. Your primary responsibilty , if I am not mistaken, is to make maths simpler and attractive . But many times I am forced to move away from many of the posts . A five minute presentation takes half an hour !

Так мало того, сама по себе задача крайне глупая.

solving very bad..to long...

Имею плохую привычку. Начав что то смотреть, досматриваю до конца, а потомстрадаю от потерянного времени.

Похоже, чуваку просто нравится пером по бумаге водить))) Продемонстрировал фантастическое владение степенями!

Only ((3)^-2)^(1/9)=3^(-2/9)

Chłopie, ty to skomplikowałeś, a nie uprościłeś !... przez logarytm o podstawie 3 szybko wychodzi 3^(-2/9)

If it is a question of Math Olympiad then everyone would succeed in that exam....even myself.

Super Academy,   A bit tedious process... (   Maybe the result here is a bit easier:     (1/9)^(1/9) = 1 / 9^(1/9) = 3^(-2/9) = 1/3 * 3^(7/9) = 1/3 * (3^7)^(1/9) = 1/3 * 3^(6/9) * 3^(1/9) = 1/3 * 3^(2/3) * 3^(1/9) = 1/3 * cbrt(9) * root(3; 9).     Here root(a; b) means a^(1/b) for a∊ℝ, b∊ℕ,     and cbrt(x) is a cube root of x∊ℝ.      What is the meaning of simplification in this case? I'm from Russia, and in our high school we have no strict rules for the result of simplification.

There is a faster and less tedious method

Herr professor, if your friend did this sea of ​​calculations and is going to give a value that needs to be calculated using a log or calculator, wouldn't it be much simpler to use the following scheme: 1) (1/9)^(1/9) = X 2) I raise both sides to the power 9 1/9 = X^9 0.111111 = X^9 X = ninth root of 0.1111111111 It will give the same value as yours, calculated, 0.78338 Hugs!

What is exactly the scholar level of students who have to solve this ? This solution is too long and if a student had to solve this during math Olympics, this would be a waste of time !!!😮

a=1/9 => a^a = x => x = cln ( a * ln(a) ) => x=0,78338... => (1/9)^(1/9)=0,78338...

I have never seen such inefficiency.

You win a Fields Medal for tedium.

This answer we may get in one row by multiplying 1/9 by 3^7. Then we obtain (3^7 / 3^9)^9

Ugh, this was hard to swallow! Slow mathematics (credits for this), just to get a result, that doesn't simplify, while the real solution doesn't need any calculation ( 1 / 3^(2/9) or 1 / 9rt(3²) ). To get close to the decimal result (0.78338103693727233...), I've tried some (2) manual Newton steps with a bad, but easy to calculate, starting value (1): N(x) = 7/9 x + 2/(27x^3.5) N(1) = 7/9 + 2/27 = 23/27 x = 23/27 = 851/999 = 0._851_ is still far away, so let us take a second step with some approximations: We need to calculate 7/9 * 23/27 + 2/(27 (23/27)^3.5) = 161/243 + 2/(27 x^3.5). Our greatest problem for manual calculation is this x^3.5. But we know, that x^4.5 = sqrt(x^9) ahould greater than the solution for (1/9)^(1/9), 1/3, let's say 1/2 (just a guess!). What we need, is x^3.5 = x^(7/2) approx.= (1/2) / x = 27/46. => 2/(27 x^3.5) approx.= (2*46)/(27*27) = (92/3)/243. Thus: N(23/27) approx.= 161/243 + (30 2/3) / 243 approx.= 161/243 + 30/243 = 191/243. (I've rounded down, because we expect to overshoot). This value of x_2 = 191/243 approx.= 0.786008 is a pretty good (rational) approximation...

Alcuni passaggi ripetuti e inutili sopratutto!

Esse trem em português já é complicado agora imagina uma explicação em grego fala português.

wowww

3^7/9x3^(-9/9)=3^(-2/9)

It doesn't make sense. You only demostrated the different ways of manipulation expression rather than simplify it.

(3^(-2))^1/9=3^(-2/9)=3^(7/9)/3 much shorter

wowwww

Is this an eye exam?

(1/9)^1/9 = 3^(-2/9) < make it simple in less than 60 sec!

Where are you from bro?

A simple calculation is made so much complicated and boring.

НУ И РЕШЕНИЕ?! ОЧЕНЬ ГЛУПО!

It's a litle bit complicated.

400 par ka ganit nahin samajh me aa raha

10:42 still continue to monitor

(1/9)^1/9= r9(1/9)=(r9(1))/(r9(9))= 1/r9(9)=1/9^1/9=9^(-1/9)=3^(-2/9)

😮 demasiados pasos redundantes, este ejercicio es para hacerlo más corto....esto fue muy latero

Simply (1/9)^(1/9) = (1/3^2)^(1/9) = 3^(-2/9) = 3^(-(9-7)/9) = 3^(7/9). 3^(-9/9) =(⁹√3⁷)/3 = (⁹√2187)/3

Почему не показал как из 3^7 получил 2187?

Why r u creating controversy 😮😮

ans : 1/[3^(2/9)]

От степени 1/9 так и не избавился)))))))

It took a long period of time to solve the simple equation,you just make us confused about your simple, topic, what a good teacher. You deserved no subcriber indeed.😀😀😀

You kill the soul of maths that is time

Un poco más y me duermo

You are taking all the roads to complicate rather than simplfying. It takes only two steps to simplfy it.

making something out of nothing ! 9 is simply 3^2 so the question becomes 3^(-2/9) which can be written with rational denominator if desired by putting (7/9 - 1 ) instead of (2/9). olympiad? ha!

He Teacher juan is best

El final es de pena

I am disagree. Answer is 1/3^(2/9) or 1/ 9√9

I feel bad for your students if you're a teacher 😂😂😂 why don't you make it easier and simple . for such easy question 😢

Tenightrootovnine 😂

A7A

I understand that the idea is to show how to use the different properties of powers. That's Ok. But I mean you rodount too much. I might say you show a way how to complicate something simple. Anyway, who teaches, learns.

Is also wrong... The correct result is 0.00000000258 while this gives out 0.783381...

Muchal kak koshka mishku

Stöpslerei!

Quanta complicação!!

So long Only three steps solution like

😮

Repite...repite y repite las igualdades.!! Interminable !

(9)⁹ ans

Boring démonstration and too complicate

1/3

Правильнее назвать это видео "усложни", а не "упрости".🤭Степень девятки 2/9≈1/3, т. е. искать надо значение, близкое к ³√3.

Don't wrote the same expression twice...You bore me with that..

It is not fair It is a joke You are sure, it is good math ?

Unnecessarily a mental fellow

(1/9) ^(1/9) = (9)√(1/9) = (9)√0.111111...

oh no, it only got worse

Oi!

좀 어리석은 풀이다

Что-то очень длинное и ни к чему хорошему не приведшее.

Tenent root of power Boring....💤💤💤💤

Зачем это все?

вялотекущая шизофрения....

Le résultat n'est pas plus simple que l'énoncé, légèrement ridicule.

What a terrible approach ... You should start by stating what “simplify” mean - radical on top, and the smallest integer on the bottom. This is the objective. This objective leads to the next logical question - what should we multiply top and bottom by, to get rid of the radical at the bottom? What we have at the bottom is 9^(1/9) which is the same as (3^2)^(1/9). And what want to see instead is (3^9)^(1/9) because it simplifies to 3. Since 9=2+7, all we need is multiply the top and bottom by (3^7)^(1/9): 1 / (9^(1/9) = (3^7)^(1/9) / ( (3^2)^(1/9) * (3^7)^(1/9) ) = (3^7)^(1/9) / (3^9)^(1/9) = (3^7)^(1/9) / 3 This is it. By clear reasoning you arrive to the plan and to the answer IMMEDIATELY, in couple of minutes. Instead of going through seemingly random transformations for 13 minutes, with no clear sense of purpose, arriving to the final answer what seems almost by accident. You give your students no help in solving similar problems on their own. Thanks god I had great math teachers who taught me - before taking a pen and starting to write - think first, state your objective, articulate the plan, and only then start writing.

This is torture!!!!! Unnecessarily repetitious....

Not sure how this has so many views or why it was recommended to me. This is terrible work. Do not do math like this, and do not try to teach this way!

Бред

To much exaggerated and complicated

This is not a solution! 🤕

STOP WRITING 9 LIKE q!!!

You are half mad...

Too much repetition

Painful to watch.

Very poor method..

Yours is no good process, I think.

Very botingand is very boring

pardon mais c'est poussif !

Are you that desperate to make a youtube video that you would make one to waste everybody's time? C'mon man - get with the program.

Sehr witzig??????

This is a complete waste of time

To write the number nine use 9 not q

USELESS

Wasting my time.

Boring

Very very poor maths teaching,every time रिपीटेशन ।it is frustrating and boring method।please don't allow this teacher in future।