Это в уме решается...

Yes, or (via exponential and trigonometric forms) ... x⁴ = -4 x⁴ + 4 = 0 (x²)² + 2² = 0 (x² - 2·i)·(x² + 2·i) = 0 --- /// case: (x² - 2·i) = 0 x² - 2·i = 0 x² = 2·i || || set 2·i on exponential form: || || z = 0 + 2·i || || |z| = √(0² + 2²) = √4 = 2 || || θ = π/2 rad || || z = 2·e^(i·(π/2)) || x² = 2·e^(i·(π/2)) x = ±(2·e^(i·(π/2)))^(1/2) x = ±2^(1/2)·e^(i·(π/2))^(1/2) x = ±2^(1/2)·e^(i·(π/2)^(1/2)) x = ±2^(1/2)·e^(i·(π/4)) x = ±√2·e^(i·(π/4)) x = ±√2·(cos(π/4) + i·sin(π/4)) x = ±√2·(√2/2 + i·√2/2) x = ±(√2·√2/2 + i·√2·√2/2) x = 1 + i and -1 - i --- /// case: (x² + 2·i) = 0 x² + 2·i = 0 x² = -2·i || || set -2·i on exponential form:: || || z = 0 - 2·i || || |z| = √(0² + 2²) = √4 = 2 || || θ = -π/2 rad || || z = 2·e^(i·(-π/2)) || x² = 2·e^(i·(-π/2)) x = ±(2·e^(i·(-π/2)))^(1/2) x = ±2^(1/2)·e^(i·(-π/2))^(1/2) x = ±2^(1/2)·e^(i·(-π/2)^(1/2)) x = ±2^(1/2)·e^(i·(-π/4)) x = ±√2·e^(i·(-π/4)) x = ±√2·(cos(-π/4) + i·sin(-π/4)) x = ±√2·(√2/2 - i·√2/2) x = ±(√2·√2/2 - i·√2·√2/2) x = 1 - i and -1 + i --- /// final results: ■ root #1: x = 1 + i ■ root #2: x = -1 - i ■ root #3: x = 1 - i ■ root #4: x = -1 + i 🙂

You can solve by inspection in a couple of seconds.

2

The very unfortunate thing with solutions like the one shown here is, that they do only solve the special case of the problem ... there is nothing to learn from it ... example: Just modify the problem to 99^x = 5/x and the shown solution leads nowhere,. A good solution would solve this modified problem as well (just the 100 needed to be replaced by the 99 in the solution formula).

There is nothing wrong with the English the RU-vidr is speaking. His accent is much clearer than that of someone from Wales or Scotland. By your reckoning, the likes of the Late Nelson Mandela didn't speak standard English. This is a case of the pot calling the kettle black. What matters here is the substance and not the style. What happened to the concept-cum-practice of constructive criiticism?

This is one long way of finding the solutions...

You are wrong...

X=2

Thank you sir

x=2

X=2

You missed x-2 so you’re wrong

2х2х2+2=10, и думать нечего

По моему он не правильно решил

{2x+2x ➖ }=4x^2 2^2x^2 1^1x^2 1x^2 (x ➖ 2x+1).

Решается в уме за 1 мин. Что же у них за олимпиады такие? И кстати, на 0 делить нельзя!

QUÉ TAL ERRORRRRR......

Personnellement, étant donné que la solution 2 est évidente, j'aurais effectué une division euclidienne par (x-2) pour pouvoir trouver l'équation sous forme (x-2)(...) où (...) est l'équation de second degré ayant un discriminant négatif et ensuite j'aurais procédé comme vous...

Pra que uma confusão dessas? Simplifica isto

Menteur que tu sois

-1+i rt3, -1-irt3, 2 Without writing one thing down

X^3+x^2=36 X^3+x^2=27+9 X+x=3√27+2√9 2x=3+3 X=6/2 X=3 3^3+3^2=36 27+9=36

Thank you

Неужели за рубежом все так запущено?))))... Уравнение на одно действие!)

Bro change your thumbnail .

Два корня квадратных из двух.

Where do these things work? It's just wast of our time. If you know operations it's enough.

2,- 2

Power 4and4 cancel then x=2

1*3*3+1=10

Merci

И это задача для олимпиады???

100^x=5/x x=1/2=0.5

Essa Eu fiz só olhando. 1×1×1+9=10

x = -W(-ln(2)/32)/ln2

X=2

Элементарное уравнение.

The principles u used aren't clear though is been a long time i tried my hands on it.

{2+2 ➖ }+{x+x ➖ }={4+x^2}=4x^2 2^2x^2 1^1x^2 1x^2 (x ➖ 2x+1).4 2^2 (x ➖ 2x+2).

Improve in terms of explanation make it more clear be aware its not everyone who understands certain concepts these are some of the things that makes some people lose interest in interesting subjects

👏👏👏

Amazing!!

Este *%#@& da muchas vueltas. Pone "solución" y vuelve a escribir el problema. ¡Qué le den!

X = -1

x✔ x=✔ 3³ x✔ x=✔ 27 x✔ x=3✔ 3 x=3 Pourquoi toute cette subtilité?

3 (the main value).

X=2

Its one BY the other factor....NOT " into "....!!!

Your thumbnail is different from the beginning of the vid. (2x²) = (2x²) is not the same as 2x² = (2x²)