Y = 8

X=0

Решила за 10 секунд. а равно двум.

Si fa a occhio in 3 secondi: sqrt{y\sqrt{y}}=y^{3/4}, perciò y=6^{4/3}.Stop

Этот способ решения таких задач с постоянным переписыванием условий и не оставляющий места для мозгов чуть-чуть поработать является прекрасным способом отупления учащихся

too simple for olympiad

Stop this please. Olympiad ?

(x/√x)=5 (x^1/x^1/2)=5 (x^1:x^1/2)= (x^1/2)=5 √x =5 x= 5^2= 25 √25 = 5 ✌🏻 Potenze con stessa base si sottraggono (1-1/2)=1/2

I loved these in High School and ate it up. Then the dream soured when i attempted final year math with extreme algebra and i couldn't keep up. It's much easier getting 100% on tests with straight forward algebra.

В уме решается за 5 сек :2 А не может быть 1 и не может быть 3. Значит : 2

0, в уме посчитала за 10 секунд.

При взгляде на пример, ответ сказала сразу- это 2.В голове со школы сидит 2 в кубк это 8 плюс 2 = 10😅 . А тут куча ненужной "воды " . У нас это называлось " метод подбора " .

Даже не брав в руки ручки, карандаша можно сказать какое число

Он верно академию наук окончил

Что он чушь несет, здесь ли лдумать не надо ---2*5=10

😂😂😂😂😂

This is a long process, this way won't work when it comes to exams due to time⏳but it is a creative method of which i respect

Блин и без всех выкладок сразу понятно что ответ 25.😂

Clue lies with 10 = 8 + 2 = 2^3+2 . Now, transfer them with the corresponding power partner. (a - 2) (a^2 + 2a +5) =0 x = 2 & two complex roots

Nice one, well understood 👏.

First to comment

Mistake at 5:46. It should be x^3 + 1^2 + 2 * (x - 1)(x + 1). Also, x3 = (-1 + sqrt[5]) / 2, and x4 = (-1 - sqrt[5]) / 2

X=25

Pazzesco in questo video è possibile assistere a come complicarsi la vita inutilmente.... Innanzitutto bastava razionalizzare e già il risultato usciva fuori da solo poi nella seconda parte quando c'era x^2-25x bastava dividere tutto per x stessa cosa .....boh non capisco perché complicarsi la vita così

Why not use a calculator from the start?🤦🏼‍♂️

-4 и 0

Let / be root /2 +/x = 2 (/2+/x)² = (2)² 2+x=4 X=4-2 ×=2 Please explain why my calculation is wrong there

(*x+2)^4-2^4=0 [(x+2)^2-2^2)((x+2)^2+2^2)]=0 [((x(x+4)((x+2)^2+2^2=0 x=0 x=-4 x=-2

±1

Simplify the fraction by sqrt(x), so you have 1/(2*sqrt(x)) on the left of the equation. Multiply both sides with 2*sqrt(x) to get 1=8*sqrt(x). Divide both sides by 8 and square both sides and you have the same answer without solving a quadratic equation.

Τhe number under the root must always be equal to zero or greater than zero, there is never a negative number under the roots odd or even. If the square root outcome is a negative number, the equation has no solution! if the outcome under the square root is zero the equation has a double solution. That is what we were taught in greek high schools.

Thank you 😊😊 Continue ur good work sir I really understand whatever you teach here

Thank you 😊😊 Continue ur good work sir I really understand whatever you teach here

X = 2

Unajua Well done 💪💪💪

This is not for me😢

2 и не надо мчитать

Math Olympiad?

Wonderful thank-you

2^y + 2^y = 8 2(2^y) = 8 2^y = 4 log2(2^y) = log2(4) y = log(4)/log(2) y = 2

is this a joke ?

(x+2)=16 (x+2)⁴=(±2) ⁴ x+2=±2 X=0 X=-4. 😂👍

It is very simple😂 2

x = 0

Coool maths

You could go so much faster. x(3x+4) = 10 3x^2 + 4x - 10 = 0 x^2 + (4/3)x - 10/3 = 0 (x+4/6)^2 = 10/3 + 16/36 = 136/36 = 34/9 x+4/6 = +-sqrt(34/9) = +-(1/3)sqrt(34) x = -2/3 +- (1/3)sqrt(34) y = 3x + 4 = -2 +- sqrt(34) + 4 = 2 +- sqrt(34)

Cross-multiply and collect terms to see that this is a quadratic in sqrt(x): x - 5 sqrt(x) = 0 Because this is quadratic in sqrt(x) there are two solutions for sqrt(x). Find them by factoring: sqrt(x) [ sqrt(x) - 5 ] = 0 So one solution for sqrt(x) comes from sqrt(x)=0 => x=0. But owing to sqrt(x) being in the denominator of the original equation, we must reject this. The other solution for sqrt(x) comes from sqrt(x)-5=0 => sqrt(x)=5 => x=25.

√(x+2)^4=√16; x^2+4x=0; D=√4^2-4*1*0=√16; X1=0; X2=-4.

25

(X+2)^4= 2^4 X+2=2 X=0, /X+2/ , -( X+2)=2 -X -2=2 X =-4