Відео

why not just use pascal's triangle?

Congratulations on being incredibly stupid! Clearly you should move the sqrt(2) to the other side.

wouldn't it be easier if you get sqrt(2) to the other side and you get sqrt(x)=2-sqrt(2)? now you square both sides and voila. x=6-4sqrt(2)

WRONG. WTF? x can not be greater than 2. Olympiad for kindergarten? Check your work before posting IDlOT.

First of all, it can't be an Olympiad question Second of all, you even solved it wrong qwq you literally introduced an extra root to the equation

Binomial expansion of the 10th power is painful; but expansion of (3 - 2√2)^5 took me less than 2 minutes with mental math as = (3^5 + 10 * 3^3 * 8 + 5 * 3* 64) - (5 * 3^4 * 2 + 10 * 3^2 * 16 + 128)√2 = (243 + 2160 + 960) - (810 + 1440 + 128)√2 = 3363 - 2378√2.

Squaring the thing twice, what can go wrong? Like, maybe, introducing extra roots that shouldn't be there? What's wrong with sqrt(x) = 2-sqrt(2) => x = (2-sqrt(2))^2 = 4 -2*2*sqrt(2)+2 = 6-4*sqrt(2)? (with the 6+4*sqrt(2) being totally wrong).

вообщето количество корней равно степени уравнения тоесть умножаем обе чести на левый делитель потом переносим всё в левую часть и решаем как уравнение 7 степени x^7-3x^6-2x^5-3x^4+x^3=0

How I solve this: 1. Calculate sqrt(2), it's ~1.414. 2. Find the lack to get 2: 2-1.414=~0.586 3. Mutiply 0.586 two times to get a value we need to sqrt, it's ~0.3434. 4. Let's try: sqrt(2)+sqrt(0.3434), it's really close to 2. The answer is x=0.3434

Does not look like an olimpiad question at all, just some school equity for beginners

= 3363- 3362.999851 = 0.000148. (0.4 to the power 10 is very close to zero)

I can also use binomial expansion. I even made an identity for (a+b)¹⁰ still this works nice method though

Thats only one of the three solutions.

?=16^4×15=15×2^16=30×2^15

Pretty clever way of finding the exact value of x. Dr. Ajit Thakur (USA).

The solutions x = 1 and x = -6 are obvious (120 = 2 × 3 × 4 × 5). So just multiply it all out to get the equation in standard quartic form, divide the quartic by (x -1)(x + 6) = x² +5x - 6, and all that's left is a quadratic to solve.

(x+4)! = 5! --> x+4 = 5 --> x = 1

The answer is 2^16×15

(1+1)•(1+2)•(1+3)•(1+5)=2•3•4•5= 120=5! <=> x=1

Sqr(121)=+11 m^2=121 ==> m^2-121=0 ==> (m+11)*(m-11)=0 ==> m=11 or m=-11 sqr(x) is only >= 0

Si se hace el logaritmo complejo en el 5^x=-√(15) se podria hallar una solucion?

Very astucious change of variable. Love it.

That was good

এত মার প্যাঁচ করলেন,ছাত্র ছাত্রীরা না ঘাবড়ে যায়।

I just immediately saw that the four sets of brackets hold a series of numbers that are incremental. And since 2x3x4x5=120 I just assumed the answer was that X=1. So within 5 seconds I thought I had the answer. I really didn't consider there may be others possible. I really need to learn to be more pedantic with these kinds of problems.

120=5! = 5x4x3x2x1 solved

Helpful, informative Algebra solution

Awesome 👏 info as usual

It was satisfying to see how everything came to a neat conclusion though the path was long

(2)+2ab(4)=8(1)+1ab(2^2) (1^2) (ab ➖ 2ab+1).

The general solution for a+b can be written as a^2/(a-11). Now you can choose different options from there. The problem has infintely many solutions.

Thanks 🙏🏻 it will be more easy for beginners if music 🎵 is off and you explain

+/_4

La respuesta es V2

lg48/lg8.

Given x + y = 5 and xy = 50, this means that x^2 + y^2 = -75. No real solution.

Answer is 8

(x-1)log5 =2log3 x = (log 5 +2log3)/log5

Result is: [ 4 * sqrt(2) ]/ [sqrt(5) - sqrt (3) ]

4-sqrt(15)={ [ sqrt(5) - sqrt(3) ] ^ 2} / 2

I'm not sure but it must have a easier way to solve that

thank you. can you tell me what app is used for this kind of drawing?

If you wanted your answer in this format, no calculator was gonna give you it.

Une mauvaise application

5^(2x)=5^(ln15/5) , 2x=log15/log5 , x=log15/log25 , test , OK , x=~ 0.841303

50000 minus 5

Позор, 4 действия на 8 минут

Why does the same variable have a different value? I mean if we put 2 in the place of every x then also we get 10

Так как численные значения логарифмов приблизительные, надо обязательно оговорить, до какого знака округляете. Или поставить знак приблизительного равенства(волнистый), а не точный. Да?

Could x=0?