Відео

K+1=K => K+1-K=0 => 1=0 and this equation is false for every K. Therefore the initial equation has no solutions. The rest is beating the foam without any sense.

I like the video ❤ The explanation was so clear Every step is clear And i loved your sweet way to talk❤

Great 👌

How come? I can't believe the ans!

x=Log[2,(3±3Sqrt[3]i)/2]

2^(Log[8,1.6875]+1 3/9+(Log[8,.0625e^π]+1 3/9)i)*2^(Log[8,1.6875]+1 3/9+(Log[8,.0625e^π]+1 3/9)i)*2^(Log[8,1.6875]+1 3/9+(Log[8,.0625e^π]+1 3/9)i)=-27 x=Log[8,-27]=Log[8,1.6875]+1 3/9+(Log[8,.0625e^π]+1 3/9)i

Nonsensical

you even don't know what is math!

When k+x=k and x not equal to 0 the only answer for k is "no solution exists", even in complex world. Both 2 method are wrong!!! 1st method: divided by 0 is always undefind but not infinity. u are doing like 1/0=2/0=0 so1=2?? 2nd method is when u take squre to both side should only take positive root. U are doing like (-1)^2=1^2=1 so -1=1?? Go back to primary school and stop posting such rubbish.

Lol 1=0 no

X= 16 its in my head.

Среди порядковых чисел( ординалов) такого нет, чтобы следующее число было бы равно предыдушему. Однако среди кардиналов такое возможно , если к бесконечности прибавить 1 то получится бесконечность.

Took me 0.2 seconds to arrive to 2.

2^(3a) = 20 3a lg(2) = lg(20) 3a = (lg(2) + 1) / lg(2) 3a = 1 + 1/lg(2) a = 1/3 + 1/lg(8)

When was this a Olympiad math problem? Your solution of the problem is sooo wrong. Please remove it.

😅 entertaining but what how much us A?

25^25=5^(25+25)=88817841970012523233890533447265625 X=25

16^16=4^(16+16)=18446744073709551616 x=16 It’s in my head.

K=2

Oh my dear, sorry but this are 8 minutes of nonsens. Subtraction of k ob both sides gives us 1=0 which has solution.

1/0 is not infinity.

Quite easy even for me. It is even simpler if you apply base-2 logs from the beginning, since 4 is 2^2 and 24 is 3x2^3

I personally think just doing 1= √X / 2 is easier in the step before logs, because every normal integer without an exponent is basically itself to the power of 1. 1= √X /2, 2= √X, X=4

It's a pity there is not (and impossible) to have a kind of moderation at UA-cam: the purpose would be to keep such boolshit out of our attention. And, call me racist, but I mostly (only?) saw such shitty math lectures from Indian lecturers.. hard to imagine why, just the fact

Nice one ❤

That is complete, utter nonsense.

Great solution, clicked the video after finding soln in a way and was surprised to see exactly same being done here too

Please stop posting my videos directly to your channel!!!

Please stop posting my videos directly to your channel!!

میشه ۸+۲تا رادیکال ۷ششم=۴+رادیکال ۷ سوم.

Got the answers x=1 and x=4 without even beggining calculating. Just by looking at thumbnail.

(Sqrt[35]+Sqrt[5])/(Sqrt[35]-Sqrt[5])=1.3 recurring+0.3 recurringSqrt[7]=(4+Sqrt[7])/3

Good work, but here's a slightly more elegant solution: sqrt(x)^sqrt(x) = x x^(0.5x^0.5) = x^1 Equate powers: 0.5x^0.5 = 1 x^0.5 = 2 x = 4

Answer to the last question would be x = 1 OR x = 27 following the same method

This is not an Olympiad question. You might’ve misunderstood it as such from some middle/early high school competition, but definitely not a math olympics question.

But I did not think 1

Bro if x is 4 then sr of 4 is 2 and 2 power 2 is 4

Way too complicate.

I don't think that's an olympiad it's too easy

oops... k^4 => 4 solutions by inspection k = 3, k = -3

because there's k^4 it means: THERE'S FOUR SOLUTIONS

Good 🎉❤❤

Hello My Dear Family😍😍😍 I hope you all are well 🤗🤗🤗 If you like this video about How to solve this math problem Please like and subscribe to my channel as it helps me a lot，🙏🙏🙏🙏 ❤❤🎉🎉

No need to calculation, x=1

So many illegal solutions in one piece! 😂 It was a task from 1st april.....

You can not put a negative number as a square root.

Gimme a break: 9 minutes for What??? 5 secs: K=2

This is nonsense to which meaning cannot be assigned. Most commenters believe you are incompetent. I think you are intentionally misleading.

What a waste !!!

My God, and this is the simple Harvard University Admission ? It´s obvious that K = 2 not necessary to do any operation.