Відео

Doesnt work

It's expired as of 4/22/2024

You cannot directly assume u=2 from u^u=4. You need to specify that the equation seeks real number solutions and rigorously prove that u has to take on the value of 2.

hello, sir, I have a question. What's your purpose to substitute x=y=0 into the equation in the first place? Or it is just a try?

P R O M O S M

Didn’t work

Doesn’t work lie

Awesome Video I recently visited your UA-cam channel and enjoyed your content. However, I noticed that your videos are not ranking well due to some issues. I believe that if these issues are addressed, your videos will rank better. Here are the issues I found with your channel: 1. Video SEO has not been properly done. 2. Videos are not being released on time. 3. The video titles are not optimized. 4. Keyword research has not been done effectively. 5. Video thumbnails are not appealing enough to attract viewers. Solving these issues will help your UA-cam channel perform better

This is not an olympiad problem...

But x=0 is also valid, Sir. Let a=7^x and b=11^x.Square both sides, simplify and get (a-b) (a-2b)=0. Hence a/b=1 or a/b=2 which yields x= 0 or x=log2/log(7/11).

A=4 B=7

I got -(9W(-2/9･ln3))/(2ln3) ≅ 1.41138174220196 where W is the Lambert W function as an answer

You dear sir, are making a simple question difficult. Simplify math. If you make it seem difficult, no one would join your channel.

It isvclassical to prove that a^2+b^2 div by 7 if and only if a div by 7 and b div by 7. This is an important point.

Is 6.4x10^99

ab=2 (1) bc=6 (2) ca=3 (3) Divide (2) by (1) --> c/a=3 c=3a Plug in (3): 3a²=3 --> a=±1 c=±3 From (2) b=±2

a+2ab+b=8 It is diophantine equation Multiply by 2: 2a+4ab+2b=16 2a(1+2b)+2b=16 Add 1 to both sides: 2a(1+2b)+(1+2b)=17 (1+2a)(1+2b)=17 Note that equation is cyclical (1+2a)=1 --> a=0 (1+2b)=17 --> b=8 (a,b)={(0,8),(8,0)} --> a+b=8

On minute 5:28 you made a mistake. You expanded 6+2*sqrt(5) to 2(3+sqrt(5) which you cannot do. It would have to be 6(2*sqrt(5)) in order to do that. Best wishes bro

It's never mentioned that a and b are whole numbers so your answer is invalid

Me: One of them equals 0 and the other equals 8, so a+b is also 8, change my mind.

I think the question was restricted to integers. Usually this should be straight forward

absolutely wrong. There are infinite solutions, why you arbitrarily chose a = 0 and a = -2 makes no sense. The actual solution is to multiply both sides by a, solve the quadratic formula for b anb you can solve that b = (16 - 2a)/(4a+2). In your example you picked a = 0, then b = (16-2*0)/(4*0 + 2) = 8. This gives a complete solution for ANY VALUE of a or b. If a = 2314242 then b = -0.4999981635 and a + 2ab + b = 8 is still satisfied. You solved for two "cases" but wrongly thought those are the only two cases, unless the question is incomplete and specifies integer values for a and b only, but that is a CRITICAL aspect of the question

a=1, b=7/3 -> a+b=10/3 i don't see any restriction on whether a,b are positive integers, or whether they can be 0, whether they can be a negative integers, whether they can be rational, etc.

A²+2ab+b²=8 (a+b)²=8 Solving for a+b would give u A+b=±√8

This is simple that we can solve root by full calculation

50⁵⁰=(2×25)⁵⁰ =2⁵⁰25⁵⁰ =4²⁵25²⁵25²⁵ =100²⁵25²⁵ Then 50⁵⁰/25²⁵=(100²⁵25²⁵)/25²⁵ =100²⁵ =1E25

Basicly : 3

or trial and error x = 2

Please write large.are you jelous

you deserve a big like because i love math than any other subject :)

Algorithm be like this guy will watch anything

you can just square the first equation, get x²= [sqrt (i)]² + 2 sqrt ((i) (-i)) + [sqrt (-i)]² => x= sqrt 2

You can argue this geometrically much easier and faster.

'Promosm' 🙋

1-1-1-1 = -2

100! or 50^100 is also another interesting problem to solve.

1/2 + 1/4 + 1/8 + 1/16 + ··· =1 7^1=7

I prepare a triangle. It is an isosceles triangle with vertices of 36° 72° 72° and two sides of length 1. At this time, the length of the base can be found to be 2sin18° using the perpendicular line. If we write the bisector of the angle of 72° again, we get a similar triangle with 2sin18° as the length of the two sides. If you use the ratio of similitude, you will get the answer of 2sin18°. Divide by 2 to get the answer.

As somewhat of a math channel myseld, this was an amazing explanation. Keep up the good work! ❤

Wasn't the task 81^x = x^81 ? You're doing it with 18, but the thumbnail says 81

A polynome of the nth has n solutions in |C. +/- Sq.root of 3 was pretty obvious +/- i.sqrt(3) came very quickly

180..i found this by factoring all of the given numbers and looking for a pattern.; 4 = 2 x 2. 18 = 2 x 3 x 3. 48 = 2 x 4 x 6, 100 = 2 x 5 x 10. Our solutions would be 2 x A x B, with A being the next number in the series ( therefore 6). To find the value of B, I noticed another pattern: if I added the last two multiplicands of 18, and used that as the subsequent B value, it worked. Therefore, the solution to the sequence would be 2 X 6 X (5 + 10), = 180.

By discarding k=1 you discarded 1^1=1^1 and 0^0=0^0=1.

X=y

Ridiculous to call this a 'Fascinating Problem'...Just say : m-7 = -m 2m = 7, m = 3.5 , lol

for this question we can simply do by converting the RHS to factors of 3 and 5 as the values are simple. though the explanation is very clear, this will be the general method to approach this question. ☺☺☺☺

Hay

log base 2 of the golden ratio?

Its ln to the base a X in second step,

1/X++1/y== Y××++X/Xy 2XY/2Xy X==y==1