C. 3/8

В.

You r not even following the rules of BODMAS . Forsake 😢

30..

Answer B

9+81=90; 3^2=9. m=2

I made it be m4^m = 1 so m must be less than 1 and m2^m2 = 1 ignoring the infinite recursion, it can only be 2

Learn this simple trick for many problems containing ±x±y=c where c is any real constant. Let ±x=c/2+z and ±y=c/2-z. Clearly ±x±y=c/2+z+c/2-z=c. Then substitute these expressions into the second equation and solve for z. Finally back-solve for x and y. In this case substitute x=3+z and y=3-z into the second equation giving (3+z)(3-z)=-z^2+9=36. Rearranging gives z^2=-27, z=±i3√3, x=3±i3√3, and y=3∓i3√3. Now how difficult was that?

You're clearly trying to write while holding a camera in your other hand. Looks uncomfortable and comes across that way too. Doesn't help that you seem to be doing it on scratch paper.

step 1: is it 1? step 2: is it 2 or more? step 3: well it can't be 0 step 4: must be 0,5 solved in 15 secs

Take ln2 of both sides

Why so difficult. Each side to the power of 1/m and that's it.

By the way, the i can't be inside of the square root, as the square root of i is different from the square root of -1.

x²(x²+1) = 20 2²(2²+1) = 20 x = 2, problem is solved 🤪

If m is an integer, then it's got to be a power such that 4^m is rational. Positive integers make 1/m too small. Negative integers make 1/m negative. 0 is bad. Then it's got to be a fraction. To remain rational, the denominator must be 1 or 2. 1 failed, so we try 1/2. It works. Imagine the graphs, it's the only solution. So m=2.

Two solutions: 8i and --8i

"1x1=2" -terrance Howard

Needs work. Writing is hard to read, you skipped some parts and went back, which confused me. You have a hard time keeping the part your writing on screen. When verifying the solution, you come up with parts I couldn't follow, like at 5:10 where you had (root(-8i))^2 + 2 - 8i X 8i + etc, where did that 2 come from?

Also by simetry 8i is a solution too.

maybe, using modular function the result can come faster

n = 1/2 is a solution

Awesome ❤❤

Without viewing the video, I solved it in my head by guessing M =3 then 4, and finally 5 which is answer. Similar problems with very large M would require many more guesses. I don't know how to solve with algebra.

Way overengeneered. m! = m * (m-1) * (m-2)! m^3 - m = m * (m^2 - 1) = m * (m - 1) * (m +1) We can tell right away that m is not 0 or 1, so we can divide by m*(m-1) and are left with (m-2)! = m+1. From here it's evident that m can't be a large number, must be larger than 3, and that m+1 can't be a prime number such as 5 or 7. So only m+1=6 -> m=5 is really an option.

It can be done simpler. Until 1:46 it is ok. But then (m-2)! >= (m-2)(m-3) = m^2 - 5m +6 = m(m-5) + 6 >= m +6 if m>= 6. Thus m can't be greater than 5. It is enough to check m=1,2,3,4,5 to see that the only solution is 5

Factorial grows faster than cubic polynomials. So the answer needs to be before the points where x³ - x goes below x! (Quite early) That means the answer is not very large, we can probably brute-force it! Lets write a few cubes 1³ = 1 2³ = 6 3³ = 9 4³ = 64 5³ = 125 And a few factorial 2! = 1*2 3! = 2*3 4! =6*4 5! = 20*5 6! = 120*6 Ok, so 5! (120) Is really close to 5³. The intersection point is near. And we can see that 4 doesn't work 20 ≠ 64 - 4. So it HAS to be either 5 or 6 (Else we overshoot the intersection of the two graphs) 5! = 120 5³ = 125 5³ - 5 = 120 Ok, so it WAS 5. m = 5 satisfies m! = m³ - m Quick and easy educated brute-force to the rescue

Saludos de Argentina!

thanks. no solution with logarithm?

I would rather go for hit and trial in the first place within 15 sec after reading the question

x [(x-1)!-1] = 3 But 3 is prime, so the only numbers wich makes this product correct is 3 . 1 = 3, wich leads to x=3. And then, m=5. Is this solution correct?

This is so crazy!

Only 1 minute in and just a primary school teacher. When you write n! = n (n-1)(n-2)(n-3)! That wouldn't work if n=3 right? Because then 3 = 3 * 2 * 1 * 0! Oh wait, should I read the last part as 0 factorial, which is 1? Than is does work. But how about n = 2 2! = 2 * 1 * 0 * -1! Is that correct? I am learning about minus factorials now. However, there is a 0 there, so it should add up to 2! = 0 Or where am I wrong.

This is an incorrect question.The identity is false

Bad lighting, bad penmanship, and mistakes in the video.

You are in the 99% my friend.

sir ::This is a foolish you are fooling. zero multiplied by anything is zero so zero =zero.

The 0/0 is maningless. Entire process is wrong.

You can not divide with 0 So 0 still equals to 0 till the end of logic!

divide by zero ... no no

C"est très amusant! mais une règle fondamentale énonce<<lorsque dans un produit un facteur est nul,tout le produit est nul<<<donc 6(3- 3)=,-7(3-3)implique 6(0) =7(0) implique 0=0 .mais,attention des cas amusants comme votre exemple peuvent induire les élèves en erreur!

No se puede dividir por 0

Your thumbnail is not the equation you performed. The denominator of the thumbnail was (x-1)! which means simplifying would give you a quartic to solve.

When one removes (3-3) on both sides of the equation, one is diving both sides by 3-3, which is zero. This is not possible. There you have the origin of 6=7. Similarly one can do x^2-x^2=x^2-x^2 and then x(x-x) = (x+x)(x-x) (so far all correct) and then take out the (x-x) and leave x=x+x or 1x = 2x or 1=2.

All you've done is rearrange terms. This is kind of a waste of time.

change the line with 6x3-3x6=7x3-3x7 to (6x3)-(3x6)=(7x3)-(3x7). the parenthases are needed to make the solution true. Now, it is not possible to conclude that the following line is true, because that would be saying that the '3' in both equations is equal to '3-3' aka 0