Відео

th th th th th :D

Did this in my head in three seconds, meanwhile 8 min later this dude is still writing equations. We are doomed .

4^x = x 1 = x/4^x 1 = x.e^(-xln(4)) ln(1/4) = xln(1/4).e^(xln(1/4) W(n, ln(1/4)) = xln(1/4) x = W(n, ln(1/4))/ln(1/4) n x ... -2 1.2484056050489365 + 5.5045741348280212i -1 0.0639598103013431 + 1.0908433765399576i 0 0.0639598103013431 - 1.0908433765399576i 1 1.2484056050489365 - 5.5045741348280212i ...

Solved this in my head before watching 👍

This is tooooooooo long for a simple equation

A very thorough approach, and I do wonder about forcing the lhs into the sum of cubes. The alternative solution most likely uses modulo arithmetic. It is enough to guess and check, which is easy. Then demonstrate that the lhs is an increasing function, therefore just one solution.

THIS IS SO EASY😊

Finding 3 in your head takes less time. Like a lot less time than this.

as many don't seem to understand there is a difference between assuming and testing a result and instead calculating a result, i want instead to point out something. this calculation still is not-so-empyrical. we're assuming n is an integer value, along with 35 factors. there are actually infinite possible values of n (before calculating its value, i mean), not just integer values, along with the factors resulting in 35. you can multiply 5/2 times 14 and the result is still 35 but this exercise is assuming (based on what?) that the 2 parenthesis are equal to integer values, along with the fact that consequently the left parenthesis is lower than the right one. if x and y are both 1/2, the left parenthesis is bigger than the right one (1 > 1/4). also, even if they are both 1, it results in 2 > 1... so i don't see why the left parenthesis should be "granted" to be < than the right one.

Why choose 3 in the first case rather anything else? Because you worked it out in your head like most. This video is BS.

Vous nous faites chier avec vos pseudo théories que vous avez créé vous même !!!!!!!!!!

(3^3)=27+(2^3)=8 27+8=35 so n=3

What the hell is he doing all the time ..!😂😂😂😂 I think he teaches us how to make simple problem into complex one.,.......

It appears that you made a mistake in case 1. You were adding the two equations not subtracting.

word for word replicate of other videos...

sir could you help with integration problems?

yeah yeah cambridge and other bs x=1. simple as that

Nul

X = -3

Log(i) =log(1)+i.arg(i) = pi/2 i

They don't ask questions on anything you haven't learned so a no brainer for anyone coming out of college (or cal 3).

Thank you sir💖

Very good 🎉

Trial and error is much easier than this

X^5 -1=0 X^5=0+1 X^5=1 X^5=1^5 X=1

35 = 27 + 8 = 2^3 + 3^3

Why would you do silly math for no reason at all.

Very nice presentation, really entertaining, keep up the good work😃

There is a MUCH faster analytical way to solve this in 3 easy steps using modular math: STEP 1 3^n + 2^n = 35 => n ≠ 0 (mod 2) 1 ≡ 1 (mod 3) [1,2,...] ≡ 2 => n = 2 m + 1 for m>=0 STEP 2 3 9^m + 2 4^m = 35 (mod 4) 3 ≡ 3 (mod 9) [2,8,5,...] ≡ 8 => m = 3 a + 1 STEP 3 27 9^3a + 8 4^3a = 35 If a>0, left side > 35 => a=0 => m = 1 => n = 3 Also works with much larger numbers that can't be guessed with trial and error ;)

Value of w ?

x^2-1=√x+1 =(x+1)(x-1)=(x+1)^1/2 (x+1)(x-1)÷(x+1)=(x+1)^1/2 ÷(x+1) (x-1)=(x+1)^1/2 . (x+1)^-1 =(x+1)^-1/2 (x+1)^2 = (x+1) x^2 +2x +1 =x +1 x^2 + 2x - x +1 - 1=0 x^2. + x=0 x(x+1). =0 either x=0 or (x+1)=0 therefore x+1=0. ; x=-1 Test if x=0. x^2 -1~√x+1. Test: if x=1 (- 1)^2 -1~√-1 +1 1 -1~ 0 therefore 0=0 ; if x= -1 0 -1~√0+1 -1~1 reject

You can do this more easily by lettling f(x) = 2^x + 3^x. We know f'(x)=2^x * log(2) + 3^x * log(3) > 0, thus f is injective. We know by trial and error that f(3)=35. But because f is injective, this is the only value of x for which f(x)=35. So n=3 is the only solution to our original problem.

Thank you sir🥳💖

Its simple hit and trial

Thank you💖

Why not use logaritmos?

=2^3+3=11, x=3

(x^2 +x^2)=9 2x^4=9 (X^2)^2. )=9 X^2=9 X=√9 X=3

Algebraic teram Qaudtic equation

thank you for reminding me of when we were in high school.

X =1

Bruh x is just 1 I did it in my head Does that mean I'm qualified for cambridge

Why even factor it out?

it so easy it 1 1 to the power of 1 is one minus 1 is 0 that soooo easy

Factoring

M=2 if given that m is int

Answer is just simply 1.

Thank you💖

If you think the answer is 1, you probably never took algebra classes. Because it is to the power of 5, there is 5 solutions. 1 is only one of them

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