What a waste!

If I put for x=-0,5, in the equation, then the solution of the equation is 1,11'. For this I try with the number -0,485 and put it for x. Then the solution of the equation is: [(-0,485+1)^2/-0,485^2] - [(-0,485+1)^2/(-0,485-1)^2 =1,006 <=> x=-0,485

I will dieeeee by solving this complex solution... 15mins long explnation for a single question..... !!!!!!!!!!! (

Mathematical thought. The square root and squaring do NOT cancel.!!The square root of x squared is the absolute value of X.Thats where the plus or minus come from!!

答案是根號3-1，非根號3+1

Solution is so Long.it takes so much time

Your quadratic equation is wrong. You need to correct it

In quadratic equation you were solving for 1/x. Not x As well as 2A is in denominator

Soluția este greșita radical din 3-1 e buna

my wife loves watching your videos

a`5=32=2 '5so a=2

The complex solutions can be written in terms of square roots; no cube or higher roots are needed.

x^( (1/ln5)lnx ) = 5⁴ ( (1/ln5)lnx )lnx = 4ln5 ln²x = 4ln²5 lnx = ± 2ln5 lnx = 2ln5 = ln25 => *x = 25* lnx = -2ln5 = ln(1/25) => *x = 1/25*

А не проще логарифмировать Получаем простое квадратное уравнение относительно log (x+3) log (x+3)*(log(x+3)+log 3)=1-log3 Я думаю что решение будет 2-3 минуты, а не 14 минут как на видео

I solved this in my head in 2 tenths of a second.

You lost half time ofrece the video saying and doing nothing important.

2 , a⁵-32=0 aεR Move 32 to the right Change 32 to 2⁵ a=2, aεr

Trivial. a = 2 AND the other four imaginary roots separated on the "unit" circle ( times 2 ) and spaced 2*pi/5 radians apart radially around the circle.

Practice self-care and self-love to overcome negative energy. Nurture your mind, body, and soul with activities that bring you joy and peace

For y2(no y1) you forgot to wrote 2 in dominator!!!

A ray of sunshine online.

(x^2 -3•x)!=15•(3!•8!) <=> 15•(3!•8!)=3.628.800 <=> (x^2-3x)!=3.628.800 <=> 10! = 3.628.800 <=> (5^2-3•5)!=10! 10!/6•40.320=15 <=>3.628.800÷241.920<=> (5^2-3•5)!/3!•8!=15 <=> x=5

Great job 👍

Inside the formula you wrote inside the square-root should be 2 square root of ab ; not 2ab

Fantastic

Y=3x+4 So x(x+4) y(y+4) =2, 8 (X^2 +4x) (y^+4y) =2, 8 Can we do like this?? ? Iam getting sol for x as -2, (-6+-2√3) /3 Is that correct? ?

It is very simple solution but you make it very long solution. Pls simplified it.

3 + 2√5 got it

3+2V5.

Teacher or barbar

This isn't as rigorous as your solution but here's an alternative. The problem suggests a solution of the form x = a + b√2. Then x^3 = a^3 + 3a^2 * b√2 + 6a * b^2 + 2√2b^3 = 55 + 63√2 Regrouping x^3 = a^3 + 6a*b^2 + √2*(3a^2*b + 2b^3) Alternately x^3 = a(a^2 + 6b^2) + √2*b(3a^2 + 2b^2) If the proposed solution is valid, the first term should equal 55 and the second should equal 63√2. Or 55 = a(a^2 + 6b^2) and 63 = b(3a^2 + 2b^2) The first equation above indicates that a > 0 since the terms inside the parentheses are positive. Similarly the second equation indicates that b>0. So the first equation factors into: a a^2 + 6b^2 1 55 5 11 11 5 55 1 b can be determined for each pair. It equals 3 for the first and is imaginary for the others, i.e. only the first pair gives a valid solution of the first equation. Substituting a = 1 and b = 3 into the second equation gives 63, which satisfies it. So the solution is 1 + 3√2

X=3

A whole Lot of arkward manipulations, senseless!

64m2-64m3=3

Your video has a unique charm that sets it apart from the rest. Captivating!

You need a roadmap to get to the solution.

Diffecult to under stand why you but y from where you bring

Instead of expanding to powers can we use log? ?

@ahmethulisiarslan8829 Votre affirmation que les eéponses sont fausses est une insulte gratuite envers "Super Academy" delta = ( log 3 - 2) ^2 donc de forme (a-b)^2 mais (a-b) ^ 2 = (b - a) ^2 car a^2 - 2ab + b^2 = b^2 - 2ba +a^2 donc ( log 3 - 2) ^ 2 = (2 - log3 ) ^ 2 et racine de delta = +/- (2 - log 3) Sachez aussi que la formule [ -b +/- sqr(Delta) ] / [2a] peut aussi s'écrire [ -b -/+ sqr(Delta) ] / [2a] Réflécissez avant de poster un commentaire !!!! Prenez le temps de réviser les formules fondamentales @ahmethulisiarslan8829 Your assertion that the answers are false is a gratuitous insult to 'Super Academy' delta = (log 3 - 2) ^2 therefore of form (a-b)^2 but (a-b) ^2 = (b - a) ^2 because a^2 - 2ab + b^2 = b^2 - 2ba +a^2 so ( log 3 - 2) ^ 2 = (2 - log3 ) ^ 2 and root of delta = +/- (2 - log 3) Also note that the formula [ -b +/- sqr(Delta) ] / [2a] can also be written [ -b -/+ sqr(Delta) ] / [2a] Think before posting a comment!!!! Take the time to review the fundamental formulas Kind regards

Fantastic

These answers are wrong because the values ​​in the square root are positive in absolute value. The value of log3-2 is negative. When the square of this value goes outside the square root, it should be 2-log3. Kind regards.. time:6:54

Thanks a lot for step-by-step solving process. By letting 3(x+3) = t, the given becomes t^(log(t)) = 10. taking log both sides log(t)•log(t) = log(10); [log(t)]^2 =1; log(t) = 1 or -1. t = 10 or 1/10 That is, 3(x+3) = 10 or 3(x+3) =1/10 will be a little simpler.

Too long and drawn out. Too many substitutions.

2^2/x•2^1/2x•2^1/4x=2^5 2^(2/x+1/2x+1/4x)=2^5 2/x+1/2x+1/4x=5 8+2+1=20x 11=20x x=11/20

Thank you for taking the time to prepare this video and sharing it. You can speed it up when one step is super easy and obvious you can accelerate through it. However I think your solution is not what should be presented on KZbin. It's much more obvious to take 25 as a common factor and then do the trick you did with much smaller numbers. In my opinion redo the video repost and compare the different feedback.

Boring, too much ON the Long side

Easy

2√5+3

[log(3x + 9)]log(x + 3) = log(10/3) log(x + 3) = u (u + log3)u = 1 - log3 u² + (log3)u + (log3 - 1) = 0 u = [-log3 ± √(log²3 - 4log3 + 4)]/2 u = [-log3 ± (log3 - 2)]/2 u = -1 or u = 1 - log3 u = -1 log(x + 3) = -1 => *x = 1/10 - 3* u = 1 - log3 = log(10/3) log(x + 3) = log(10/3) => *x = 10/3 - 3*

The problem has been made to be sight soluble. Relating to x ^ log 27 + 9^log x =36. 1. log is base 10. 2. 27 + 9 = 36. 3. If 9 = 9 ^ 1 = 9 ^ log x gives 1 = log x (base 10) gives x =10. 4. Substituting x = 10 into x ^ log 27 gives 10 ^ (log 27 (base 10)) = 27. Giving a. x = 10 gives 10 ^ log 27 + 9^log 10 =36 where 27 + 9 = 36. b. real values of x <> 10 give values < 36 for x < 10, and > 36 for x > 10. c. If the LHS was left unchanged, and required value of x was 100, the RHS required would be 27 ^ 2 + 9 ^ 2 = 729 + 81 = 810, or ( x ^ log 27 + 9^log x = 810 ).