If you rewrite the square root to the power with exponentt 1/2, you can simply apply the rules a=a^1 and a^n*a^m=a^(n+m) to dothe calculation .... You may simplify the result by applying (a^n)^m=a^(n*m) and 8=2^3.

Très facile

10^x=20 x=Log(2)+1

The complex solutions are wrong and introduced by the squaring, just plug it into x*sqrt(x) and you will get ×- i instead of 1

3.4142

Which Programm do you use for your videos?

Great! Thanks a lot!

sum i=0 to n of 3i = 3n(n+1)/2. For n=33 this sum is 99*17. Then add on 34 1's to get 34+99*17=1717

Thanks so much!

1+4+7+10+……+100=1717

Xpand

Thank you so much.

I'm not watching the whole thing but thr answer is basically there 1:30 in and it's x = 1

I did it like, x . √x - 1 = 0 x¹ . x½ - 1 = 0 x¹ᐩ½ - 1 = 0 x³⁄₂ - 1 = 0 x³⁄₂ = 1 x = (√1³) x = ±(√1) S: {-1, 1} what did I do wrong? idk if I did some dumb basic maths mistake but it seems everything is alright (alr nvm, just saw that the (-1) root would give me a complex equation 😅)

x = 1

Without method the answer is 16

Thanks so much!

Sqrt[6+Sqrt[32]]=2+Sqrt[2]

I rewrote all 8s as 2^3 early on thinking it would simplify. At the end I got 2^(21/8) which I guess is the same as 8^(7/8) if you again rewrite as 2^3=8 by factoring the 21=3×7.

you squared both sides so you have to check the solutions, the 2 complex ones give you -1 so they are not good

Try using complex method

Great work! Thanks a lot!

Thanks so much!

Exactly. Merci Everybody. 😊

Для чого такий довгий шлях? Зроби спочатку все можливе під коренем.

Why bother with complex numbers? Are real numbers not enough?

Кто мне объяснил нахуя переписывать исходное задание, тем более, что решение очевидно.... ☹️🥴

Well done. Thanks so much!

Always look for substitutions to make the problem solution easier. Substitute x=y+5/2 into the given equation and rearrange to: (y+5/2)^4-(y-5/2)^4=2*4[y^3(5/2)+y(5/2)^3]=0 Divide by 8(5/2)y to obtain y^2+(5/2)^2=0, which has roots y=±i5/2. Also y=0 is a root. Thus, x=y+5/2=(5±5i)/2 or x=5/2

2.5; (5+5i)/2; (5-5i)/2

Thanks a bunch. The result is 81

Можно очень коротко.

bro.. thank youb so uch for this helpul content'' i appreciate it subscribing to see more as it..

It was very easy, I did it at once. Thank' s for tranning ! 😊

(Sqrt[Sqrt[Sqrt[3]]])^32=81 It’s in my head without scratch work.

(⁸√3)³² = (3^⅛)³² = 3⁴ = 81

((16(16(16)^1/2)^1/2)^1/2)^16=((16(16×4)^1/2)^1/2)^16=((16×8)^1/2)^16=((2^4×2^3)^1/2)^16 =2^7×16/2=2^56

Both numerator and denominator have a common factor: square root of 3.

1

Bro, 2 ^ 3 + 2 = 2 × 2 × 2 + 2 so x = 2 lol

2:23 where did the plus 1 come from

1.?

I solved it instantly because 2 was the first number i checked

x*ln(x/5)=5^2 * ln5 , x*ln(x/5) = 25* ln5 , /:5 , (x/5)*ln(x/5) = 5* ln5 , ln(x/5)*e^ln(x/5) = 5* ln5 , ln(x/5)*e^ln(x/5) = ln5*e^ln5 , ln(x/5)=ln5 , x/5=5 , x=25 , test , (25/5)^25=2.98023*10^17 , 5^5^2=2.98023*10^17 , same , OK ,

Cubic equations ALWAYS have three solutions - find all three.

OH PLEASE, just stop running circles, just solve it already

Really ? The equation is solved in the First couple of seconds and then the Video just continues for some more 6 minutes.

Good solutions. I thought for sure Lambert W would be used to solve this in at least one method, but you successfully avoided it.

Излишне подробно и затянуто. Каждое решение сопровождается изобретением колеса и палки-копалки.

hey as this is a math channel can u help me with something. Can u tell me all the different names of the three main nature of roots like bsquared -4ac>0 and b^2-4ac<0 and bsquared-4ac=0. Sometimes they are called real and distinct, sometimes real and equal, sometimes complex, not real, repeating, tangent to the curve, intersecting the curve and other stuff. can u pls tell me all names and explain which one each is. plsss urgent..................