วีดีโอ

Euler formula is simply wrong. One side is the exp function which tends to infinity and the other side is a sum of 2 trig functions each ranging from -1 to 1, ie it's finite. So, these 2 sides can't be equal.

Two variables in one equation can not be solved separately.🎉🎉

писят два

1/14+1/（14×13）=1/13 則14+182=196

your answer is one of the infinitely-many values. this is because -1 = exp[i*(pi + 2*pi*n)] where n = ...,-2,-1,0,1,2,... therefore, the final answer is exp[i*pi^3/2 (1+2*n)] where n = ...,-2,-1,0,1,2,...

You need to write the pi symbol more clearly as it often appears as though you are writing the variable x!

438

-3 y. Course y -1 √>>t =202 D

how do you just manage to come up with 1681 as a factor in ur brain? is there any sort of way to find a factor of numbers give you a proper set? plz help

A questão é interessante, mas é preciso especificar melhor até que ponto deve- se resolvê-la. BRASIL Dezembro de 2024.

How i root of i = 4.81

Too slow in writing.😂

Решил за 25 секунд Ответ: e^(π/2)

I think this question is wrong I = exp(I *[pi/2 +2npi] ) So 1/I pow Gives Exp(pi/2*[1 + 4n]) So many boobs and noobs I the exponent 😊

Really use less

when I studied math, this was called rationalizing. And since you know where you want to go, don't go thru all this. Once you place everything under the 5th root. multiply them by 8/8 and be done with a rational denominator 2.

{2+4}=6 2^3 (x ➖ 3x+2).

1/(20)+1/(21)=1/41 /(10)+/(10^11) /(5^6) /(5^2^3)/(2^3^1^1) (2^3) (ab ➖ 3ab+2).

How to know to add 169

Hello ,everybody ! Greetings! Thanks , respects and considerations to all of you ...! Let me say that , here everything is just wonderful but uncomplete ( sorry to underline) !!! We have an infinity of solutions for this equation ( or equality ) . Let : a not null , b not null, k not null ,1/a+1/b=1/13 and a=k*b (or b= k*a) . Solving the system : 1/a+1/b=1/13 and a=k*b we get : 13(a+b)=a*b 13(k*b+b)=k*b*b 13(k+1)b- k*b*b=0 b(13*k+13-k*b)=0 and b not null 13*k+13-k*b=0 From here , let find k K*b-13*k=13 K=(13)/(b-13) Fron there: a =k*b=(13b)/( b-13) b not egal to 13 , b >0 and a>0 , that means that b belongs to the interval ] +13; + infinity [ Now for any b belonging to the given interval we have : a+b=b(k+1)=( b*b)/(b-13) Thanks ! Fron a former African and Senegalese Student in Moscow, Russia!

x^6/729=1(x ➖ 1x+1).

(5 ➖ 2)^2^3 (2^3 ➖ 1)^1^1^1 (2^3) (x ➖ 3x+2).

(x^6+18)=18x^6 3^6x^6 3^2^3x^2^3 1^2^1x^1^3 2x^3 (x ➖ 3x+2).

(2)+(2)=4(y ➖ 2x+2). 4^4 2^2^2^2 1^2^1^1 21(xy ➖ 2xy+2).

Just use the exponential form and divide the exponent by 2 😁

(x ➖ 1ix+1i).

So the answer to a meaningless problem is something so beautiful!!!😋

What is the purpose for the step of multiplying and dividing by 2?

Nice

A+b 82

분모통일시키면 ab=13 a+b=1바로나오자나 a=1-b b(1-b)=13 -b^2+b-13=0 이차방정식풀어서 b값 뽑으면 끝아님?

Well explained

Define "I" as X Squared. So then the square root of X Squared is X. Answer is X if you define "I" as X^2. But that is just not making the assumption that "I" means imaginary number.

But, what is the point of the "solution". You will recurr to the calculator anyway, and the solution is as hard as the problem. Nonsense, to me.

Mathematica gives this, 0.0000255089026236085942824535846310. Excel gives right the first 18 decimals. Is the answer of the vid.

Thank you professor for good solution.

You do realise, right, that the problem can be solved in 10 seconds?

Why not cancel the power on both sides X/3 = 2 X = 3*2 = 6 Check (6/3)⁶ = 2⁶ = 64 Done

Класс

asnwer= a+b=15 isit

PRINCIPAL ROOT: 64 = 2^6 (x/3)^6= 2^6 x/3 = 2 x = 6 OTHER ROOTS: degree 6 polynomial, therefore we should convert this into Polar and rotate by π/3 so: 6∠0, 6∠π/3, 6∠2π/3, 6∠π, 6∠4π/3 and 6∠5π/3 are the solutions! if ya want rectangular just use rcos(θ) + i*rsin(θ), for r∠θ ±6 and ±3±3i√3

can we assume that if there is even power (same) on both LHS and RHS as in this case '6' then whatever will be value of x then it always well have same value in negative? Like on solving value of x was 6 so can we assume since it had even power(6) then it also will have same value but in negative(-6)? without actually solving the equation?

ASNWER=4

x/13+y/13=1/a+1/b x+y=1 x/13=......

0.5/13+0.5/13=1/a+1/b

Minus sign should be plus in next to last step

Your solution seems right. In fact, e (exp) pi/2 is a real number 4.810495. This number appears intuitively very strange. Can you please provide a proof that this solution is correct?

14

8:36 . . . (-2)(-23814) = +46368

(5^(1/2)-2)^8. let 5^(1/2) = a & 2= b then equation become (a-b)^8=[{(a-b)^2}^2]^2= [{(5^(1/2)-2)^2}^2]^2 using (a-b)^2=a^2+b^2-2ab {(5+4-4[5^(1/2)])^2}^2 {(9-4√5)^2}^2 again (a-b)^2=a^2+b^2-2ab (81+80-72√5)^2 (161-72√5)^2 using (a-b)^2=a^2+b^2-2ab (51841-23184√5)