The equation has two solutions integer. If we take b=na with n positive integer there is solutions for a and b integer with n=1,13. If n=1 a=b=26 and if n=13 a=14 and b=14*13. Then other solution is a+b=56.

While this presented an interesting way to show that (1+i*sqrt[3])/2 is the cube root of -1, it might have been quicker to represent it in complex polar form as exp[i*pi/3], and recognise that 7777 powers is exp[i *(7776+1) * pi/3] or exp[i*(1296*2*pi +pi/3)] so result is just exp[i*pi/3]. Admittedly, this does assume that one notices that (1+i*Sqrt[3])/2 is unimodular and recognises the 30-60-90 triangle to get the phase.

(1+3/2)^3^4^3^4^3^4^3^4 (1+1/2)^1^2^2^1^2^2^1^2^2^3^2^2 (/1)^1^1^1^1^1^1^3^1^2 ()^3^2 ( x ➖ 3x+2).

This isn't good bro what quality is this this question is so far from olympia level Olympiads are 100x tougher and you are putting olympiad in the thumbnail and a picture of einstein! this is as funny as worse your thumbnail is 😂😂😂

Beautiful.

Why 169 '

Very elegant method.

can you learn us sth about methamatics for example making a video about what is the imaginary number and how is the multiplication works on them.

That one is perfect

The expression involves raising to a non-integer power, which generally results in a complex number. Here's the breakdown: 1. is the square root of , approximately . 2. Raising to a non-integer power is defined in the complex plane and involves Euler's formula: (-1)^x = e^{i \pi x} where . 3. Substituting : (-1)^{\pi^{0.5}} = e^{i \pi (1.77245385091)} Simplifying: e^{i \pi (1.77245385091)} = e^{i (\pi + 0.77245385091\pi)} = e^{i \pi} \cdot e^{i (0.77245385091\pi)} Since : (-1)^{\pi^{0.5}} = -1 \cdot e^{i (0.77245385091\pi)} This gives a complex result, with magnitude 1 and an angle of in radians.

Can we find a general strategy to solve 1/a+1/b=k which is a+b?

Is this video a joke? Taking 7th root gives you 7 different complex solutions on the left side and 7 on the right side of the equation. So the guy arbitrarily chose 1 if them ignoring remaining 48 and claims it a solution :)

it might be faster to convert the square roots to exponents & distribute. Then add them up. You are quickly left with m^(7/8) = 2^7 fraction rule to rewrite will get you 8th root of m = 2 then you can be left with m = 2^8 Either way, it was an excellent video. thank you for it!

m=256

Nice 🎉

Nice 🎉

why is it not 78? 1 2 3 1 -- + -- = -- = -- 39 39 39 13 so 39 + 39 = 78??? Did this video show a minimum solution or am I just getting frustrated over nothing?

m^(7/8) = 2^7 Taking 7th root both sides m^(1/8) = 2 Taking both sides to the 8th power m = 2^8 aka 256

e^1*e^2=e^3

2^7 2^7^1 2^1^1 2^1 (m ➖ 2m+1).

This looks crazy. The answer is simply -i and it is 5 seconds of mental arithmetic. Keep the Argand diagram in mind.

This can be further expand using De Morvier's theorem

√(m) √(m) √(m) =128 Or m√(m) √(m) =128*128 Or m^2*m*m√m=128^4 Or m^2×m^2×√m=128^4 Or( m^4) ^2×(√m) ^2= 128^4 Or m^9=2*2*2*2*2*2*2*2*2 =2^9 Now m=2

128= 2^7 , the m power , 1/2 , 3/2 , 3/4 , 7/4 , 7/8 , m^(7/8)=2^7 , / ()^(8/7) , m=(2^7)^(8/7) , m=2^8 , m=256 ,

Лажа. Автор - обманщик, ворует время у зрителей

Extremely difficult

1/a = 0.5/13 a= 13/0.5 a= 26 Similarly , b=26 Therefore , a+b=52

x^3 = e^i2nπ x = e^(i2nπ/3), n = 0, ±1 x1 = 1 x2 = cos -2π/3 + i sin -2π/3 = -1/2 - i√3/2 x3 = cos 2π/3 + i sin 2π/3 = -1/2 + i√3/2 or x^3 - 1 = 0 (x - 1)(x^2 + x + 1) = 0 x = 1, (-1 ± √(1 - 4))/2 = (-1 ± i√3)/2

3^x = 6 x = ln 6 / ln 3 = 1 + ln 2 / ln 3

e^iπ/4 + e^-iπ/4 = cos π/4 + i sin π/4 + cos -π/4 + i sin -π/4 = 2 cos π/4 = 2/√2 = √2

x^3 = e^(iπ(1/2 + 2n)) x = e^(iπ(1/6 + 2n/3)), n = 0, -1, 1 x1 = cos π/6 + i sin π/6 = √3/2 + i/2 x2 = cos -π/2 + i sin -π/2 = -i x3 = cos 5π/6 + i sin 5π/6 = -√3/2 + i/2

divide both sides by b: a/b + 1 = 5√(a/b) let √(a/b) = u, thus u > 0 u^2 - 5u + 1 = 0 u = (5 ± √21)/2 a/b = (5 ± √21)^2/4 = (23 ± 5√21)/2

8^9/5^5 8^3^2/2^3^2^3 2^3^1^1/1^1^1^3 1^1^2/1^3 1^2/1^3 2/3 (abc ➖ 3abc+2).

c/a = 2, c/b = 3 c = 2a = 3b b = 2a/3 a + b + c = a + 2a/3 + 2a = 11a/3 but ab = 200 => 2a^2/3 = 200 => a^2 = 300 => a = ±√300 => a + b + c = ±11√300/3

(1 + i)√a = 12 √a = 12/(1+i) = 6(1-i) a = 36(1 - 2i - 1) = -72i similarly (1+i)√(-a) = 12 leads to -a = -72i so a = ±72i

I really liked that you gave 2 methods. Your first method would have been the one most people would start with. The second method was beautiful.

i = 90° => sqrt(i) = 45° (solution one: 2 * 45° = 90°) or sqrt(i) = 225° (solution two: 2 * 225° = 450° ~ 90°)

awesome

Downvoted for wasting my time. Write faster or speed up the video.

There is a logically mistake in the Lets Verify. Each case you write first LHS = RHS i.e. ( 2/2 · 2/2 = 2/2 · 2/2 ... ) If so, then you do not need to show the steps, because you have shown us both sides are equal. Therefore, you should calculate LHS and then calculate RHS and then show us both sides are equal. Don't prove or verify both sides at the same time. then RHS

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A few more examples would be very helpful. Thank you

You're just rewriting the quantity into a more complex expresion. It's absurd!

Its irritating 🤬🤬🤬🤬🤬🤬🤬🤬

{e^x+e^x ➖}{ e^2x+e^2x ➖ }={e^x^2+e^4x^2}=e^4x^4 e^2^2x^2^2 e^1^2x^1^1 e^2x^1 (ex ➖ 2ex+1).

(2+i^2)/(3 ➖ i) (x ➖ 2i^2x +2i^2)/(x ➖ 3ix+3i) (x ➖ 1i1x+2i^1)/(x ➖ 3ix+1)(x+2i)/x ➖ 3i) (x ➖ 3ix+2i).

I was amazed he was able to get an answer to an equation with two unkowns then I realized there were infinite solutions. Nice pencil moving though.

Very simple, 6^x (36 -1) = 60 6^x = 60/35 = 12/7 Apply logerthms then find x value. xlog6 = log(12/7) We should know logerthm, log inverse values from 1 to 20 while preparing for such exams.

The golden Ratio ! Again this one ! ( in fact the logarithm of golden ratio ) . So i guess we can solve any exp(x) + A * exp(2x ) = exp(3x) with A integer , and we ll find as solution the logarithm of metallic number ( silver ratio , etc .. ) ; a kind of relation with . exp(x) + A * exp(2x ) = exp(3x) and the sequences u(n+1) = A u(n) + u(n-1) . Thank you for the share .

With this account, it does not matter if the parameters a and b are changed a=14 b=184 equally a=184 b=14