Let x>=y WLOG as we can exchange x and y by symmetry. So: x=1/2+u and y=x1/2-u; u>=0 xy= 1/4-u^2=1 ==>u^2=3/4...u=sqr(3)/2 as we assumed that u>=0 Then (1/2*(1+sqr(3));1/2*(1-sqr(3))) Exchanging...(1/2*(1-sqr(3));1/2*(1+sqr(3)))

Cambridge or not Cambridge, it's total nonsense! Sorry for the debunking!

Sorry Sir, but the best (and universal) way to solve such a system is (with very FEW steps) ... | x + y = 1 | | xy = 1 recall (very important): • consider S = sum and P = product • then roots of k² - Sk + P = 0 are x and y => we have to solve k² - k + 1 = 0 • x = k₁ = [-(-1) + √((-1)² - 4·1·1)]/(2·1) = (1 + 3i)/2 = 1/2 + (i√3)/2 • y = k₂ = [-(-1) - √((-1)² - 4·1·1)]/(2·1) = (1 - 3i)/2 = 1/2 - (i√3)/2 final result: ■ x = 1/2 + (i√3)/2 and ■ y = 1/2 - (i√3)/2 or (recall: addition and multiplication are commutative) ■ x = 1/2 - (i√3)/2 and ■ y = 1/2 + (i√3)/2 🙂

M=4

❤🎉❤🎉

(x•x•x)/{2•2•2}=(2•2•2)/{x•x•x} By cross multiplication.... x⁶=2⁶ Since powers are same ,therefore, x=2.....

m=4😊

Excellent 👏🏻👏🏻

√m+√-m=36 √m+i√m=36 (√m+i√m)^2=36^2 m+2i√m*√m+i^2*√m^2=36^2 m+2im-m=36^2 2im=36^2 36^2 m= _______ 2i 36^2 i m= ______ * _____ 2i i 36^2*i -1296i m= ______ = ___________ 2(-1) 2 m= - 648i

2^a=x; 3^b=y -->x^2+y^2=7; x^3+y^3=10 ---> (x+y)(x^2-xy+y^2)=10 ---> (x+y)(7-xy)=10=1*10 or 2*5 Then x+y=1, 7-xy=10, or all different combinations leading to real solutions

Easy. x=2

2and7

My favorite "first choice substitution" for similar equations cam be used here. u=x+y:, v=x-y {x^2 + y^2 = 7, (x^3 + y^3) = 10} {(u+v)^2+(u-v)^2=7, (u+v)^3+(u-v)^3=10} {2 u^2 + 2 v^2 = 7, 2 u^3 + 6 u v^2 = 10} We calculate 2v^2 from the first equation and substitute it into the second equation { 2 v^2 = 7 -2 u^2, 2 u^3 + 3 u (7 -2 u^2) = 10} -4 u^3 + 21 u - 10 = 0

沒有解。

Tetration equation? Seeing and hearing of this the first time. Thanks for sharing this video, sir. Much respect.

Lovely math problem form Onlinemaths Tv...😅😂❤

Nice work piece. Thanks for the video, sir

Great explanation. Thanks.

x^6 = 2^6 suggests right away that you use logarithms. 6.logx = 6.log2 ÷ 6 so that logx = log2, therefore x = 2... Which you can see just by observation.

Can somebody explain the part when he added 2n? How can he even do something like that? And can we make some simpler steps like: 1^x =5 xln1 = ln5 x=ln5/ln1=log1(5) and the answer is compex infinity

✅ 🤚 ☑ 👍 ✔ 👏

Thank you for the video

Nice maths

Nice question, I love this problem and the solution applied. Thanks man.

Thanks! Much ❤❤❤

x= 2exp(i*pi/3),2exp(2i*pi/3),2exp(-i*pi/3),2exp(-2i*pi/3),2,-2

Why so many steps ? The 6 roots of this equation can be solved with few (detailed) steps: (x·x·x)/(2·2·2) = (2·2·2)/(x·x·x) x⁶ = 2⁶ x⁶ - 2⁶ = 0 (x³)² - (2³)² = 0 (x³ - 2³)·(x³ + 2³) = 0 (x - 2)·(x² + 2x + 4)·(x + 2)·(x² - 2x + 4) = 0 x - 2 = 0 => x = +2 x + 2 = 0 => x = -2 x² + 2x + 4 => x₁ et x₂ = (-2 ± 2i√3)/2 = -1 ± i√3 x² - 2x + 4 => x₁ et x₂ = (+2 ± 2i√3)/2 = +1 ± i√3 --- /// final result: 6 solutions (roots): ■ x = +2 ■ x = -2 ■ x = -1 + i√3 ■ x = -1 - i√3 ■ x = +1 + i√3 ■ x = +1 - i√3 :-)

N equal 2,571, n = 2,571 has degreed from xy3 those 2,571 x 2,571 x 2,571 equal final quiteful goals to target seventeen. Hai japam former BPUPKI of indonesian republic, my king on jogja the majesty HB X was have met caesar naruhito. Naruhito still love indonesia. My big salutation of affect from naruhito.. he knows earn from BPUPKI is peeneeteea 9 . And laksamana maeda give circle table in his home to teenagers force affordable of soekarno to proclamation Indonesian Republic. Thanks japan. My eyes always was blues fall of japan kindness.

I like this video because I realised that to find x, we should use all remaining english alphabets from a to z and finally say x is equal to all other alphabet of English. If it is not enough, we can also use greek latin and Portuguese alphabet. This is algebra 😂😂

解が実数でなさそうであることはすぐにわかりましたが......

wait, is that i = √(-1); or i² = -1; ?

x=2

2/2=2/2 So X=2.

Degree 6 so we can expect six solutions. Let DOS = "difference of squares". By DOS, (x^3-2^3)(x^3+2^3)=0. For x^3-2^3, clearly x=2 is a root, so x-2 is a factor. By PD, the remaining polynomial is x^2+2x+4. By CTS, x^2+2x+1 = -4+1 => (x+1)^2=-3 => x = -1+-sqrt(3)i. For x^3+2^3, clearly -2 is a root, so x+2 is a factor. By PD, the remaining polynomial is x^2-2x+4. By CTS, x^2-2x+1 = -4+1 => (x-1)^2=-3 => x = 1+-sqrt(3)i.

I knew that it will have 2 and -2 as hi solution but didn't expect that Imaginary Solutions will also arrive.

Thats so easy

idiot olympiad fr

Great energy, great explanation

Great video

(X*X*X)/(2*2*2)=(2*2*2)/(X*X*X) X= ±2 X= ±1±Sqrt[3]i It’s in my head.

That is not x^3, but another mistake made it correct 😅

Learn to work with quantities in the complex plane and easily and rapidly obtain complex solutions. Any complex number z=a+b*i may be written as z=|z|e^{iθ} where |z|=√(a^2+b^2) and θ=tan^{-1}(b/a), where θ must be adjusted to fall into the correct quadrant. If x^n=a+b*i=|z|e^{iθ}, it may also be expressed as x^n=|z|e^{iθ+i2jπ}, where j is any integer. Taking the nth root of this expression gives x_j=|z|^{1/n}*e^{iθ/n+i2jπ/n}=|z|^{1/n}[cos(θ/n+2jπ/n)+i*sin(θ/n+2jπ/n)],j=0,1,...,n-1. These n roots lie equally spaced on a radius |z|^{1/n} circle centered on the origin of the complex plane.

Can you make me understand please

But I don't understand this math

it is possible to generalize? 1^x = y x = ln(y)/2npii

1

This happens whenever you use a branch of the logarithm such that ln(1) isn't zero, but some pure imaginary non zero. Of course, this equation is impossible over reals and any brach that "agrees" with the reals such that ln(1) =0.

Amazing video 👏

Amazing! Transcendental and complex numbers together make math magic.

Amazing!