Видео

X=1 is one of the roots.

(x^4 )^2 ➖ (16)^2{x^16 ➖ 256}=240 2^120 2^60 2^30 2^15 2^3^5 2^3^5^1 2^3^1^1 2^3:(x ➖ 3x+2)( x^2)^2 ➖ (4)^2={x^4 ➖ 16}=123^4 3^2^2 1^1^2 1^2 (x ➖ 2x+1) .

Well explained man. Thanks for this detailed work, sir.

Well explained man. Thanks for this detailed work, sir.

❤

Factor to m(x - √3)(x + √3) = 0 then set each factor equal to 0. m = 0 m = +√3 m = -√3 simple

The original looks like m decimal m decimal m on the right side. In that case any non-negative integer will suffice for m.

{x^8+x^8 ➖ }+{z^3+z^3 ➖ }={x^16+z^6}=xz^22 xz^16^6 4^4^3^2 xz^2^2^2^2^3^2 xz^1^1^1^1^3^2 x^3^2(xz ➖ 3xz+2) {y^3+y^3 ➖ }+{x^3+x^3 ➖ }={y^6+x^6}=yx^12 yx^6^6yx^2^3^2^3 yx^2^1^1^3 yx^2^3(yx ➖ 3yx+2) {z^3+z3 ➖ }+{y^3+y^3 ➖ }={z^6+y^6}=zy^12 zy^6^6 zy^2^3^2^3 zy^2^1^1^3 zy^2^3 (zy ➖ 3zy+2).

(x)^3 ➖ (1)^3={x^3 ➖ 1}={x^0+x^0 ➖}=x^1 (x ➖ 1x+1).x^3(x^3)^2 ➖ (3x^2)^2=x^3{x9 ➖ 9x^4}=(x^3 )^2 ➖ 9x^5={x^9 ➖ 9x^5}=9x^4 3^2x^2^2 3^1x^1^2 3x^2 (x ➖ 3x+2)

First to comment...😂😂😂

(2)+(2)=4 (y ➖ 2x+2) 3^2/2 1^1/2 1/2 {xy➖ 2xy+1).

X=887, I just solved it in my head in a few seconds 😅😅😅

👍👍👏👏🙏

X=5, thanks

x^(x)^2 ➖ (2)^2=x^{x^2 ➖ 4}=x^{0+x^0 ➖ }=x^1 (x ➖ 1x+1) 2^22^2 1^1^1^2 1^2 (x ➖ 2x+1).

{x+x+x x ➖ x ➖ x}=x^3 +(x^3 )^2➖( x)^^2 ➖ 16=x^3+ {x^9 ➖ x^2} ➖ 16 =x^3+x^7 ➖ (16)^2=x^3+{x^7 ➖ 256}={x^3+249}=249x^3 10^20^49x^3 10^20^7^7x^3 10^2^10^1^1x^3 2^5^2^2^5x^3 1^1^1^2^1x^3 2x^3(x ➖ 3x+2).(x^3 )^2➖ (x )^2➖ 8={x^9 ➖ x^2 }➖ 8=x^7 ➖ (8)^2={x^7 ➖ 64}= 57 3^19 3^19^1 3^1^1 3^1 (x ➖ 3x+1).

❤🎉

❤❤❤❤❤❤

✅

Thank u

Good one man.

I don't think this works. (2i)^8=256, and the 4th root of 256 is 4. I mean, if you wanted the roots of x^4=256, you'd have {4, -4, 4i, -4i}, but the radical indicates the principal root, which is 4.

Uncle J.J

Solution overcomplicated! √(7-2√40) =√(7-2√(5·2)) =√(5-2√(5·2)+2)=√((√5-√2)^2 )=√5-√2 or √2-√5

Use the shortcut. √(7 - 2√10), 5 + 2 = 7 and 5 x 2 = 10. Automatic: √5 - √2

1

❤

Good 🎉🎉🎉

or.. just do sqrt of 16 which is 4 and sqrt of 4 which is 2...

you re missing a square on your cosine

how is sqrt(cos^4(x)) = |cosx|

🥱🥱🥱🤦🏻‍♂️🤦🏻‍♂️And now get the all solutions to x^1024=1... Good luck. LAAAAME

This is a very nice question solved in a clear and unique way.

I am in 7th grade about to do algebra 2 any tips?

2 real solutions, because -1/e < -ln4/5 < 0 1.5271834618689137843439681976847183206615191189475680905718280703... or 7.0378231590674309036420394728342539963399273768491109730261293405...

No real solutions

Kudos man...😂😂😢

{4x+4x ➖ }{9x+9x ➖ }={8x^2+18x^2}=26x^4 2^13x^4 2^13^1x^4 2^1^1x^2^2 1x1^2 x^1^2 (x ➖ 2x+1). {6x+6x ➖ }=12x^2 3^4x^2 3^2^2x^2 3^1^1x^2 3x^2 (x ➖ 3x+2).

This beautiful man. Keep running this channel sir and the sky is yours.....❤

👍👍

Cool one 😅😅

Cheap proof.

I pove this in my head....😂😂😂

Nice work

Nice proof man.

+1 or -1

Where do you get x_1 = sqrt(2+sqrt(2))/2 + (sqrt(2-sqrt(2))/2 * i from? Same with the other x_(odd number). Not explained. Hint: use the half-angle formula.