MIND-BLOWING Radical Math Challenge | Can You Solve This?

MIND-BLOWING Radical Math Challenge | Can You Solve This?

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In this video, we explore an intriguing algebra problem involving radicals, perfect for those preparing for Math Olympiad. This question will enhance the understanding of radical expressions and skills of problem-solving. Watch as we break down the solution step-by-step, providing clear explanations and insights along the way.

If you're a Math Olympiad participant or simply enjoy tackling competitive math problems, this video is for you. Make sure to like, subscribe, and hit the notification bell to stay updated with more exciting math challenges. Let's solve this radical algebra problem together!

🔍 In this video:

Detailed walkthrough of a challenging algebra problem.
Tips and tricks for solving complex radical expression.
Encouragement to enhance your problem-solving skills and mathematical thinking.

📣 Call to Action:

Have a go at the problem yourself before watching the solution!
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[ManojkantSamal]

(506-201.*21)/6250 (may be ), I answered as per 5×(x^)6
X=(*21-1)/10
*=read as square root
Explain later

[АндрейПергаев-зн]

Зачем лишние действия
Когда получили
5х=(sqrt 21-1)/2
Проще всего три раза возвести в квадрат и все решение

[gregevgeni]

Nominator :
11 + √11 + √21 + √231 = (√11)^2 + √11 + √21 + √21 • √11 = (√11 + √21)(1 + √11) (*)
because 231 =21 • 11 .
Denominator :
1 + √11 + √21 +√11 = (1 +√11) + (√11 +√21) (**).
Put w = 1 + √11 and u = √11 +√21 .
Then the (*) and (**) written
Nominator = w • u
Denominator = w + u => the given contition written 1/x = (w • u) / (w + u) => x = (w + u)/(w •u) = 1/w + 1/u (***).
But 1/w = 1/(1+√11) = (√11 -1)/10 and
1/u = 1/(√21 +√11) = (√21 - √11)/10 and the (***) : x = (√11 - 1 + √21 - √11)/10 =>
x = (√21 - 1)/10 => (5x) = (√21 - 1)/2
We want to found this
(5x)^6 = (1/2)^6 • (√21 - 1)^6 = ?
Let √21 - 1 = t => √21 = t + 1 =>
21 = t^2 + 2t +1 => t^2 = 20 - 2t =>( •t), t^3 = 20t - 2t^2 = 20t - 2(20- 2t) = 16t - 40.
t^6 =(t^3)^2 = (16 -4t)^2 =
256 - 128t + 16t^2 =
256 - 128t + 16(20 - 2t) =
256 - 128t + 320 - 32t = 576 - 160t=
576 - 160(√21 -1) = 736 - 160√21.
So (5x)^6 = (1/2)^2 • (736 - 160√21).

[RealQinnMalloryu]

{1x+1x ➖ }/{x+x+}=2x^2/x^2=2x (x ➖ 2x+2){ 11x+11x ➖ }+{11x+11x ➖ }+{21x+21x ➖ }+{231x+231x ➖ }/{1x+1x ➖ }+{11x+11x ➖ }+{21x +21x ➖ }+{11x+11x ➖ }= 462x^2}=528x^8 /{2x^2+22x^2}={24x^4 22x^2}{46x^6+42x^2}={108x^8 +22x^2}= 132x^10 528x^8/132x^10 =44x^2^1 2^22x^2^1 2^2^11x^2^1 1^1^11^1x^2^1 1^1x^2^1 x^2^1 (x ➖ 2x+1) 30x^6 3^10x^6 3^2^5x^3^2 1^1^1x^3^2 x^3^2 (x ➖ 3x+2).