Brazilian Math Olympiad | A Nice Algebra Challenge

Brazilian Math Olympiad | A Nice Algebra Challenge
Welcome to another exciting math challenge! In this video, we'll tackle a fascinating algebra problem from the Brazilian Math Olympiad. If you love solving complex equations and pushing your problem-solving skills to the limit, this is the perfect challenge for you. Join me as we work through this intricate algebra problem step-by-step. Don't forget to like, comment, and subscribe for more math challenges and solutions!
🔍 In this video:
Detailed walkthrough of a challenging algebra problem from the Brazilian Math Olympiad.
Tips and tricks for solving complex algebraic equations.
Encouragement to enhance your problem-solving skills and mathematical thinking.
📣 Call to Action:
Have a go at the problem yourself before watching the solution!
Share your solutions and approaches in the comments below.
If you enjoyed this challenge, give it a thumbs up and subscribe for more intriguing math problems!
Time-stamps:
00:00 Introduction
01:30 Factorization
03:32 Rationalization
07:02 Algebraic identities
10:44 Evaluating expression
11:20 Answer
🔗 Useful Links:
• Thailand Math Olympiad...
• A Nice Algebra Simplif...
• A Nice Algebra Problem...
• A Nice Simplification ...
#matholympiad #brazil #algebra #math #algebratricks #education #problemsolving #mathematics #expression #simplification
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Thank You for Watching!!

[@maxwellarregui814]

Sres. Gracias es un buen ejercicio, comprobada su respuesta por otro procedimiento, hice con ( a + b ) ^ 5, mucho más largo pero llegué al mismo resultado; 38804 * 6 ^ (1/2). Éxitos.

[@pietergeerkens6324]

Chase out the perfect squares!!! That's pretty fundamental for a final answer, giving 38804 * sqrt(6).

[@user-kp2rd5qv8g]

Note that x= 5+2sqrt(6) and y = 5-2sqrt(6). Again, x^5-y^5 = (x-y)[(x+y)^4-3xy(x+y)^2 +x^2y^2]. Here, x-y = 4sqrt(6), x+y = 10, xy=1. So, x^5-y^5 = 4sqrt(6)[10,000-300+1] = 38804sqrt(6).

[@alighsh6075]

xy = 1

[@ald6980]

Let's use some naive theory and simplify. x and y are roots of t^2-10t+1=0; a(n) =( x^n - y^n)/sqrt(6). It is widely known [and can be easily proved or checked] that a(n) follows the recurrent equation a(n) = 10a(n-1)-a(n-2). a(0) = 0, a(1) = 4, a(2)=4*10-0=40; a(3)=40*10-4=396; a(4)=396*10-40 = 3920; a(5) = 3920*10-396 = 38804. x^5-y^5 = sqrt(6)a(5) = 38804*sqrt(6).

[@walterwen2975]

Brazilian Math Olympiad: x = (√6 + √2 + √3 + 2)/(√6 - √2 + √3 - 2), y = (√6 + √2 - √3 - 2)/(√6 - √2 - √3 + 2); x⁵ - y⁵ = ? x = (√6 + √2 + √3 + 2)/(√6 - √2 + √3 - 2) √6 + √2 + √3 + 2 = √3(√2 + 1) + √2(√2 + 1) = (√2 + 1)(√3 + √2) √6 - √2 + √3 - 2) = √3(√2 + 1) - √2(√2 + 1) = (√2 + 1)(√3 - √2) x = (√3 + √2)/(√3 - √2) = (√3 + √2)²/(3 - 2) = 3 + 2 + 2√6 = 5 + 2√6 y = (√6 + √2 - √3 - 2)/(√6 - √2 - √3 + 2) √6 + √2 - √3 - 2 = √3(√2 - 1) - √2(√2 - 1) = (√2 - 1)(√3 - √2) √6 - √2 - √3 + 2 = √3(√2 - 1) + √2(√2 - 1) = (√2 - 1)(√3 + √2) y = (√3 - √2)/(√3 + √2) = (√3 - √2)²/(3 - 2) = 3 + 2 - 2√6 = 5 - 2√6 Let: a = 5, b = 2√6; x⁵ - y⁵ = (a + b)⁵ - (a - b)⁵ (a ± b)⁵ = a⁵ ± 5a⁴b + 10a³b² ± 10a²b³ + 5ab⁴ ± b⁵ (a + b)⁵ + (a - b)⁵ = 2b(5a⁴ + 10a²b² + b⁴) = 2b[5(a⁴ + 2a²b² + b⁴) - 4b⁴] = 2b[5(a² + b²)² - 4b⁴] = 2(2√6)[5(25 + 24)² - 4(2√6)⁴] = (4√6)(12005 - 2304) = 9701(4√6) = 38804√6; x⁵ - y⁵ = (a + b)⁵ - (a - b)⁵ = 38804√6

[@RealQinnMalloryu4]

3^2+1+1+1/3^2 ➖ 1+1 ➖ 1 1^1/3^2 3^2 (x ➖ 3x+2) 3^2+1+1+1/3^2 ➖ 1 ➖ 1+1 1^1/3^2 3^2 (y ➖ 3y+2)

[@RealQinnMalloryu4]

x^1y^1 (y ➖ 1x+1)

[@RealQinnMalloryu4]

(x^5)^2=x^25 (y^5)^2=y^25,{x^25 ➖ y^25}=xy^0,{xy^0+xy^0 ➖}= xy^1 (xy ➖ 1xy+1)

[@vacuumcarexpo]

Why didn't you simplify √96?

[@kassuskassus6263]

You can do it and get 38804sqrt6.

[@gnanadesikansenthilnathan6750]

I got the answer.

[@user-ny6jf9is3t]

95050

[@user-ny6jf9is3t]

Υπολογισα κατα λαθος το χ^5+y^5.

[@johnstanley5692]

alternative? x &y simplify to x= 5+2*sqrt(6), y= 5-2*sqrt(6), also x^5-y^5 = (x-y)*( (x^4+y^4) + x*y*(x^2+y^2) + (x*y)^2) ) x - y = 4* sqrt(6), , x + y = 10, x*y =1=> x^5-y^4 = 4* sqrt(6)*( (x^4+y^4) + (x^2+y^2) + 1 ) (x+y)^2 = 100 =(x^2+y^2 +2) => x^2+y^2 = 98, (x+y)^4 = ( (x^4+y^4) +98+1)=> x^4+y^4 = 9602=> x^5-y^5 = 38804*sqrt(6)