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An Intriguing Radical Problem | Algebra Challenge
Рет қаралды 1,131
Күн бұрынAn Intriguing Radical Problem | Algebra Challenge
Dive into this intriguing radical problem and challenge your algebra skills! In this video, we'll explore a fascinating radical equation that will test problem-solving abilities. Whether you're preparing for a math competition or just love solving algebra problems, this challenge is perfect for you. Watch the video, try to solve the problem, and let us know your solution in the comments below. Happy solving!
🔍 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!
🔗 Useful Links:
• Thailand Math Olympiad...
• A Nice Algebra Simplif...
• A Nice Algebra Problem...
• A Nice Simplification ...
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@abcekkdo3749 Ай бұрын
E=(√3-1)/16
@user-ny6jf9is3t Ай бұрын
Ε=(ριζα3 -1)/16
@tejpalsingh366 Ай бұрын
? = (√3-1)/ 16
@user-kp2rd5qv8g Ай бұрын
Let t = 3^1/8. Then, x = (t+1)(t^2+1)(t^4+1) = [(t^4-1)(t^4+1)]/(t-1) = [t^8-1)/(t-1) = 2/(t-1). So, 1/x = (t-1)/2. Thus, f(x)= 1/x^4+2/x^3+3/2x^2 +1/2x = 1/16[t^4-1] = 1/16(√3 -1).
@user-ee7nw2rx9s 6 күн бұрын
Искомое выражение Возьмем 1 и 3 слагаемое, а также 2 и 4 Тогда (1/х^2)*(1/х^2+3)+(1/х)*(2/х^2+1/2) Надо только полученное 1/х возвести в квадрат и никаких 4 степеней нет
@ABHISHEKKUMAR-01024 Ай бұрын
We have, x = (√3 + 1)(⁴√3 + 1)(⁸√3 + 1) = (²√3 + 1)(⁴√3 + 1)(⁸√3 + 1) = (3¹/² + 1)(3¹/⁴ + 1)(3¹/⁸ + 1) = {(3¹/⁸)⁴ + 1} {(3¹/⁸)² + 1} (3¹/⁸ + 1) = (a⁴ + 1) (a² + 1) (a + 1), where a = 3¹/⁸ = {(a⁴ + 1) (a² + 1) (a + 1) (a - 1) } / (a - 1) = { (a⁴ + 1) (a² + 1) (a² - 1) } / (a - 1) = [ (a⁴ + 1) { (a²)² - 1²} ] / (a - 1) = { (a⁴ + 1) (a⁴ - 1) } / (a - 1) = { (a⁴)² - 1²} / (a - 1) = (a⁸ - 1) / (a - 1) = { (3¹/⁸)⁸ - 1} / (a - 1) = (3 - 1) / (a - 1) = 2 / (a - 1) Therefore, x = 2 / (a - 1) or, x (a - 1) = 2 or, a - 1 = 2 / x or, (2 / x) + 1 = a or, {(2 / x) + 1}⁴ = a⁴ or, (2 / x)⁴ + 4 (2 / x)³ + 6 (2 / x)² + 4 (2 / x) + 1 = (3¹/⁸)⁴ or, (16 / x⁴) + 4 (8 / x³) + 6 (4 / x²) + 4 (2 / x) + 1 = 3¹/² or, (16 / x⁴) + (32 / x³) + (24 / x²) + (8 / x) + 1 = √3 or, (16 / x⁴) + (32 / x³) + (24 / x²) + (8 / x) = √3 - 1 or, 8 { (2 / x⁴) + (4 / x³) + (3 / x²) + (1 / x) } = √3 - 1 or, (2 / x⁴) + (4 / x³) + (3 / x²) + (1 / x) = (√3 - 1) / 8 or, 2 { (1 / x⁴) + (2 / x³) + (3 / (2x²) ) + (1 / (2x) ) } = (√3 - 1) / 8 or, (1 / x⁴) + (2 / x³) + (3 / (2x²) ) + (1 / (2x) ) = (1/2) [ (√3 - 1) / 8 ] = (√3 - 1) / 16
Super Academy
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Super Academy
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