A Nice Algebra Simplification | Math Olympiad | Give It a Try!
A Nice Algebra Simplification | Math Olympiad | Give It a Try!
Welcome to another exciting Math Olympiad challenge! In this video, we present a nice algebra simplification problem that's perfect for testing your skills. Whether you're preparing for a competition or just love math, this problem is sure to challenge and engage you.
Watch the video, try to simplify the expression, and see if you can find the correct solution. Don't forget to share your answers and thoughts in the comments below!
Problem: Simplify the given algebraic expression.
Difficulty Level: Intermediate to Advanced
Tools Needed: Just your brain and a piece of paper!
🔢 What You'll Learn:
Key strategies for simplifying complex expressions
Tips and tricks to approach difficult simplification problems
Step-by-step walkthrough of the solution
🧠 Challenge Yourself:
Pause the video, try to solve the problem on your own, and then watch as we break down the solution. Share your approach and answers in the comments below!
👍 Don't Forget to:
1. Like the video if you found it helpful.
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4. Join us and enhance your problem-solving skills. Can you master this simplification problem?
Time-stamps:
00:00 Introduction
00:29 Method-1
01:05 Exponent rules
02:26 Binomial expansion
05:16 Algebraic manipulations
08:15 Evaluating expression
09:15 Answer
09:20 Method-2
Additional Resources:
• Nice Algebra Challenge...
• A Nice Simplification ...
• A Nice Algebra Problem...
• Chinese | Math Olympia...
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Пікірлер: 10
Let t=2^1/6. Then, x=1+t+t^2+t^3 +t^4 +t^5 = [t^6-1]/(t-1) = 1/(t-1). So x+1 = t/(t-1). Now, 1/x+2/x^2+1/x^3 = 1/x^3(x+1)^2 = [t^2(t-1)^3]/(t-1)^2 = t^2(t-1) = t^3-t^2 = 2^(1/2) - 2^(1/3).
Isn't x the sum of a finite geometric series?
We have, (1/x) + (2/x²) + (1/x³) = (x² + 2x + 1) /x³ = (x + 1)² /x³ x = 1 / { (2)⅙ - 1 } x + 1 = [ 1 / { (2)⅙ - 1 } ] + [ { (2)⅙ - 1 } / { (2)⅙ - 1 } ] = (2)⅙ / { (2)⅙ - 1 } And, 1 /x = (2)⅙ - 1 Therefore, (x + 1)² /x³ = [ (2)⅙ / { (2)⅙ - 1 } ]² { (2)⅙ - 1 }³ = (2)⅓ { (2)⅙ - 1 } = (2)⅓(2)⅙ - (2)⅓ = (2)½ - (2)⅓ = √2 - (2)⅓
Evaluating x using Method 2 is more direct. From 10:51, let a = 2^(1/6) for simplicity and let u = 1/x = a - 1. (Don't bother to simplify x by rationalizing the denominator.). Then, note that the desired expression can be written as: u(1 + u)^2 = (a - 1)(a^2) = [2^(1/6) - 1][2^(2/6)] = 2^(3/6) - 2^(2/6) = sqrt2 - 2^(1/3)
let a=2^(1/6), now you have obtained (1/x)=(a-1) ==>(1/x) +(2/x^2)+(1/x^3) = a^3-a^2 = 2^(1/2)-2^(1/3). You seemed to have taken an unnecessary detour once (1/x) was determined
(2^(1/3)/2^(1/6) -1)
E=√2-³√2
2^(1/2)-2^(1/3). χ=(1-α^6)/(1-α) οπου α=2^(1/6) Ε=(1/χ)(1+1/χ)^2
{2x^2+24x^2}=26x^4 48x^2 {26x^4+48x^2}=74x^6 96x^2 {74x^6+96x^2}= 170x8 124x^2 {170x^8+124x^2}= 294x^10 10^20 2^37 x^2^5 2^55^4 2^37^1x^2^5 1^1^11^2^2^2^1^1x^2^1 1^1^1^1x^2^1 x^2^1 (x ➖ 2x+1) 1x+1x ➖/x+x ➖ +2x+2x ➖/x^2+x ^2➖ =4x+4x ➖/x^2+x^2 ➖ 2x^2/x^2+4x^2/x^4+8x^2/x^4 14x^10/x^10.=:1.4x^1 1^1.2^2x1^1 1^2x^1^1 2x^1 (x ➖ 2x+1)