Nice Algebra Challenge from Olympiads | Can You Crack It?
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- Опубліковано 30 вер 2024
- Nice Algebra Challenge from Olympiads | Can You Crack It?
Think you have what it takes to solve a Math Olympiad challenge? In "Nice Algebra Challenge from Olympiads | Can You Crack It?" we delve into essential algebraic techniques that will help you simplify complex problems with ease. This video is perfect for those looking to sharpen their algebra skills and prepare for competitive exams like the Math Olympiad.
Watch as we break down the problem step-by-step, offering valuable insights and techniques to help you crack similar challenges. Whether you're a seasoned competitor or just a math enthusiast, this video will push your problem-solving abilities to new heights.
Don't forget to like, share, and subscribe for more math challenges and tutorials. Are you ready to take on the challenge?
🔢 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!
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Additional Resources:
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#matholympiad #algebrachallenge #mathsimplification #mathpuzzles #problemsolving #mathtips #mathtutorial #algebra
Thanks for Watching !
x + 1/x = a
x² = ax - 1
x³ = ax² - x = a(ax - 1) - x
x³ = (a² - 1)x - a
x⁴ = (a² - 1)x² - ax = (a² - 1)(ax - 1) - ax
x⁴ = (a³ - 2a)x - (a² - 1)
x⁵ = (a³ - 2a)x² - (a² - 1)x
x⁵ = (a³ - 2a)(ax - 1) - (a² - 1)x
x⁵ = (a⁴ - 3a² + 1)x - (a³ - 2a)
x⁶ = .....
Do you have a video on defining if bigger numbers (say 3-digit numbers are prime)? should I check all the known rules as for if it divisable by all simple numbers smaller than a squzre root of that number? Of course don't mean using internet or any apps for that reason, I mean if it's an exam or an olympiad, where you can't use computers for that
x⁷ + 1/x⁷ = 843
find
x³ + 1/x³ + x⁵ + 1/x⁵
x + 1/x = a
x² + 1/x² = a² - 2
(x² + 1/x²)(x + 1/x) = x³ + 1/x³ + x + 1/x
x³ + 1/x³ = a(a² - 2) - a
x³ + 1/x³ = a³ - 3a
(x³ + 1/x³)(x + 1/x) = x⁴ + 1/x⁴ + x² + 1/x²
x⁴ + 1/x⁴ = a(a³ - 3a) - (a² - 2)
x⁴ + 1/x⁴ = a⁴ - 4a² + 2
(x⁴ + 1/x⁴)(x + 1/x) = x⁵ + 1/x⁵ + x³ + 1/x³
(x⁴ + 1/x⁴)(x³ + 1/x³) = x⁷ + 1/x⁷ + x + 1/x
..... I don't know how to continue..
141
Let a=x+1/x and calcualte x^7+1/x^7. We find a^7-7a^5+14a^3-7a and this is equal to 843. Reolve that for a. We get a=3. Since x^3+1/x^3=a^3-3a, it is equal to 18 and x^5+1/x^5 =123, then x^3+x^5+1/x^3+1/x^5=18+123=141.
Let x+1/x=t. Then, x^3+1/x^3 = t^3-3t and x^5+1/x^5 = t^5-5t^3+5t and x^7+1/x^7 = t^7-7t^5+14t^3-7t = 343. t=3 solves this. Thus, the expression we have to evaluate, x^5+1/x^5 + x^3+1/x^3 = t^5-4t^3+2t = 141.
Thank you Sir 🙏 ...for sharing
141