An Interesting Olympiad Challenge | Can You Solve This?
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- Опубліковано 18 вер 2024
- An Interesting Olympiad 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.
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Have a go at the problem yourself before watching the solution!
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Respected Sir, Good evening
We are given that (x^6+1)/x^3 = 52. So, x is positive and x^3+1/x^3=52. Now x^3+1/x^3 = (x^3/2 + x^-3/2)^2 -2 > x^3/2 + x^-3/2 = [52 + 2}^1/2 = 3√6. So, [x^3/2 + x^-3/2 ]^3 = x^9/2 + x^-9/2 + 3 (x^3/2 + x^-3/2 ) > x^9/2 + x^-9/2 = (3√6)^3 - 9√6 = 153√6. We have to evaluate E = x^9/2/(1+x^9) = [x^9/2 + x^-9/2]^-1 = 1/153√6. So, E = 1/(153√6).
x=1/153sqrt6 or x= sqrt6/918
?= √1/(140454)= 1/(153√6)
Writing z for x^ (3/2) one gets
z ^2 + 1/z^2 = 52
(z + 1/z)^2 - 2 = 52
z + 1/z = √ (54)
Hereby
z^3 + 1/z^3
= ( z + 1/z)^3 - 3 z * (1/z)( z + 1/z)
= 54 ^ (3/2) - 3 √ (54)
= 51 * √ (54)
= 153 * √6
Hereby
z^3 /( 1 + z^6) = 1/( 153 √6)
x^9=t^6... t^2/(1+t^4)=1/52...t^2+1/t^2=52... t^3/(1+t^6)=?...t^3+1/t^3=1/?=(t+1/t)(t^2+1/t^2-1)=(t+1/t)*51... (t+1/t)^2-2=52...t+1/t=√54=3√6...?=1/51*3√6
{x^3+x^3 ➖ }/{1+1 ➖}{x^6+x^6 ➖}=x^6/{2+x^12}=x^6/2x^12=2x^6 2x^3^2 1x1^2 x^1^2 (x ➖ 2x+1).{x^9+x^9 ➖ }/{1+1 ➖ }+{x^9+x^9 ➖ }=x^18/{2+x^18}=x^18/2x^18=2x^1 (x ➖ 2x+1).
x^3/(1+x^6)=1/52 ->(1+x^6)/x^3=52 ->x^3+1/x^3=52
Let sqrt(x^9)/(1+x^9)=t ->x^9/(1+2*x^9+x^18)=t^2 ->x^9+1/x^9+2=1/t^2
(x^3+1/x^3)^3-3*x^3*(1/x^3)*(x^3+1/x^3)=1/t^2
52^3-3*52=1/t^2 -->t^2=1/140454 t=1/(153*sqrt(6)) discarding the negative for t is positive because x^3 is positive and so is 1+x^9
x^3=(52+sqrt(52^2-4))/2 or (52-sqrt(52^2-4))/2 both of which are positive.
問題に対してもっと簡素な解答になるようにしろよ。