Japan | A nice Math Olympiad Algebra Question | Ratio Simplification Problem | Can you solve this ?

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Japan | A nice Math Olympiad Algebra Question | Ratio Simplification Problem | Can you solve this ?
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Комментарии : 15

@walterwen2975 8 дней назад

Ratio Simplification Problem: a⁴ + b⁴ = 10a²b²; (a + b)/(a - b) = ? (a² + b²)² = a⁴ + b⁴ + 2a²b² = 10a²b² + 2a²b² = 12a²b², a² + b² = ± (2√3)ab (a + b)² = a² + b² + 2ab = 2ab ± (2√3)ab = 2ab(1 ± √3) (a² - b²)² = a⁴ + b⁴ - 2a²b² = 10a²b² - 2a²b² = 8a²b², a² - b² = ± (2√2)ab (a + b)/(a - b) = [(a + b)²]/(a² - b²) = [2ab(1 ± √3)]/[± (2√2)ab] = ± (1 ± √3)/√2 = ± [(√2)(1 ± √3)]/(√2)² = ± (√2 ± √6)/2 Final answer: (a + b)/(a - b) = (√2 + √6)/2; (a + b)/(a - b) = (2 - √6)/2; (a + b)/(a - b) = (- √2 - √6)/2 or (a + b)/(a - b) = (- 2 + √6)/2

@johnstanley5692 9 дней назад

Alternative a^4+b^4=10*a^2*b^2 => (a/b)^2+(b/a)^2 = 10 = (a/b + b/a)^2-2 => a/b + b/a = +/- 12^(1/2) Hence a/b = 2^(1/2)+3^(1/2) or a/b= 2^(1/2)-3^(1/2). Now (a + b)/(a - b) = ((a/b) + 1)/((a/b) - 1) -> (2^(1/2) +/- 6^(1/2))/2

@superacademy247 9 дней назад

Awesome 💯 method

@nalapurraghavendrarao6324 8 дней назад

When you know the value of a/b,use componando and dividendo to get (a+b)/(a-b)

@vacuumcarexpo 8 дней назад

√(a^2)=｜a｜ and √(b^2)=｜b｜ Why did you consider the case such that a

@ettienkoffi5781 9 дней назад

too much details

@luismauro937 6 дней назад

I also think.

@sy8146 8 дней назад

Thank you for explaining. There is a special case that is (a, b)=(0, 0). [ It satisfies (a^4)+(b^4)=10(a^2)(b^2) . ] Therefore, if (a, b)=(0, 0), (a+b)/(a-b) cannot be defined. [ And, if (a, b)≠(0, 0), (a+b)/(a-b) can be calculated by the way of the video or of other comments. ]

@superacademy247 8 дней назад

You're welcome 😊

@BRUBRUETNONO 7 дней назад

Hi, Thanks for your interesting problem. Here is the way I solved it. Of course I didn't look at your solution. Tell me if you like my solution. Greetings and keep up the good work. Recall. If a^4+b^4=10a^2.b^2 then calculat (a+b)/(a-b)=... Begin Let u=(a+b)/(a-b) Then u is not defined for a=b (so also for a=b=0) Then for the value u, a0 and b0 and ab Let's assume that a,b € R Let's make the calculations below First we can write, a^4+b^4=(a^2+b^2)^2-2a^2.b^2=10a^2.b^2 => (a^2+b^2)^2-2a^2.b^2=10a^2.b^2 (i) => (a^2+b^2)^2=12a^2.b^2 => (a^2+b^2)=+sqrt(12)a.b or (a^2+b^2)=-sqrt(12)a.b => (a+b)^2-2ab=+sqrt(12)a.b or (a+b)^2-2ab=-sqrt(12)a.b => (a+b)^2=(2+sqrt(12))a.b or (a+b)^2=(2-sqrt(12))a.b As a,b € R and a > b and sqrt(12)>2 then Case (i.i) (a+b)^2=(2+sqrt(12))a.b on the condition that a and b has got the same sign or Case (i.ii) (a+b)^2=(2-sqrt(12))a.b on the condition that a and b has got the opposite sign Second we can write a^4+b^4=(a^2-b^2)^2+2a^2.b^2=10a^2.b^2 => (a^2-b^2)^2+2a^2.b^2=10a^2.b^2 (ii) => (a^2-b^2)^2=8a^2.b^2 => (a^2-b^2)=+sqrt(8)a.b or (a^2-b^2)=-sqrt(8)a.b => (a+b)(a-b)=+sqrt(8)a.b or (a+b)(a-b)=-sqrt(8)a.b By assuming that a > b As a,b € R then Case (ii.i) (a+b)(a-b)=+sqrt(8)a.b on the condition that a and b has got the same sign Case (ii.ii) (a+b)(a-b)=-sqrt(8)a.b on the condition that a and b has got the opposite sign Still assuming a > b, if we remark that (a+b)^2/[(a+b)(a-b)]=(a+b)/(a-b)=u then we can calculate the requested value on the considered two sub condition cases. Condition i: a and b has got the same sign (and a > b) then u=(a+b)/(a-b)=[(2+sqrt(12))a.b]/[+sqrt(8)a.b]=(2+sqrt(12))/sqrt(8) u=(a+b)/(a-b)=(2+2sqrt(3))/2sqrt(2)=(1+sqrt(3))/sqrt(2)=(sqrt(2)+sqrt(3))/2 Or Condition ii: a and b has got the opposite sign (and a > b) then u=(a+b)/(a-b)=[(2-sqrt(12))a.b]/[-sqrt(8)a.b]=(-2+sqrt(12))/sqrt(8) u=(a+b)/(a-b)=(-2+2sqrt(3))/2sqrt(2)=(-1+sqrt(3))/sqrt(2)=(-sqrt(2)+sqrt(3))/2 Newly assuming that b > a As a,b € R then Case (ii.iii) (a+b)(a-b)=+sqrt(8)a.b on the condition that a and b has got the opposite sign Case (ii.iv) (a+b)(a-b)=-sqrt(8)a.b on the condition that a and b has got the same sign Newly assuming b > a, if we remark that (a+b)^2/[(a+b)(a-b)]=(a+b)/(a-b)=u then we can calculate the requested value on the considered two sub condition cases. Condition iii: a and b has got the same sign (and b > a) then u=(a+b)/(a-b)=[(2+sqrt(12))a.b]/[-sqrt(8)a.b]=-(2+sqrt(12))/sqrt(8) u=(a+b)/(a-b)=-(2+2sqrt(3))/2sqrt(2)=-(1+sqrt(3))/sqrt(2)=-(sqrt(2)+sqrt(3))/2 Or Condition iv: a and b has got the opposite sign (and b > a) then u=(a+b)/(a-b)=[(2-sqrt(12))a.b]/[+sqrt(8)a.b]=(2-sqrt(12))/sqrt(8) u=(a+b)/(a-b)=(2-2sqrt(3))/2sqrt(2)=(1-sqrt(3))/sqrt(2)=(sqrt(2)-sqrt(3))/2 To conclude, on the conditions that a,b are real numbers condition i : a>b and a and b has got the same sign => u=(a+b)/(a-b)=(+sqrt(2)+sqrt(3))/2 condition ii : a>b and a and b has got opposite sign => u=(a+b)/(a-b)=(-sqrt(2)+sqrt(3))/2 condition iii: a u=(a+b)/(a-b)=(-sqrt(2)-sqrt(3))/2 condition iv : a u=(a+b)/(a-b)=(+sqrt(2)-sqrt(3))/2 END

@superacademy247 7 дней назад

Thanks 👍 💯 😊 for your input and support

@davidshen5916 4 дня назад

Let K=（A+B）/（A-B）， A/B=（K+1）/（K-1），（K+1）^4+（K-1）^4=10（K+1）^2*（K-1）^2， 2（K^4+6K^2+1）=10（K^4-2K^2+1）， 8K^4-32K^2+8=0， K^4-4K^2+1=0，

@user-ee7nw2rx9s 8 дней назад

Решение не верно. а=b=0, эта пара чисел удовлетворяет условиям. Можете подставить в выражение и получить 0=0 А теперь попробуйте найти значение искомого выражения.

@superacademy247 8 дней назад

The solutions are correct. Go and find out...do more research.

@user-ee7nw2rx9s 8 дней назад

@@superacademy247 Числа а=b=0 удовлетворяют условиям., а значение искомого выражения 0/0. Объясните почему такое возможно.