Japanese Math Olympiad Question | You should know this trick!

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[@rdeviltron]

I love your videos so much and I've subscribed since you had 70.000 subscribers. Keep the great work going to get your 100k subs.

[@Zarunias]

With these small numbers I actually was way faster by just doing the calculations with the actual numbers.

[@guitartommo2794]

9:29 kiddy giggling in the background😊

[@Evgeniy_Akhmetshin]

Здесь получается комплексное число с мнимой единицой, иначе ни квадратный корень, ни квадрический из отрицаиельного числа не вытащить.

[@MrHaiopei]

Vielen Dank, hat Spaß gemacht

[@learncommunolizer]

Thank you very much

[@walterwen2975]

Japanese Math Olympiad Question: √[(100² + 81² + 19⁴)/2] = ? 100² + 81² + 19⁴ = 10⁴ + 9⁴ + 19⁴; Let: a = 10, b = 9 100² + 81² + 19⁴ = a⁴ + b⁴ + (a + b)⁴, a⁴ + b⁴ = (a² + b²)² - 2a²b² (a + b)⁴ = [(a + b)²]² = (a² + b² + 2ab)² = (a² + b²)² + 4ab(a² + b²) + 4a²b² a⁴ + b⁴ + (a + b)⁴ = 2(a² + b²)² + 4ab(a² + b²) + 2a²b² = 2[(a² + b²)² + 2ab(a² + b²) + a²b²] = 2(a² + b² + ab)² √[(100² + 81² + 19⁴)/2] = √{[2(a² + b² + ab)²]/2} = a² + b² + ab = 10² + 9² +(10)(9) = 100 + 81 + 90 = 271

[@ПашаПитецкий]

Вумничка 😊❤

[@lthakur92411]

I think this is very long method,

[@angelomarella5983]

Really too long....

[@angelomarella5983]

Solution: sr 100^2/2 + sr 81^2/2+ sr 19^4/2= 10/sr 2+ 9/sr2 + 19^2/ sr2= 100/2 + 81/2 + 19^2/2= (100+81+361)/2= 271 sr= square root Bye bye