China | Math Olympiad | Nice Algebra Radical Problem 

Nice solution and explanation. There is another way to solve the problem after finding out that: 2x - 1 = √3 and therefore by squaring: 2x^2 = 2x + 1 By multiplying by x we obtain: 2x^3 = 2x^2 + x = 3x + 1 By taking to the third power we obtain: 8x^9 = 27x^3 + 1 + 9x (3x + 1) = 27x^3 + 27x^2 + 9x + 1 Therefore: 16x^9 = 27(2x^3 ) + 27(2x^2 ) + 9(2x) + 2 = 27(3x + 1) + 27(2x + 1) + 18x + 2 And therefore: 16x^9 = 153x + 56 The rest is known.

16*4x^2 is not equal to 16(4x+2).

Very nice math solution, but I genuinely thought this was recorded in 2006. You should consider using better equipment. It really makes a difference.

((sqrt,(3)+1)³)³ (a+b)³=a³+3a²b+3ab²+b³ twice. without x. Simple.

The solution steps do not need this complexity , Which has no need Answer the question: no more than 4 steps,, and with ease

Gracias a estos ejercicios los alumnos se alejan de las matemáticas. Ejercicios que no tienen un objetivo, ni aplicación práctica. Los jóvenes de hoy en día aprenden diferente a los jóvenes que fuimos nosotros. El chico de hoy agarra la calculadora y resuelve en 2 minutos

La verdad me confunde

The idea is very deep, but if ((((SQRT 3 +1)/2) ^2)^2)^2 and Result multiply by (SQRT 3 +1)/2) the result will be more faster, for my opinion.

Compared to direct calculation, your way takes triple efforts at least.

Very nice. 2 things I didn't understand: root 2 was replaced with 1, and 4x² was rewritten as (4x + 2).

ليست هذه رياضيات

C'est quoi ça?

(Sqrt(6)+Sqrt(2)/Sqrt(8))^9=8.28125+4.78125Sqrt(3)

Many thanks; good health to you.

Aplicando el triángulo de Pascal

Простое решение. Все упрощаем и вычисляем методом устного счета. 1. Сокращаем все выражение в скобках на √2. 2. Складываем числитель √ 3+1 округляем, получаем 2,8. 3. Делим числитель на знаменатель 2,8 : 4 = 0,7. 4. Возводим в 9 степень 0,7 х 0,7= 0,49 округляем до 0,5. 0,5 х 0,5 х 0,5 х 0,5= 0,0625 округляем до 0,06 0,7 х 0,06=0,042 Если нужна большая точность, используем калькулятор.

Newton's binomial is much easier and direct

Շատ բարդ է լուծած , կարելի արմատ 2-ով կրճատելուց հետո, 2անգամ խորանարդ բարձրացնել։

You solve that mathematic become complex

It's interesting, but maybe the direct calculation was even simpler. But anyway, nice exercise!

= { (√6 + √2) / √8 }^(9) = { [(√6 + √2).√8] / 8 }^(9) = { (√48 + √16) / 8 }^(9) = { (4√3 + 4) / 8 }^(9) = { (1 + √3) / 2 }^(9) = { (1 + √3) }^(9) / 2^(9) First = (1 + √3)^(9) = (1 + √3)².(1 + √3)².(1 + √3)².(1 + √3)².(1 + √3) = (1 + 2√3 + 3).(1 + 2√3 + 3).(1 + 2√3 + 3).(1 + 2√3 + 3).(1 + √3) = (4 + 2√3).(4 + 2√3).(4 + 2√3).(4 + 2√3).(1 + √3) = 16.(2 + √3).(2 + √3).(2 + √3).(2 + √3).(1 + √3) = 16.(4 + 4√3 + 3).(4 + 4√3 + 3).(1 + √3) = 16.(7 + 4√3).(7 + 4√3).(1 + √3) = 16.(49 + 56√3 + 48).(1 + √3) = 16.(97 + 56√3).(1 + √3) = 16.(97 + 97√3 + 56√3 + 168) = 16.(265 + 153√3) = (265 + 153√3) * 2^(4) Recall = { (1 + √3) }^(9) / 2^(9) = (265 + 153√3) * 2^(4) / 2^(9) = (265 + 153√3) / 2^(5) = (265 + 153√3) / 32 ≈ 16.5626

Показатель в выражении -- это буква q или цифра 9? Научитесь, пожалуйста, грамотно писать.

х=(√3+1)/2 х^2=(√3+2)/2 х^4=(4√3+7)/4 х^8=(56√3+97)/16 х^9=(153√2+265)/32

I did not understand anything !