Відео

Too many unnecessary intermediate steps with repeats. Dr. Ajit Thakur (USA).

13.928

x=(1/3) log(4) 20 x=0.7203

x=log(5) (20/3) x=0.5894

X=2+3e^(i 2n pi/6), n=0,1,2,3,4,5

Today is my birthday! Well done, Math Hunter ;)

=(3vʼ3(2+vʼ3))/(3vʼ3(2-vʼ3))= =((2+vʼ3)(2+vʼ3))/((2-vʼ3)(2+vʼ3))= =(2+vʼ3)²/(2²-(vʼ3)²)= =7+4vʼ3≈13.9282

Nice solution and answer of math problem! Cool, Math Hunter >:D

Amazing !

(5^x * 5^x * 5^x)/(5^x + 5^x) = 10/3 ... divide numerator and denominator of LHS by 5^x ... (5^x * 5^x)/(1 + 1) = 10/3 (25^x) = 20/3 x = ln(20/3)/ln(25) x = [ln(20) - ln(3)]/[2*ln(5)] x = [ln(5) + ln(4) - ln(3)]/[2*ln(5)] x = (1/2) + [ln(4) - ln(3)]/[2*ln(5)] x = (1/2) + [2*ln(2) - ln(3)]/[2*ln(5)] x = (1/2) + ln(2)/ln(5) - ln(3)/[2*ln(5)] x = ½ + log₅(2) - ½*log₅(3)

Thanks so much !

(√108 + √81)/(√108 - √81) = = (6√3 + 9)/(6√3 - 9) = (2√3 + 3)/(2√3 - 3) = (2 + √3)/(2 - √3) = [(2 + √3)*(2 + √3)]/[(2 - √3)*(2 + √3)] = [ 4 + 4√3 + 3 ] / [ 4 - 3 ] = [ 7 + 4√3 ] / [ 1 ] = 7 + 4√3

Done

Write 4 exactly 😮

Conjugate method.. Done 👍

I wish you could give this exercise some flavour by calculating the square root of the number we call A, where A=7+4√3=(2+√3)² and hence√A=2+√3

Standard approach ... (sqr(108)+sqr(81))/(sqr(108)-sqr(81)) =(sqr(108)+sqr(81))^2/(108-81) =(108+81+2*sqr(108)*sqr(81))/27 =(27*7+2*6*sqr(3)*9)/27 = (27*7+27*4*sqr(3)/27 = 7+4*sqr(3)

That was way too easy

(x-6)(x+6)(x^2+36)=0 , x= 6 , -6 , 6i , -6i ,

6

x/6 . x/6 = 6/x . 6/x x^2 / 36 = 36 / x^2 x^2 . x^2 = 36 . 36 x^4 = 1296 x^4 = 6^4 x = 6

Well done ! Thanks so much !

x=(3ln2)/(ln5-ln2)-2≈0.2694

Like old-fashioned courtship, slow but inevitable.

Nice !)

Wait a minute, I have seen this somewhere before, but I don't know where and when ... Anyway, good solution of this problem, Math Hunter.

Wait a minute, I have seen this somewhere before, but I don't know where and when ... Anyway, good solution of this problem, Math Hunter.

Teşekkürler

Wiki Imaginary unit There are two complex square roots of −1: i and −i, just as there are two complex square roots of every real number other than zero (which has one double square root).

The uploader probably showed two methods to show that there is more than one method to solve the problem. However, I think that only a few people solved the problem using the 1st method. The 2nd method should be recommended because the calculations of 1st method are complicated, it takes more time, and there is a risk of making mistakes. For such an easy problem, it is important to solve it quickly, not just to get the answer right.

簡単そうで奥深いですね。毎回興味深く拝見しています。

Japanese Math Olympiad: √(3 + √8) + √(3 - √8) =? 3 = √9 > √8, √(3 + √8) + √(3 - √8) > 0 3 ± √8 = 3 ± 2√2 = (√2)² ± 2√2 + 1 = (√2 ± 1)² √(3 + √8) + √(3 - √8) = √[(√2 + 1)²] + √[(√2 - 1)²] = (√2 + 1) + (√2 - 1) = 2√2

Gostei do exercício.

Solution: (10^x+10^x+10^x)/(5^x+5^x) = 100 ⟹ 3*10^x/(2*5^x) = 100 |*2/3 ⟹ (10/5)^x = 200/3 ⟹ 2^x = 200/3 |lb() ⟹ x = lb(200/3) ≈ 6,0589 ⟹

Great work ! Thanks a lot !

J'ai bien apprécié ta méthode qui est élégante. Mais j'ai eu l'impression qu'elle demandait un peu trop de temps. Alors j'ai essayé la méthode "bourrin" qui consiste à se lancer dans des calculs. Et j'ai trouvé qu'elle était sensiblement plus courte. Dans le cadre d'un concours c'est toujours appréciable. Je te propose donc de l'effectuer par curiosité. Posons X=(rac(15)+rac(5))/rac(20), la quantité que l'on cherche à élever à la puissance 9. On obtient facilement X = (1+rac(3))/2 après avoir simplifié par rac(5). Puis X² = (2+rac(3))/2 en développant l'identité remarquable (a+b)²=(1+rac(3))² et en simplifiant. Puis X⁴ = X².X² = (7+4rac(3))/4 en développant l'identité remarquable (a+b)²=(2+rac(3))² et en simplifiant. Puis X^8 = X⁴. X⁴ = (97+56rac(3))/16 en développant l'identité remarquable (a+b)²=(7+4rac(3))² et en simplifiant. Et enfin X^9= X^8.X = (265+153rac(3))/32. En faisant le produit de (97+56rac(3))/16 par X. Simple et efficace. Il suffit de faire 5 lignes de calculs élémentaires. 😉👍

The problem appears to have been wrongly structured from two angles. One, one of the variables has to be a negative number. Two, product is not exactly 100. Accordingly, X = 16.1803399 and Y = - 6.1803399 Verification X + Y = 10 16.1803399 + (-6.1803399) = 10 XY = 16.1803399 x (-6.1803399) = - 100. 0000001 Please correct me if I am wrong.

Корень числа 2 равен ~1,4. 1.4-1 равно 0.4. Чем выше степень, в которую возводится число меньше 1, тем все меньше и меньше число и в конце концов стремится к нулю. Вот такие бухгалтеры и работают в правительстве!!

Great work ! Thanks so much !

(5^2/2^2)^1/5= (5^2x5^3/2^2x5^3)^1/5=5/500^1/5

Keep going

Simplistic and "talkative"

In fact, sqrt(-1) has 2 roots: -i and +i.

(25/4)^1/5=(200/32)^1/5=(200^1/5)/2.

a/(i*b+a)=a/(i*b+a)*(a-i*b)/(a-i*b)=(a^2-i*a*b)/(a^2+b^2)=(a^2)/(a^2+b^2)-i*(a*b)/(a^2+b^2)= .... and show result for example for n=0: x=([ln(5)]^2-i*ln(5)*PI)/([ln(5)]^2+[PI]^2) This reduces the complex number in the denominator. Nice work.

Congratulations with 10K subscribers, Math Hunter. I'm your old subscriber >:D Good luck on your good future !

And why is that? You still have two square roots to calculate.

After you decided to use Euler's identity as a solution to the problem, I got confused - how to understand this??? Good solution, Math Hunter ;)

x= log(2) 200/3 x=6.0589

Let x = (√2 - 1) ==> x+1 = √2 (x+1)² = 2 x² + 2x + 1 = 2 x² = 1 - 2x x⁴ = (1 - 2x)² = 1 - 4x + 4x² = 1 - 4x + 4(1 - 2x) = 5 - 12x x⁸ = (5 - 12x)² = 25 - 120x + 144x² = 25 - 120x + 144(1 - 2x) = 169 - 408x x¹⁰ = x⁸ * x² = = (169 - 408x) * (1 - 2x) = (169 - 408x) - (338x - 816x²) = 169 - 746x + 816x² = 169 - 746x + 816(1-2x) = 985 - 2378x ... x = (√2 - 1) ... = 985 - 2378(√2 - 1) = 985 - 2378√2 + 2378 = 3363 - 2378√2