Math Olympiad | Can you solve this ? | Simplify Square Root Problem

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Math Olympiad | Can you solve this ? | Simplify Square Root Problem

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Күн бұрын

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Пікірлер: 28

@user-hd9do6tc8k Ай бұрын

Умножить числитель и знаменатель дроби на 2, в знаменателе будет формула квадрат разности корня из 5 и корня из трех. А дальше только упростить! 😊

@paulortega5317 Ай бұрын

let x = the answer. u = x/4. Then u^2 = 4+sqrt(15) and 1/u^2 = 4-sqrt(15). u^2+1/u^2 = 8 and then u+1/u=sqrt(10). u^2+1=u*sqrt(10). 5+sqrt(15)=u*sqrt(10). u=(5+sqrt(15))/sqrt(10). u=sqrt(2)*(sqrt(5) + sqrt(3))/2. x = 4u = 2*sqrt(2)*(sqrt(5) + sqrt(3)).

@kenrutherford7607 Ай бұрын

splendid!!!!

@muratyasaryldz8168 Ай бұрын

Çok güzel bir paylaşım.Muzik ve matematik öyle muhteşem bilim ki konuşmaları anlamasam da ne yaptığınızı anliyorum.Cok teşekkürler

@gnanadesikansenthilnathan6750 Ай бұрын

Clear. Splendid and excellent.

@learncommunolizer Ай бұрын

Thanks. Your welcome

@betto63mx92 11 күн бұрын

Para no obtener un resultado como el que esperaba, propongo esta alternativa mas directa y con un resultado como el que tampoco me convence . . A partir de Sq(64 + 16 sq15) Sq( 16 ( 4 + sq15 )) 4sq( 4 + sq15)

@key_board_x 3 күн бұрын

= √[16 / (4 - √15)] = √[16.(4 + √15) / {(4 - √15).(4 + √15)} ] = √[16.(4 + √15) / {16 - 15} ] = √[16.(4 + √15)] = 4√(4 + √15) → you need to find a number x sush as x² = 4 + √15 Suppose that: x = √a + √b x = √a + √b x² = (√a + √b)² x² = (√a)² + 2.(√a).(√b) + (√b)² x² = a + 2.√(ab) + b x² = (a + b) + 2.√(ab) → by identification to (4 + √15) a + b = 4 2.√(ab) = √15 √(ab) = (1/2).√15 √(ab) = √[(1/4) * 15] ab = (1/4) * 15 ab = 15/4 Resume a + b = 4 ← this is the sum S ab = 15/4 ← this is the product P a & b are the solution of the equation: x² - Sx + P = 0 x² - 4x + (15/4) = 0 Δ = (- 4)² - [4 * (15/4)] = 16 - 15 = 1 x = (4 ± 1)/2 x = 5/2 ← this is a x = 3/2 ← this is b We can see that: [ √(3/2) + √(5/2) ]² = (3/2) + 2√[(3/2) * (5/2)] + (5/2) [ √(3/2) + √(5/2) ]² = (8/2) + 2√(15/4) [ √(3/2) + √(5/2) ]² = 4 + √15 Restart: = 4√(4 + √15) = 4√[ √(3/2) + √(5/2) ]² = 4.[√(3/2) + √(5/2)] = 4√(3/2) + 4√(5/2) = √(48/2) + √(80/2) = √24 + √40 = 2√6 + 2√10 = 2.(√6 + √10)

@user-ee7nw2rx9s Ай бұрын

Умножить на 2 Sqrt (32/(8-2sqrt 15)) Таким образом 32=2*4^2 8-2sqrt 15=(sqrt 5-sqrt 3)^2 Извлечение квадратного корня 4 *sqr2 /(sqrt 5-sqrt 3) Избавиться от иррациональности в знаменателе 4 sqrt 2*(sqrt 5+sqrt3) /2=2 sqrt 10+2sqrt6 2 минуты, а не 8 как на видео

@arturchirki Ай бұрын

Бессмысленная задача, и никакой красоты в решении и ответа.Задача смотрится проще, чем ответ.

@mauriziograndi1750 28 күн бұрын

You right here we see taking pleasure in writing plenty of small steps. I believe the shorter is the smartest.

@lifewatery7472 Ай бұрын

√[16/(4-√15)] =√[32/(8-2√15)] =√{32/[(√5)²-2√5•√3+(√3)²]} =√[32/(√5-√3)²] =4√2/(√5-√3) =4√2(√5+√3)/[(√5-√3)•(√5+√3)] =4√2(√5+√3)/2 =2√2(√5+√3)

@lifewatery7472 Ай бұрын

√[16/(4-√15)] =4√[2/(8-2√15)] =4√2/√{[(√5)²-2√5•√3+(√3)²]} =4√2/√(√5-√3)² =4√2/(√5-√3) =4√2(√5+√3)/[(√5-√3)•(√5+√3)] =4√2(√5+√3)/2 =2√2(√5+√3)

@user-ud3cg2vp5s Ай бұрын

Thank u ❤

@learncommunolizer Ай бұрын

You're welcome 😊

@aeroeng22 25 күн бұрын

that's funny, Harris begging for your hard earned cash at the beginning of this video, in an ad. I wouldn't show her the restroom if her hair was on fire.

@user-gr3ut4gk9w 18 күн бұрын

Как же бесят переписывания по 100 раз одно и тоже

@mathshindustani6216 6 күн бұрын

There is easier way to solve this bro U made it lengthy But nice try

@reinhardtristaneugen9113 Ай бұрын

also ich habe da 11 vor dem Komma, aber weil das natürlich zum Quadrat 121 ergibt, musste ich weiterrechnen, und kam auf 11,09. Wie ich darauf kam? Ich zog die Wurzel von 15 im Kopf, was mich zu 3,87 führte und subtrahierte und das brachte mich zu 13 / 100, wobei man ja dividiert, indem man mit dem Kehrwert multipliziert, sodass ich 1600 durch 13 hatte und das führt zu 123 und eben zu 11.09, so man die Wurzel zieht. Le p'tit Daniel