Элементарно. Числитель и знаменатель умножаем на sqrt 2. Тогда в числителе будет 6*sqrt(2), а в знаменателе [sqrt(2)]^2=2. Ну а дальше 6 и 2 сокращаем на 2. Получается 3*sqrt(2)
@@yurenchu what I intended to arrived was a quick and mistakenly wrote. To clarify it should have been as follows: 6 √(6 +√ 35) √6.√(36 +6√35) √6 √(√21 +√15) ^2 √6(√21 +√15) Thanks for pinpointing the same.
What do you mean? Did you derive the answer as (6√6)/(√21 - √15) , with an non-reduced denominator? (Note that both numerator and denominator also contain a factor √3 , so the numerator and the denominator aren't even in co-prime terms.)
@@ALEX.1R yeah but if x^2=36 then that would be correct but if you just take a sqrt you will get positive solution not negative because that’s what sqrt finds but if it’s a equation which your finding x then yes it is also negative
Thank you this is very helpful!
Like old-fashioned courtship, slow but inevitable.
Excellent ! Thank you so much !
And what’s the point? The final form is no simpler than the beginning.
What is tha correct answar ..
What's the name of the app that you write?
Very nice congratulations ❤❤❤❤❤❤
Conjugate method.. Done 👍
Amazing !
Can you enlighten us 6/√2 become 3 √2
Элементарно. Числитель и знаменатель умножаем на sqrt 2. Тогда в числителе будет 6*sqrt(2), а в знаменателе [sqrt(2)]^2=2. Ну а дальше 6 и 2 сокращаем на 2. Получается 3*sqrt(2)
√( 36/(6 - √35) ) =
... multiply both numerator and denominator within square root operator by (6+√35) ...
= √( 36(6+√35) / [(6 - √35)(6+√35)] )
= √( 36(6+√35) / [(36 - 35)] )
= √( 36(6+√35) / [1] )
= √( 9*4*(6+√35) )
= √( 9*(24+4√35) )
= √( 9*(24+2√140) )
= √( 9*( 14 + 10 +2(√14)(√10) ) )
= √( 3²*(√14 + √10)² )
= 3*(√14 + √10)
= 3√14 + 3√10
@@yurenchu what I intended to arrived was a quick and mistakenly wrote. To clarify it should have been as follows: 6 √(6 +√ 35)
√6.√(36 +6√35)
√6 √(√21 +√15) ^2
√6(√21 +√15)
Thanks for pinpointing the same.
Interesting surd puzzle which ends with ( √21 - √15 )
What do you mean? Did you derive the answer as (6√6)/(√21 - √15) , with an non-reduced denominator? (Note that both numerator and denominator also contain a factor √3 , so the numerator and the denominator aren't even in co-prime terms.)
Very nice congratulations ❤❤❤❤❤❤❤
Wonder UA-cam require video to be long, the author seem to make the video long on purpose
36
خطأ في جدر ستة وثلاثين وطريقة طويلة .
3(√14 + √10)=20.712
√14 + √10 = 6.904x3=20.712
36 : 0,099= 3.000 😮🎉😅😂❤
Very nice congratulations ❤❤❤❤❤❤❤
sqrt (36) has 2 roots: +6 and -6.
Nun uh sqrt takes only positive
@@ΕντιΚαιτααζι what?! (-6)^2 = 36, (+6)^2 = 36.
@@ALEX.1R yeah but if x^2=36 then that would be correct but if you just take a sqrt you will get positive solution not negative because that’s what sqrt finds but if it’s a equation which your finding x then yes it is also negative
Square root is not an equation to have roots.
Интересно, но финал почти такой же, как старт. Преобразование ради преобразования...
And why is that? You still have two square roots to calculate.