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Take the conjugate as sqrt5 + sqrt 3 - sqrt8 to get 2*sqrt 15 in denominator which cancels with 30 to give sqrt 15 in numerator to be multiplied with the conjugate to give answer in fewer steps
Thanks for sharing your alternative approach! 🔥🔥🔥I’m glad you found a more efficient way to solve the problem! 💯🙏
Input30/(sqrt(8) + sqrt(5) + sqrt(3)) = 5 sqrt(3) + 3 sqrt(5) - 2 sqrt(30)=5Sqrt[3]+3Sqrt[5]-2Sqrt[30]ResultTrueLogarithmic formlog(2 sqrt(2) + sqrt(3) + sqrt(5), 30) - log(2 sqrt(2) + sqrt(3) + sqrt(5), sqrt(8) + sqrt(5) + sqrt(3)) = log(2 sqrt(2) + sqrt(3) + sqrt(5), 5 sqrt(3) + 3 sqrt(5) - 2 sqrt(30))
5 digits against 7 but simplified 🙂
Nice!
Thanks for watching! 🙏😊I’m glad you found it interesting! 💯💖
3^10/2^3+2^3+3 1^2^5/1^1^1+2^1+3 1^1/2^1+3 1/2^1+3 /2+3 (x ➖ 3x+2) .
7,5 my answer
I certainly hope that this kind of pointless exercise isn't really something Oxford University is using on their entrance exams.
Take the conjugate as sqrt5 + sqrt 3 - sqrt8 to get 2*sqrt 15 in denominator which cancels with 30 to give sqrt 15 in numerator to be multiplied with the conjugate to give answer in fewer steps
Thanks for sharing your alternative approach! 🔥🔥🔥I’m glad you found a more efficient way to solve the problem! 💯🙏
Input
30/(sqrt(8) + sqrt(5) + sqrt(3)) = 5 sqrt(3) + 3 sqrt(5) - 2 sqrt(30)=5Sqrt[3]+3Sqrt[5]-2Sqrt[30]
Result
True
Logarithmic form
log(2 sqrt(2) + sqrt(3) + sqrt(5), 30) - log(2 sqrt(2) + sqrt(3) + sqrt(5), sqrt(8) + sqrt(5) + sqrt(3)) = log(2 sqrt(2) + sqrt(3) + sqrt(5), 5 sqrt(3) + 3 sqrt(5) - 2 sqrt(30))
5 digits against 7 but simplified 🙂
Nice!
Thanks for watching! 🙏😊I’m glad you found it interesting! 💯💖
3^10/2^3+2^3+3 1^2^5/1^1^1+2^1+3 1^1/2^1+3 1/2^1+3 /2+3 (x ➖ 3x+2) .
7,5 my answer
I certainly hope that this kind of pointless exercise isn't really something Oxford University is using on their entrance exams.