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Simply: (2/3)^(2/3) = 3√(2/3)^2 = 3√(4/9) = 3√[(4*3)/(9*3)] = 3√(12/27) = 3√12 /3 NB! In this example, 3√ means cube root
Exactly, simple and no need for twaddle. Btw, no such problem from Harvard exists, that is just nonsense.
1:33 Multiplying the both the numerator and denominator by cube root of 3 in this step saves the trouble.
(2/3)^(2/3) = (4*3/(9*3))^(1/3) = 12^(1/3)/3
Finally I could quickly solve one of your challenges 😀
Nice work! Great work! Well done!🙏✅😎💕😍🔥
My god, how much more can you drag a 1 minute answer out for? I missed the end as I fell asleep
I came up with cubed root of (4/9), which ironically is the same answer.
Define 'simplify'. It is simple enough as it is.
I could have stated rationalize the denominator
Useful review of exponent math.
{1.1+2}}=1.3 (x ➖ 3x+1).
Need to provide your arbitrary definition of "simplify". x^x looks about as simplified as x^x can be.
I appreciate you raising the point about simplification. It's all about finding the most suitable representation.
@@superacademy247 Back when I was in school, we used the term "canonical" for a specific representation. We were already simple.
@@superacademy247Quite. And suitable for what? The exercice has no purpose.
What you did was rationalize the denominator. It is NOT any simpler than the original function!!
It's all about finding the most suitable representation.
Do you really think that this is simplifying? I don't!
It's the natural way of rationalizing the denominator
@superacademy247 but it's a points exercise when the answer is as complicated as the initial mathematical expression. As a graduate of Mathematics I would leave this expression as it is & move on!
(4/9)^(1/3)
Very long winded way of doing it, At cube root of 4/9 change 4/9 to 12/27 and you have 1/3 times the cube root of 12. done
Thanks for a positive feedback. I appreciate your approach to this fractional exponents.💕🙏✅😎
Please dedicate this episode to STD V students.
Unfortunately, this question is beyond the scope of STD V learners
Cómo complicar algo ridiculamente fácil para hacerlo parecer difícil.
Thanks for your feedback! I'm always open to different perspectives on mathematical elegance. 😎
Two lines answer has been dragged on for nothing.
Illustration k
Simply: (2/3)^(2/3) = 3√(2/3)^2 = 3√(4/9) = 3√[(4*3)/(9*3)] = 3√(12/27) = 3√12 /3 NB! In this example, 3√ means cube root
Exactly, simple and no need for twaddle. Btw, no such problem from Harvard exists, that is just nonsense.
1:33 Multiplying the both the numerator and denominator by cube root of 3 in this step saves the trouble.
(2/3)^(2/3) = (4*3/(9*3))^(1/3) = 12^(1/3)/3
Finally I could quickly solve one of your challenges 😀
Nice work! Great work! Well done!🙏✅😎💕😍🔥
My god, how much more can you drag a 1 minute answer out for? I missed the end as I fell asleep
I came up with cubed root of (4/9), which ironically is the same answer.
Define 'simplify'. It is simple enough as it is.
I could have stated rationalize the denominator
Useful review of exponent math.
{1.1+2}}=1.3 (x ➖ 3x+1).
Need to provide your arbitrary definition of "simplify". x^x looks about as simplified as x^x can be.
I appreciate you raising the point about simplification. It's all about finding the most suitable representation.
@@superacademy247 Back when I was in school, we used the term "canonical" for a specific representation. We were already simple.
@@superacademy247Quite. And suitable for what? The exercice has no purpose.
What you did was rationalize the denominator. It is NOT any simpler than the original function!!
It's all about finding the most suitable representation.
Do you really think that this is simplifying? I don't!
It's the natural way of rationalizing the denominator
@superacademy247 but it's a points exercise when the answer is as complicated as the initial mathematical expression. As a graduate of Mathematics I would leave this expression as it is & move on!
(4/9)^(1/3)
Very long winded way of doing it,
At cube root of 4/9 change 4/9 to 12/27 and you have 1/3 times the cube root of 12. done
Thanks for a positive feedback. I appreciate your approach to this fractional exponents.💕🙏✅😎
Please dedicate this episode to STD V students.
Unfortunately, this question is beyond the scope of STD V learners
Cómo complicar algo ridiculamente fácil para hacerlo parecer difícil.
Thanks for your feedback! I'm always open to different perspectives on mathematical elegance. 😎
Two lines answer has been dragged on for nothing.
Illustration k