To simplify
![\sqrt[4]{\dfrac{24x^6y}{128x^4y^5}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cdfrac%7B24x%5E6y%7D%7B128x%5E4y%5E5%7D%7D)
we need to use the fact that
![\sqrt[4]{x^4}=|x|](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E4%7D%3D%7Cx%7C)
Why the absolute value? It's because 
.
We start by rewriting as
![\sqrt[4]{\dfrac{2^23x^6y}{2^6x^4y^5}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cdfrac%7B2%5E23x%5E6y%7D%7B2%5E6x%5E4y%5E5%7D%7D)
![\sqrt[4]{\dfrac{2^23x^4x^2y}{2^42^2x^4y^4y}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cdfrac%7B2%5E23x%5E4x%5E2y%7D%7B2%5E42%5E2x%5E4y%5E4y%7D%7D)
Since 
, we have 
, and the above reduces to
![\sqrt[4]{\dfrac{3x^2y}{2^4y^4y}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cdfrac%7B3x%5E2y%7D%7B2%5E4y%5E4y%7D%7D)
Then we pull out any 4th powers under the radical, and simplify everything we can:
![\dfrac1{\sqrt[4]{2^4y^4}}\sqrt[4]{\dfrac{3x^2y}{y}}](https://tex.z-dn.net/?f=%5Cdfrac1%7B%5Csqrt%5B4%5D%7B2%5E4y%5E4%7D%7D%5Csqrt%5B4%5D%7B%5Cdfrac%7B3x%5E2y%7D%7By%7D%7D)
![\dfrac1{|2y|}\sqrt[4]{3x^2}](https://tex.z-dn.net/?f=%5Cdfrac1%7B%7C2y%7C%7D%5Csqrt%5B4%5D%7B3x%5E2%7D)
where 
allows us to write 
, and this also means that 
. So we end up with
![\dfrac{\sqrt[4]{3x^2}}{2y}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B4%5D%7B3x%5E2%7D%7D%7B2y%7D)
making the last option the correct answer.
Answer:
7√6 - 5√7
Step-by-step explanation:
5√7 + 12√6 - 10√7 - 5√6 (rearranging)
= (5√7 - 10√7) + (12√6 - 5√6) (factor out √6 and √7 respectively)
= (5 - 10) √7 + (12 - 5)√6
= -5√7 + 7√6 (rearrange)
= 7√6 - 5√7
Stop being like that I helped u
Answer:
what's the question ndndn
