Answer:
your work is correct
Step-by-step explanation:
[tex]\sqrt{\dfrac{16x^4y^8}{5}}=\sqrt{\left(\dfrac{4x^2y^4}{5}\right)^2\cdot5}=\boxed{\dfrac{4x^2y^4\sqrt{5}}{5}}[/tex]
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Additional comment
You can multiply by √5/√5 as you have done, and as conventionally taught, or you can multiply by 5/5 under the radical to make the denominator a perfect square.
