`sqrt(64xy^5)/(sqrt8x)`

To simplify reduce the 64 and the 8 to prime bases (ie the lowest number to which they are a power. ie 2)and convert the square root to a power - a square root is the reciprical of a square (`^2` ) so, as an exponent it is the reciprocal of 2 (or `2/1` ) which is `1/2`

`(2^6 xy^5)^(1/2)/(2^3x)^(1/2)`

When working with exponents we know that the powers must be multiplied:

`(2^(6 times 1/2) x ^(1/2) y^(5 times 1/2))/( (2^(3 times 1/2) x^(1/2))`

`therefore( 2^(6/2)x^(1/2) y^(5/2))/ (2^(3/2) x^(1/2))`

`therefore (2^3 x^(1/2)y^(5/2))/(2^(3/2) x^(1/2))`

Further using the rules of exponents we know that divide indicates that we must subtract the exponents when the base is the same :

`therefore 2^(3-3/2) x^(1/2-1/2) y^(5/2)`

`therefore = 2^(3/2) x^0 y^(5/2)`

We know that anything which has a power (exponent) of 0 is 1

`a^0=1`

`therefore = 2^(3/2) y^(5/2)`

This can be converted back to square roots or left like it is. Thus

`= sqrt(2^3 y^5)` or `sqrt(8y^5)` or `(8y^5)^(1/2)`