Please help me solve this!?

√(75x^2y^-4)

We know that √(a*b*c) = √a * √b * √c, so let's rearrange the problem in a similar manner

√(75x^2y^-4) = √75 * √x^2 * √y^-4 and solve each separately

√75 = √25 * √3 = 5√3

√x^2 = x

√y^-4 = y^-2 = 1/y^2

..and put them back together

√(75x^2y^-4) = 5x√3/y^2

sqrt(A*B) = sqrt(A) * sqrt(B)

So, all of those things under the radical can be dealt with individually

sqrt(75x^2y^-4) = sqrt(75) * sqrt(x^2) * sqrt(y^-4)

Note that 75 can be rewritten as 3*25, so we can rewrite the expression one more time like this:

sqrt(75x^2y^-4) = sqrt(3) * sqrt(25)) * sqrt(x^2) * sqrt(y^-4)

Now do the square roots wherever you can:

sqrt(75x^2y^-4) = sqrt(3) * 5 * x * y^-2

We're left with only one thing under the radical that we can't get rid of.

sqrt(75x^2y^-4) = sqrt(3) * 5 * x * y^-2

or, if you want all positive exponents:

sqrt(75x^2y^-4) = 5* x * sqrt(3)/(y^2)

By convention, the whole values are written before the radical