First note that #x >= 0# - otherwise the square roots will not have real values.

That having been said:

#5sqrt(16x^3)-2xsqrt(9x)#

#=5sqrt((4x)^2*x) - 2xsqrt(3^2*x)#

#=5sqrt((4x)^2)sqrt(x) - 2xsqrt(3^2)sqrt(x)#

#=5*4x sqrt(x) - 2x*3sqrt(x)#

#=20x sqrt(x) - 6x sqrt(x)#

#=14x sqrt(x)#

This derivation mostly uses #sqrt(a*b) = sqrt(a)*sqrt(b)#

Notice that I was able to say that #sqrt((4x)^2) = 4x# because I had already established that #x >= 0#.

If I had not established that #x >= 0# then I would only have been able to say #sqrt((4x)^2) = 4 abs(x)#