Chapter 5 - Section 5.7 - Negative Exponents and Scientific Notation - Exercise Set - Page 409: 55

$\dfrac{1}{16x^{12}}$

Using the laws of exponents, the given expression, $ \left( \dfrac{4x^5}{2x^2} \right)^{-4} ,$ simplifies to \begin{array}{l}\require{cancel} \left( (4\div2)x^{5-2} \right)^{-4} \\\\= \left( 2x^{3} \right)^{-4} \\\\= 2^{1(-4)}x^{3(-4)} \\\\= 2^{-4}x^{-12} \\\\= \dfrac{1}{2^{4}x^{12}} \\\\= \dfrac{1}{16x^{12}} .\end{array}