Answer:
see below
Explanation:
For simplifying expressions of this sort, there are four rules of exponents that come into play;
![a^ba^c=a^(b+c)\\\\(ab)^c=a^cb^c\\\\(a^b)^c=a^(bc)\\\\(1)/(a^b)=a^(-b)\ \quad\text{which means}\quad a^b=(1)/(a^(-b))](https://img.qammunity.org/2020/formulas/mathematics/high-school/hi12rl5sw46jtxop8rr2xtn449hexgc8er.png)
__
I find it convenient to eliminate the fractions by adding the exponents, then rewrite any negative exponents as denominator factors.
![23. \quad(1)/(x^(-2))=x^2\\\\24. \quad(mn)/(m^2n^3)=m^(1-2)n^(1-3)=m^(-1)n^(-2)=(1)/(mn^2)\\\\25. \quad(k^(-2))/(k^(-3))=k^(-2-(-3))=k\\\\26. \quad\left((m^2)/(n^3)\right)^3=(m^(2\cdot 3))/(n^(3\cdot 3))=(m^6)/(n^9)\\\\27. \quad(x^2y^3z^4)/(x^(-3)y^(-4)z^(-5))=x^(2-(-3))y^(3-(-4))z^(4-(-5))=x^5y^7z^9](https://img.qammunity.org/2020/formulas/mathematics/high-school/9h2z7c7h29quglfdemppalzo2o4x4cnztn.png)