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