We are asked to simplify the expressions $15n^{-10}p^{-5}$, $(\frac{2}{3})^{-2}$, $\frac{4t^2u^{-1}}{t^{-5}u^7}$, $(\frac{5}{2})^{-1}$, $\frac{5^{-1}}{2^{-1}}$, $(\frac{5}{2x})^{-1}$, $(\frac{5x}{2})^{-1}$.

1. Problem Description

We are asked to simplify the expressions $15n^{-10}p^{-5}$, $(\frac{2}{3})^{-2}$, $\frac{4t^2u^{-1}}{t^{-5}u^7}$, $(\frac{5}{2})^{-1}$, $\frac{5^{-1}}{2^{-1}}$, $(\frac{5}{2x})^{-1}$, $(\frac{5x}{2})^{-1}$.

2. Solution Steps

Let's simplify each expression separately:

* $15n^{-10}p^{-5}$

Recall that $a^{-n} = \frac{1}{a^n}$. Therefore, $n^{-10} = \frac{1}{n^{10}}$ and $p^{-5} = \frac{1}{p^5}$.

Thus, $15n^{-10}p^{-5} = 15 \cdot \frac{1}{n^{10}} \cdot \frac{1}{p^5} = \frac{15}{n^{10}p^5}$.

* $(\frac{2}{3})^{-2}$

Recall that $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$.

So, $(\frac{2}{3})^{-2} = (\frac{3}{2})^2 = \frac{3^2}{2^2} = \frac{9}{4}$.

* $\frac{4t^2u^{-1}}{t^{-5}u^7}$

Using the property $a^{-n} = \frac{1}{a^n}$, we have $u^{-1} = \frac{1}{u}$ and $t^{-5} = \frac{1}{t^5}$.

So, $\frac{4t^2u^{-1}}{t^{-5}u^7} = \frac{4t^2 \cdot \frac{1}{u}}{\frac{1}{t^5} \cdot u^7} = \frac{4t^2}{u} \cdot \frac{t^5}{u^7} = \frac{4t^{2+5}}{u^{1+7}} = \frac{4t^7}{u^8}$.

* $(\frac{5}{2})^{-1}$

Using the property $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$, we have $(\frac{5}{2})^{-1} = \frac{2}{5}$.

* $\frac{5^{-1}}{2^{-1}}$

Using the property $a^{-n} = \frac{1}{a^n}$, we have $5^{-1} = \frac{1}{5}$ and $2^{-1} = \frac{1}{2}$.

So, $\frac{5^{-1}}{2^{-1}} = \frac{\frac{1}{5}}{\frac{1}{2}} = \frac{1}{5} \cdot \frac{2}{1} = \frac{2}{5}$.

* $(\frac{5}{2x})^{-1}$

Using the property $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$, we have $(\frac{5}{2x})^{-1} = \frac{2x}{5}$.

* $(\frac{5x}{2})^{-1}$

Using the property $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$, we have $(\frac{5x}{2})^{-1} = \frac{2}{5x}$.

3. Final Answer

$15n^{-10}p^{-5} = \frac{15}{n^{10}p^5}$

$(\frac{2}{3})^{-2} = \frac{9}{4}$

$\frac{4t^2u^{-1}}{t^{-5}u^7} = \frac{4t^7}{u^8}$

$(\frac{5}{2})^{-1} = \frac{2}{5}$

$\frac{5^{-1}}{2^{-1}} = \frac{2}{5}$

$(\frac{5}{2x})^{-1} = \frac{2x}{5}$

$(\frac{5x}{2})^{-1} = \frac{2}{5x}$