The problem is to simplify the expression $1 \div 2 \cdot \sqrt{x^{-3}}$.

1. Problem Description

The problem is to simplify the expression $1 \div 2 \cdot \sqrt{x^{-3}}$.

2. Solution Steps

First, we rewrite the division as multiplication by the reciprocal:

$1 \div 2 = 1 \cdot \frac{1}{2} = \frac{1}{2}$

So the expression becomes:

$\frac{1}{2} \cdot \sqrt{x^{-3}}$

Recall that $x^{-n} = \frac{1}{x^n}$. Then $x^{-3} = \frac{1}{x^3}$. The expression becomes:

$\frac{1}{2} \cdot \sqrt{\frac{1}{x^3}}$

We can rewrite the square root as:

$\sqrt{\frac{1}{x^3}} = \frac{\sqrt{1}}{\sqrt{x^3}} = \frac{1}{\sqrt{x^3}}$

Then the expression is:

$\frac{1}{2} \cdot \frac{1}{\sqrt{x^3}} = \frac{1}{2\sqrt{x^3}}$

We can rewrite $\sqrt{x^3}$ as $x^{\frac{3}{2}}$.

Then the expression is:

$\frac{1}{2x^{\frac{3}{2}}}$

We can also write $x^3 = x^2 \cdot x$, so $\sqrt{x^3} = \sqrt{x^2 \cdot x} = \sqrt{x^2}\cdot\sqrt{x} = x\sqrt{x}$.

Then the expression is:

$\frac{1}{2x\sqrt{x}}$

To rationalize the denominator, multiply by $\frac{\sqrt{x}}{\sqrt{x}}$:

$\frac{1}{2x\sqrt{x}} \cdot \frac{\sqrt{x}}{\sqrt{x}} = \frac{\sqrt{x}}{2x \cdot x} = \frac{\sqrt{x}}{2x^2}$

3. Final Answer

$\frac{\sqrt{x}}{2x^2}$