Question

1. \\(\frac{x^{4}y^{-3}}{3xy^{-3} \cdot 3xy}\\)

Explanation:

Step1: Simplify denominator constants

Multiply the constants: $3 \cdot 3 = 9$

Step2: Combine x terms in denominator

Add exponents of $x$: $x^1 \cdot x^1 = x^{1+1} = x^2$

Step3: Combine y terms in denominator

Add exponents of $y$: $y^{-3} \cdot y^1 = y^{-3+1} = y^{-2}$

Step4: Rewrite the entire fraction

$$\frac{x^4 y^{-3}}{9x^2 y^{-2}}$$

Step5: Simplify x variable terms

Subtract exponents of $x$: $\frac{x^4}{x^2} = x^{4-2} = x^2$

Step6: Simplify y variable terms

Subtract exponents of $y$: $\frac{y^{-3}}{y^{-2}} = y^{-3 - (-2)} = y^{-1} = \frac{1}{y}$

Step7: Combine all simplified parts

Multiply the results: $x^2 \cdot \frac{1}{9y} = \frac{x^2}{9y}$

Answer:

$\frac{x^2}{9y}$