Simplify the following using the laws of exponents: (Leave the final answer with positive exponents and in surd form where applicable.) 1.2.1 \( \frac{\sqrt{16 p^{4}} \times\left(p^{4} q^{4}\right)^{0}}{(2 p)^{2}} \) \( 1.2 .2 \quad \frac{\left(p^{2} q^{2}\right)^{2 r} \times\left(p q^{3}\right)^{3 r}}{\left(p^{3} q^{3}\right)^{2 r}} \)

Alright, let's dive into the world of exponents and simplify these expressions together! 1.2.1: First, let’s simplify the expression: We have \( \frac{\sqrt{16 p^{4}} \times (p^{4} q^{4})^{0}}{(2 p)^{2}} \). Since anything raised to the power of 0 is 1, \( (p^{4} q^{4})^{0} = 1 \). Next, \( \sqrt{16 p^{4}} = \sqrt{16} \times \sqrt{p^{4}} = 4p^{2} \). Now let's simplify the denominator: \( (2p)^{2} = 4p^{2} \). Putting it all together, we get: \[ \frac{4p^{2} \times 1}{4p^{2}} = 1 \] So, our final answer is simply \( 1 \). 1.2.2: Now, let’s tackle the second expression: We need to simplify \( \frac{(p^{2} q^{2})^{2r} \times (pq^{3})^{3r}}{(p^{3} q^{3})^{2r}} \). Starting with the numerator, we have \( (p^{2} q^{2})^{2r} = p^{4r} q^{4r} \) and \( (pq^{3})^{3r} = p^{3r} q^{9r} \). Multiplying these gives us: \[ p^{4r} q^{4r} \times p^{3r} q^{9r} = p^{7r} q^{13r} \] Now, simplifying the denominator, we find \( (p^{3} q^{3})^{2r} = p^{6r} q^{6r} \). Putting everything together for our fraction: \[ \frac{p^{7r} q^{13r}}{p^{6r} q^{6r}} = p^{7r - 6r} q^{13r - 6r} = p^{1r} q^{7r} \] Thus, our simplified expression is: \[ pq^{7r} \] So, the final answers are: 1.2.1: \( 1 \) 1.2.2: \( pq^{7r} \)