Vereenvoudig (toon ALLE berekeninge) die volgende sonder om 'n sakrekenaar te gebruik: 3.1.1 $igg( 2a^{3} \bigg)^{3}$ 3.1.2 $\log_{p} + \log_{1}$ 3.1.3 $\frac{\sqrt{48} - \sqrt{12}}{2\sqrt{75}}$ - NSC Technical Mathematics - Question 3 - 2018 - Paper 1

Using the properties of logarithms, we know that:

$\log_{1} = 0$

Thus, the expression simplifies to:

$\log_{p} + 0 = \log_{p}$

First, we simplify the terms under the square root:

$\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3}$ $\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}$

So, we have:

$\sqrt{48} - \sqrt{12} = 4\sqrt{3} - 2\sqrt{3} = 2\sqrt{3}$

Next, we address the denominator:

$\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}$

Now substituting back in, our expression is:

$\frac{2\sqrt{3}}{2 \cdot 5\sqrt{3}} = \frac{2\sqrt{3}}{10\sqrt{3}} = \frac{2}{10} = \frac{1}{5}$