Sure, let's rationalize the denominators step by step for each given problem.
### Part (a): [tex]\(\frac{3}{\sqrt{3}}\)[/tex]
To rationalize the denominator, we multiply both the numerator and the denominator by [tex]\(\sqrt{3}\)[/tex]:
[tex]\[
\frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3 \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \frac{3 \sqrt{3}}{3}
\][/tex]
Simplifying the fraction:
[tex]\[
\frac{3 \sqrt{3}}{3} = \sqrt{3}
\][/tex]
So, [tex]\(\frac{3}{\sqrt{3}} = \sqrt{3}\)[/tex].
### Part (b): [tex]\(\frac{4}{\sqrt{5}}\)[/tex]
To rationalize the denominator, we multiply both the numerator and the denominator by [tex]\(\sqrt{5}\)[/tex]:
[tex]\[
\frac{4}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{4 \cdot \sqrt{5}}{\sqrt{5} \cdot \sqrt{5}} = \frac{4 \sqrt{5}}{5}
\][/tex]
So, [tex]\(\frac{4}{\sqrt{5}} = \frac{4 \sqrt{5}}{5}\)[/tex].
### Part (c): [tex]\(\frac{3}{2 \sqrt{3}}\)[/tex]
To rationalize the denominator, we multiply both the numerator and the denominator by [tex]\(\sqrt{3}\)[/tex]:
[tex]\[
\frac{3}{2 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3 \cdot \sqrt{3}}{2 \sqrt{3} \cdot \sqrt{3}} = \frac{3 \sqrt{3}}{2 \cdot 3}
\][/tex]
Simplifying the fraction:
[tex]\[
\frac{3 \sqrt{3}}{6} = \frac{\sqrt{3}}{2}
\][/tex]
So, [tex]\(\frac{3}{2 \sqrt{3}} = \frac{\sqrt{3}}{2}\)[/tex].
### Part (d): [tex]\(\frac{3 \sqrt{5}}{2 \sqrt{7}}\)[/tex]
To rationalize the denominator, we multiply both the numerator and the denominator by [tex]\(\sqrt{7}\)[/tex]:
[tex]\[
\frac{3 \sqrt{5}}{2 \sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{3 \sqrt{5} \cdot \sqrt{7}}{2 \sqrt{7} \cdot \sqrt{7}} = \frac{3 \sqrt{35}}{2 \cdot 7}
\][/tex]
Simplifying the fraction:
[tex]\[
\frac{3 \sqrt{35}}{14}
\][/tex]
So, [tex]\(\frac{3 \sqrt{5}}{2 \sqrt{7}} = \frac{3 \sqrt{35}}{14}\)[/tex].
