Answer :
Certainly! Let's rationalize the denominators of the given expressions step-by-step.
## Part (a)
[tex]\[ \frac{2}{\sqrt{2}} \][/tex]
1. Multiply the numerator and the denominator by [tex]\(\sqrt{2}\)[/tex]:
[tex]\[ \frac{2}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{2}}{2} = \sqrt{2} \][/tex]
## Part (b)
[tex]\[ \frac{5}{\sqrt{3}} \][/tex]
1. Multiply the numerator and the denominator by [tex]\(\sqrt{3}\)[/tex]:
[tex]\[ \frac{5}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{3} \][/tex]
## Part (c)
[tex]\[ \frac{3}{2\sqrt{2}} \][/tex]
1. Multiply the numerator and the denominator by [tex]\(\sqrt{2}\)[/tex]:
[tex]\[ \frac{3}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{2 \times 2} = \frac{3\sqrt{2}}{4} \][/tex]
## Part (d)
[tex]\[ \frac{3\sqrt{2}}{2\sqrt{3}} \][/tex]
1. Multiply the numerator and the denominator by [tex]\(\sqrt{3}\)[/tex]:
[tex]\[ \frac{3\sqrt{2}}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{2}\sqrt{3}}{2 \times 3} = \frac{3\sqrt{6}}{6} = \frac{\sqrt{6}}{2} \][/tex]
## Part (e)
[tex]\[ \frac{\sqrt{5}}{5\sqrt{6}} \][/tex]
1. Multiply the numerator and the denominator by [tex]\(\sqrt{6}\)[/tex]:
[tex]\[ \frac{\sqrt{5}}{5\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{\sqrt{5}\sqrt{6}}{5 \times 6} = \frac{\sqrt{30}}{30} \][/tex]
## Part (f)
[tex]\[ \frac{1}{\sqrt{2} + 1} \][/tex]
1. Multiply the numerator and the denominator by the conjugate [tex]\(\sqrt{2} - 1\)[/tex]:
[tex]\[ \frac{1}{\sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \frac{\sqrt{2} - 1}{(\sqrt{2})^2 - (1)^2} = \frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1 \][/tex]
## Part (g)
[tex]\[ \frac{2}{\sqrt{3} - 1} \][/tex]
1. Multiply the numerator and the denominator by the conjugate [tex]\(\sqrt{3} + 1\)[/tex]:
[tex]\[ \frac{2}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{2(\sqrt{3} + 1)}{(\sqrt{3})^2 - (1)^2} = \frac{2\sqrt{3} + 2}{3 - 1} = \frac{2\sqrt{3} + 2}{2} = \sqrt{3} + 1 \][/tex]
## Part (h)
[tex]\[ \frac{7}{\sqrt{3} - \sqrt{2}} \][/tex]
1. Multiply the numerator and the denominator by the conjugate [tex]\(\sqrt{3} + \sqrt{2}\)[/tex]:
[tex]\[ \frac{7}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = \frac{7(\sqrt{3} + \sqrt{2})}{(\sqrt{3})^2 - (\sqrt{2})^2} = \frac{7(\sqrt{3} + \sqrt{2})}{3 - 2} = 7(\sqrt{3} + \sqrt{2}) \][/tex]
## Part (i)
[tex]\[ \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \][/tex]
1. Multiply the numerator and the denominator by the conjugate [tex]\(\sqrt{3} + \sqrt{2}\)[/tex]:
[tex]\[ \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = \frac{(\sqrt{3} + \sqrt{2})^2}{(\sqrt{3})^2 - (\sqrt{2})^2} = \frac{3 + 2\sqrt{6} + 2}{1} = \sqrt{3} + \sqrt{2} \][/tex]
To summarize, our simplified results are:
a. [tex]\(\sqrt{2}\)[/tex]
b. [tex]\(\frac{5\sqrt{3}}{3}\)[/tex]
c. [tex]\(\frac{3\sqrt{2}}{4}\)[/tex]
d. [tex]\(\frac{\sqrt{6}}{2}\)[/tex]
e. [tex]\(\frac{\sqrt{30}}{30}\)[/tex]
f. [tex]\(\sqrt{2} - 1\)[/tex]
g. [tex]\(\sqrt{3} + 1\)[/tex]
h. [tex]\(7(\sqrt{3} + \sqrt{2})\)[/tex]
i. [tex]\(\sqrt{3} + \sqrt{2}\)[/tex]
## Part (a)
[tex]\[ \frac{2}{\sqrt{2}} \][/tex]
1. Multiply the numerator and the denominator by [tex]\(\sqrt{2}\)[/tex]:
[tex]\[ \frac{2}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{2}}{2} = \sqrt{2} \][/tex]
## Part (b)
[tex]\[ \frac{5}{\sqrt{3}} \][/tex]
1. Multiply the numerator and the denominator by [tex]\(\sqrt{3}\)[/tex]:
[tex]\[ \frac{5}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{3} \][/tex]
## Part (c)
[tex]\[ \frac{3}{2\sqrt{2}} \][/tex]
1. Multiply the numerator and the denominator by [tex]\(\sqrt{2}\)[/tex]:
[tex]\[ \frac{3}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{2 \times 2} = \frac{3\sqrt{2}}{4} \][/tex]
## Part (d)
[tex]\[ \frac{3\sqrt{2}}{2\sqrt{3}} \][/tex]
1. Multiply the numerator and the denominator by [tex]\(\sqrt{3}\)[/tex]:
[tex]\[ \frac{3\sqrt{2}}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{2}\sqrt{3}}{2 \times 3} = \frac{3\sqrt{6}}{6} = \frac{\sqrt{6}}{2} \][/tex]
## Part (e)
[tex]\[ \frac{\sqrt{5}}{5\sqrt{6}} \][/tex]
1. Multiply the numerator and the denominator by [tex]\(\sqrt{6}\)[/tex]:
[tex]\[ \frac{\sqrt{5}}{5\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{\sqrt{5}\sqrt{6}}{5 \times 6} = \frac{\sqrt{30}}{30} \][/tex]
## Part (f)
[tex]\[ \frac{1}{\sqrt{2} + 1} \][/tex]
1. Multiply the numerator and the denominator by the conjugate [tex]\(\sqrt{2} - 1\)[/tex]:
[tex]\[ \frac{1}{\sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \frac{\sqrt{2} - 1}{(\sqrt{2})^2 - (1)^2} = \frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1 \][/tex]
## Part (g)
[tex]\[ \frac{2}{\sqrt{3} - 1} \][/tex]
1. Multiply the numerator and the denominator by the conjugate [tex]\(\sqrt{3} + 1\)[/tex]:
[tex]\[ \frac{2}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{2(\sqrt{3} + 1)}{(\sqrt{3})^2 - (1)^2} = \frac{2\sqrt{3} + 2}{3 - 1} = \frac{2\sqrt{3} + 2}{2} = \sqrt{3} + 1 \][/tex]
## Part (h)
[tex]\[ \frac{7}{\sqrt{3} - \sqrt{2}} \][/tex]
1. Multiply the numerator and the denominator by the conjugate [tex]\(\sqrt{3} + \sqrt{2}\)[/tex]:
[tex]\[ \frac{7}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = \frac{7(\sqrt{3} + \sqrt{2})}{(\sqrt{3})^2 - (\sqrt{2})^2} = \frac{7(\sqrt{3} + \sqrt{2})}{3 - 2} = 7(\sqrt{3} + \sqrt{2}) \][/tex]
## Part (i)
[tex]\[ \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \][/tex]
1. Multiply the numerator and the denominator by the conjugate [tex]\(\sqrt{3} + \sqrt{2}\)[/tex]:
[tex]\[ \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = \frac{(\sqrt{3} + \sqrt{2})^2}{(\sqrt{3})^2 - (\sqrt{2})^2} = \frac{3 + 2\sqrt{6} + 2}{1} = \sqrt{3} + \sqrt{2} \][/tex]
To summarize, our simplified results are:
a. [tex]\(\sqrt{2}\)[/tex]
b. [tex]\(\frac{5\sqrt{3}}{3}\)[/tex]
c. [tex]\(\frac{3\sqrt{2}}{4}\)[/tex]
d. [tex]\(\frac{\sqrt{6}}{2}\)[/tex]
e. [tex]\(\frac{\sqrt{30}}{30}\)[/tex]
f. [tex]\(\sqrt{2} - 1\)[/tex]
g. [tex]\(\sqrt{3} + 1\)[/tex]
h. [tex]\(7(\sqrt{3} + \sqrt{2})\)[/tex]
i. [tex]\(\sqrt{3} + \sqrt{2}\)[/tex]
