Worksheet for rationalizing denominators in algebraic fractions, showing examples with square roots.

Problem Overview:

The task is to rationalize the denominators of the given expressions. Rationalizing the denominator means eliminating any radicals (square roots, cube roots, etc.) from the denominator by multiplying both the numerator and the denominator by an appropriate expression.

We will solve each part step by step:

---

Part 1: Rationalizing by multiplying by one term

#### A) \( \frac{2 + \sqrt{3}}{\sqrt{3}} \)

To rationalize, multiply the numerator and denominator by \( \sqrt{3} \):

\[
\frac{2 + \sqrt{3}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{(2 + \sqrt{3}) \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \frac{2\sqrt{3} + 3}{3}
\]

So, the rationalized form is:

\[
\boxed{\frac{2\sqrt{3} + 3}{3}}
\]

#### B) \( \frac{5 + \sqrt{2}}{\sqrt{2}} \)

Multiply the numerator and denominator by \( \sqrt{2} \):

\[
\frac{5 + \sqrt{2}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{(5 + \sqrt{2}) \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{5\sqrt{2} + 2}{2}
\]

So, the rationalized form is:

\[
\boxed{\frac{5\sqrt{2} + 2}{2}}
\]

#### C) \( \frac{\sqrt{5} + 3}{\sqrt{5}} \)

Multiply the numerator and denominator by \( \sqrt{5} \):

\[
\frac{\sqrt{5} + 3}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{(\sqrt{5} + 3) \cdot \sqrt{5}}{\sqrt{5} \cdot \sqrt{5}} = \frac{5 + 3\sqrt{5}}{5}
\]

So, the rationalized form is:

\[
\boxed{\frac{5 + 3\sqrt{5}}{5}}
\]

#### F) \( \frac{\sqrt{3} - 2}{\sqrt{3}} \)

Multiply the numerator and denominator by \( \sqrt{3} \):

\[
\frac{\sqrt{3} - 2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{(\sqrt{3} - 2) \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \frac{3 - 2\sqrt{3}}{3}
\]

So, the rationalized form is:

\[
\boxed{\frac{3 - 2\sqrt{3}}{3}}
\]

#### G) \( \frac{3 + 2\sqrt{5}}{\sqrt{5}} \)

Multiply the numerator and denominator by \( \sqrt{5} \):

\[
\frac{3 + 2\sqrt{5}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{(3 + 2\sqrt{5}) \cdot \sqrt{5}}{\sqrt{5} \cdot \sqrt{5}} = \frac{3\sqrt{5} + 10}{5}
\]

So, the rationalized form is:

\[
\boxed{\frac{3\sqrt{5} + 10}{5}}
\]

#### H) \( \frac{\sqrt{10} + 3}{\sqrt{2}} \)

Multiply the numerator and denominator by \( \sqrt{2} \):

\[
\frac{\sqrt{10} + 3}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{(\sqrt{10} + 3) \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{\sqrt{20} + 3\sqrt{2}}{2} = \frac{2\sqrt{5} + 3\sqrt{2}}{2}
\]

So, the rationalized form is:

\[
\boxed{\frac{2\sqrt{5} + 3\sqrt{2}}{2}}
\]

---

Part 2: Rationalizing by multiplying by two terms within a bracket

#### K) \( \frac{5}{\sqrt{2} + 3} \)

Multiply the numerator and denominator by the conjugate of the denominator, \( \sqrt{2} - 3 \):

\[
\frac{5}{\sqrt{2} + 3} \cdot \frac{\sqrt{2} - 3}{\sqrt{2} - 3} = \frac{5(\sqrt{2} - 3)}{(\sqrt{2} + 3)(\sqrt{2} - 3)}
\]

Simplify the denominator using the difference of squares:

\[
(\sqrt{2} + 3)(\sqrt{2} - 3) = (\sqrt{2})^2 - 3^2 = 2 - 9 = -7
\]

So, the expression becomes:

\[
\frac{5(\sqrt{2} - 3)}{-7} = \frac{5\sqrt{2} - 15}{-7} = \frac{-(5\sqrt{2} - 15)}{7} = \frac{15 - 5\sqrt{2}}{7}
\]

So, the rationalized form is:

\[
\boxed{\frac{15 - 5\sqrt{2}}{7}}
\]

#### L) \( \frac{8}{1 + \sqrt{3}} \)

Multiply the numerator and denominator by the conjugate of the denominator, \( 1 - \sqrt{3} \):

\[
\frac{8}{1 + \sqrt{3}} \cdot \frac{1 - \sqrt{3}}{1 - \sqrt{3}} = \frac{8(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})}
\]

Simplify the denominator using the difference of squares:

\[
(1 + \sqrt{3})(1 - \sqrt{3}) = 1^2 - (\sqrt{3})^2 = 1 - 3 = -2
\]

So, the expression becomes:

\[
\frac{8(1 - \sqrt{3})}{-2} = \frac{8 - 8\sqrt{3}}{-2} = \frac{-(8 - 8\sqrt{3})}{2} = \frac{8\sqrt{3} - 8}{2} = 4\sqrt{3} - 4
\]

So, the rationalized form is:

\[
\boxed{4\sqrt{3} - 4}
\]

#### M) \( \frac{1}{\sqrt{5} - 2} \)

Multiply the numerator and denominator by the conjugate of the denominator, \( \sqrt{5} + 2 \):

\[
\frac{1}{\sqrt{5} - 2} \cdot \frac{\sqrt{5} + 2}{\sqrt{5} + 2} = \frac{1(\sqrt{5} + 2)}{(\sqrt{5} - 2)(\sqrt{5} + 2)}
\]

Simplify the denominator using the difference of squares:

\[
(\sqrt{5} - 2)(\sqrt{5} + 2) = (\sqrt{5})^2 - 2^2 = 5 - 4 = 1
\]

So, the expression becomes:

\[
\frac{\sqrt{5} + 2}{1} = \sqrt{5} + 2
\]

So, the rationalized form is:

\[
\boxed{\sqrt{5} + 2}
\]

#### P) \( \frac{5\sqrt{2}}{2\sqrt{5} - 3} \)

Multiply the numerator and denominator by the conjugate of the denominator, \( 2\sqrt{5} + 3 \):

\[
\frac{5\sqrt{2}}{2\sqrt{5} - 3} \cdot \frac{2\sqrt{5} + 3}{2\sqrt{5} + 3} = \frac{5\sqrt{2}(2\sqrt{5} + 3)}{(2\sqrt{5} - 3)(2\sqrt{5} + 3)}
\]

Simplify the denominator using the difference of squares:

\[
(2\sqrt{5} - 3)(2\sqrt{5} + 3) = (2\sqrt{5})^2 - 3^2 = 4 \cdot 5 - 9 = 20 - 9 = 11
\]

So, the expression becomes:

\[
\frac{5\sqrt{2}(2\sqrt{5} + 3)}{11} = \frac{10\sqrt{10} + 15\sqrt{2}}{11}
\]

So, the rationalized form is:

\[
\boxed{\frac{10\sqrt{10} + 15\sqrt{2}}{11}}
\]

#### Q) \( \frac{3\sqrt{7}}{5\sqrt{3} - 4} \)

Multiply the numerator and denominator by the conjugate of the denominator, \( 5\sqrt{3} + 4 \):

\[
\frac{3\sqrt{7}}{5\sqrt{3} - 4} \cdot \frac{5\sqrt{3} + 4}{5\sqrt{3} + 4} = \frac{3\sqrt{7}(5\sqrt{3} + 4)}{(5\sqrt{3} - 4)(5\sqrt{3} + 4)}
\]

Simplify the denominator using the difference of squares:

\[
(5\sqrt{3} - 4)(5\sqrt{3} + 4) = (5\sqrt{3})^2 - 4^2 = 25 \cdot 3 - 16 = 75 - 16 = 59
\]

So, the expression becomes:

\[
\frac{3\sqrt{7}(5\sqrt{3} + 4)}{59} = \frac{15\sqrt{21} + 12\sqrt{7}}{59}
\]

So, the rationalized form is:

\[
\boxed{\frac{15\sqrt{21} + 12\sqrt{7}}{59}}
\]

#### R) \( \frac{\sqrt{3} + 2}{\sqrt{3} - 1} \)

Multiply the numerator and denominator by the conjugate of the denominator, \( \sqrt{3} + 1 \):

\[
\frac{\sqrt{3} + 2}{\sqrt{3} - 1} \cdot \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} + 2)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}
\]

Simplify the denominator using the difference of squares:

\[
(\sqrt{3} - 1)(\sqrt{3} + 1) = (\sqrt{3})^2 - 1^2 = 3 - 1 = 2
\]

Expand the numerator:

\[
(\sqrt{3} + 2)(\sqrt{3} + 1) = \sqrt{3} \cdot \sqrt{3} + \sqrt{3} \cdot 1 + 2 \cdot \sqrt{3} + 2 \cdot 1 = 3 + \sqrt{3} + 2\sqrt{3} + 2 = 5 + 3\sqrt{3}
\]

So, the expression becomes:

\[
\frac{5 + 3\sqrt{3}}{2}
\]

So, the rationalized form is:

\[
\boxed{\frac{5 + 3\sqrt{3}}{2}}
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
&\text{A: } \frac{2\sqrt{3} + 3}{3}, \quad \text{B: } \frac{5\sqrt{2} + 2}{2}, \quad \text{C: } \frac{5 + 3\sqrt{5}}{5}, \quad \text{F: } \frac{3 - 2\sqrt{3}}{3}, \\
&\text{G: } \frac{3\sqrt{5} + 10}{5}, \quad \text{H: } \frac{2\sqrt{5} + 3\sqrt{2}}{2}, \quad \text{K: } \frac{15 - 5\sqrt{2}}{7}, \\
&\text{L: } 4\sqrt{3} - 4, \quad \text{M: } \sqrt{5} + 2, \quad \text{P: } \frac{10\sqrt{10} + 15\sqrt{2}}{11}, \\
&\text{Q: } \frac{15\sqrt{21} + 12\sqrt{7}}{59}, \quad \text{R: } \frac{5 + 3\sqrt{3}}{2}.
\end{aligned}
}
\]