Math exercises for rationalizing denominators and standardizing fractions with radicals.

Problem Overview:

The task involves rationalizing denominators and standardizing fractions. Rationalizing the denominator means eliminating any square roots or other radicals from the denominator by multiplying both the numerator and the denominator by a suitable expression. Standardizing fractions often involves simplifying them to their most basic form.

We will solve each part step by step.

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Section A: Rationalize each denominator

#### 1) \( \frac{5}{2\sqrt{6}} \)

To rationalize the denominator, multiply the numerator and the denominator by \( \sqrt{6} \):

\[
\frac{5}{2\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{5\sqrt{6}}{2 \cdot 6} = \frac{5\sqrt{6}}{12}
\]

Answer: \( \frac{5\sqrt{6}}{12} \)

---

#### 2) \( \frac{\sqrt{3}}{4\sqrt{6}} \)

Multiply the numerator and the denominator by \( \sqrt{6} \):

\[
\frac{\sqrt{3}}{4\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{\sqrt{3} \cdot \sqrt{6}}{4 \cdot 6} = \frac{\sqrt{18}}{24}
\]

Simplify \( \sqrt{18} \):

\[
\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}
\]

So,

\[
\frac{\sqrt{18}}{24} = \frac{3\sqrt{2}}{24} = \frac{\sqrt{2}}{8}
\]

Answer: \( \frac{\sqrt{2}}{8} \)

---

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

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

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

Answer: \( \frac{5\sqrt{3} - 3}{3} \)

---

#### 4) \( \frac{3\sqrt{6} + 4}{2\sqrt{6}} \)

Multiply the numerator and the denominator by \( \sqrt{6} \):

\[
\frac{3\sqrt{6} + 4}{2\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{(3\sqrt{6} + 4)\sqrt{6}}{2 \cdot 6} = \frac{3\sqrt{6} \cdot \sqrt{6} + 4\sqrt{6}}{12} = \frac{3 \cdot 6 + 4\sqrt{6}}{12} = \frac{18 + 4\sqrt{6}}{12}
\]

Simplify the fraction:

\[
\frac{18 + 4\sqrt{6}}{12} = \frac{18}{12} + \frac{4\sqrt{6}}{12} = \frac{3}{2} + \frac{\sqrt{6}}{3}
\]

Answer: \( \frac{3}{2} + \frac{\sqrt{6}}{3} \)

---

Section B: Standardize these fractions

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

Rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator, \( 5 - \sqrt{2} \):

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

Simplify the denominator using the difference of squares:

\[
(5 + \sqrt{2})(5 - \sqrt{2}) = 5^2 - (\sqrt{2})^2 = 25 - 2 = 23
\]

So,

\[
\frac{2(5 - \sqrt{2})}{23} = \frac{10 - 2\sqrt{2}}{23}
\]

Answer: \( \frac{10 - 2\sqrt{2}}{23} \)

---

#### 2) \( \frac{5}{7 - \sqrt{5}} \)

Rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator, \( 7 + \sqrt{5} \):

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

Simplify the denominator using the difference of squares:

\[
(7 - \sqrt{5})(7 + \sqrt{5}) = 7^2 - (\sqrt{5})^2 = 49 - 5 = 44
\]

So,

\[
\frac{5(7 + \sqrt{5})}{44} = \frac{35 + 5\sqrt{5}}{44}
\]

Answer: \( \frac{35 + 5\sqrt{5}}{44} \)

---

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

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

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

Simplify the denominator using the difference of squares:

\[
(5 - \sqrt{3})(5 + \sqrt{3}) = 5^2 - (\sqrt{3})^2 = 25 - 3 = 22
\]

So,

\[
\frac{\sqrt{3}(5 + \sqrt{3})}{22} = \frac{5\sqrt{3} + 3}{22}
\]

Answer: \( \frac{5\sqrt{3} + 3}{22} \)

---

#### 4) \( \frac{\sqrt{6}}{\sqrt{6} + 3} \)

Rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator, \( \sqrt{6} - 3 \):

\[
\frac{\sqrt{6}}{\sqrt{6} + 3} \cdot \frac{\sqrt{6} - 3}{\sqrt{6} - 3} = \frac{\sqrt{6}(\sqrt{6} - 3)}{(\sqrt{6} + 3)(\sqrt{6} - 3)}
\]

Simplify the denominator using the difference of squares:

\[
(\sqrt{6} + 3)(\sqrt{6} - 3) = (\sqrt{6})^2 - 3^2 = 6 - 9 = -3
\]

So,

\[
\frac{\sqrt{6}(\sqrt{6} - 3)}{-3} = \frac{6 - 3\sqrt{6}}{-3} = \frac{6}{-3} - \frac{3\sqrt{6}}{-3} = -2 + \sqrt{6}
\]

Answer: \( -2 + \sqrt{6} \)

---

Section C: Rationalize each denominator

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

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

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

Simplify the denominator using the difference of squares:

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

Expand the numerator:

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

So,

\[
\frac{1 + 2\sqrt{2}}{7}
\]

Answer: \( \frac{1 + 2\sqrt{2}}{7} \)

---

#### 2) \( \frac{2 - \sqrt{6}}{4 - \sqrt{6}} \)

Rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator, \( 4 + \sqrt{6} \):

\[
\frac{2 - \sqrt{6}}{4 - \sqrt{6}} \cdot \frac{4 + \sqrt{6}}{4 + \sqrt{6}} = \frac{(2 - \sqrt{6})(4 + \sqrt{6})}{(4 - \sqrt{6})(4 + \sqrt{6})}
\]

Simplify the denominator using the difference of squares:

\[
(4 - \sqrt{6})(4 + \sqrt{6}) = 4^2 - (\sqrt{6})^2 = 16 - 6 = 10
\]

Expand the numerator:

\[
(2 - \sqrt{6})(4 + \sqrt{6}) = 2 \cdot 4 + 2 \cdot \sqrt{6} - \sqrt{6} \cdot 4 - \sqrt{6} \cdot \sqrt{6} = 8 + 2\sqrt{6} - 4\sqrt{6} - 6 = 2 - 2\sqrt{6}
\]

So,

\[
\frac{2 - 2\sqrt{6}}{10} = \frac{2(1 - \sqrt{6})}{10} = \frac{1 - \sqrt{6}}{5}
\]

Answer: \( \frac{1 - \sqrt{6}}{5} \)

---

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

Rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator, \( 5 - \sqrt{3} \):

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

Simplify the denominator using the difference of squares:

\[
(5 + \sqrt{3})(5 - \sqrt{3}) = 5^2 - (\sqrt{3})^2 = 25 - 3 = 22
\]

Expand the numerator:

\[
(3 - \sqrt{3})(5 - \sqrt{3}) = 3 \cdot 5 + 3 \cdot (-\sqrt{3}) - \sqrt{3} \cdot 5 - \sqrt{3} \cdot (-\sqrt{3}) = 15 - 3\sqrt{3} - 5\sqrt{3} + 3 = 18 - 8\sqrt{3}
\]

So,

\[
\frac{18 - 8\sqrt{3}}{22} = \frac{2(9 - 4\sqrt{3})}{22} = \frac{9 - 4\sqrt{3}}{11}
\]

Answer: \( \frac{9 - 4\sqrt{3}}{11} \)

---

#### 4) \( \frac{3 + 3\sqrt{7}}{\sqrt{7} - 2} \)

Rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator, \( \sqrt{7} + 2 \):

\[
\frac{3 + 3\sqrt{7}}{\sqrt{7} - 2} \cdot \frac{\sqrt{7} + 2}{\sqrt{7} + 2} = \frac{(3 + 3\sqrt{7})(\sqrt{7} + 2)}{(\sqrt{7} - 2)(\sqrt{7} + 2)}
\]

Simplify the denominator using the difference of squares:

\[
(\sqrt{7} - 2)(\sqrt{7} + 2) = (\sqrt{7})^2 - 2^2 = 7 - 4 = 3
\]

Expand the numerator:

\[
(3 + 3\sqrt{7})(\sqrt{7} + 2) = 3 \cdot \sqrt{7} + 3 \cdot 2 + 3\sqrt{7} \cdot \sqrt{7} + 3\sqrt{7} \cdot 2 = 3\sqrt{7} + 6 + 3 \cdot 7 + 6\sqrt{7} = 3\sqrt{7} + 6 + 21 + 6\sqrt{7} = 27 + 9\sqrt{7}
\]

So,

\[
\frac{27 + 9\sqrt{7}}{3} = 9 + 3\sqrt{7}
\]

Answer: \( 9 + 3\sqrt{7} \)

---

Final Answers:

\[
\boxed{
\begin{array}{lll}
\text{A1: } & \frac{5\sqrt{6}}{12} & \text{B1: } \frac{10 - 2\sqrt{2}}{23} & \text{C1: } \frac{1 + 2\sqrt{2}}{7} \\
\text{A2: } & \frac{\sqrt{2}}{8} & \text{B2: } \frac{35 + 5\sqrt{5}}{44} & \text{C2: } \frac{1 - \sqrt{6}}{5} \\
\text{A3: } & \frac{5\sqrt{3} - 3}{3} & \text{B3: } \frac{5\sqrt{3} + 3}{22} & \text{C3: } \frac{9 - 4\sqrt{3}}{11} \\
\text{A4: } & \frac{3}{2} + \frac{\sqrt{6}}{3} & \text{B4: } -2 + \sqrt{6} & \text{C4: } 9 + 3\sqrt{7}
\end{array}
}
\]