Worksheet for practicing rationalizing denominators in algebra.

The task involves rationalizing the denominator of various fractions. Rationalizing the denominator means eliminating any radicals (square roots, cube roots, etc.) from the denominator of a fraction. This is typically done by multiplying both the numerator and the denominator by an appropriate expression that removes the radical.

Let's solve each problem step by step:

---

1. $\frac{1}{\sqrt{2}}$

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{2}$:
$$
\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}
$$
So, the answer is:
$$
\boxed{\frac{\sqrt{2}}{2}}
$$

---

2. $\frac{3}{\sqrt{3}}$

Multiply the numerator and denominator by $\sqrt{3}$:
$$
\frac{3}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}}{3} = \sqrt{3}
$$
So, the answer is:
$$
\boxed{\sqrt{3}}
$$

---

3. $\frac{12}{\sqrt{8}}$

First, simplify $\sqrt{8}$:
$$
\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}
$$
Now, rewrite the fraction:
$$
\frac{12}{\sqrt{8}} = \frac{12}{2\sqrt{2}} = \frac{6}{\sqrt{2}}
$$
Rationalize by multiplying by $\sqrt{2}$:
$$
\frac{6}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{6\sqrt{2}}{2} = 3\sqrt{2}
$$
So, the answer is:
$$
\boxed{3\sqrt{2}}
$$

---

4. $\frac{2}{\sqrt{3}}$

Multiply the numerator and denominator by $\sqrt{3}$:
$$
\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3}
$$
So, the answer is:
$$
\boxed{\frac{2\sqrt{3}}{3}}
$$

---

5. $\frac{10\sqrt{2}}{\sqrt{5}}$

Multiply the numerator and denominator by $\sqrt{5}$:
$$
\frac{10\sqrt{2}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{10\sqrt{2} \cdot \sqrt{5}}{5} = \frac{10\sqrt{10}}{5} = 2\sqrt{10}
$$
So, the answer is:
$$
\boxed{2\sqrt{10}}
$$

---

6. $\frac{7}{\sqrt{72}}$

Simplify $\sqrt{72}$:
$$
\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}
$$
Rewrite the fraction:
$$
\frac{7}{\sqrt{72}} = \frac{7}{6\sqrt{2}}
$$
Rationalize by multiplying by $\sqrt{2}$:
$$
\frac{7}{6\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{7\sqrt{2}}{6 \cdot 2} = \frac{7\sqrt{2}}{12}
$$
So, the answer is:
$$
\boxed{\frac{7\sqrt{2}}{12}}
$$

---

7. $\frac{5\sqrt{3}}{\sqrt{15}}$

Simplify $\sqrt{15}$:
$$
\sqrt{15} = \sqrt{3 \cdot 5}
$$
Rewrite the fraction:
$$
\frac{5\sqrt{3}}{\sqrt{15}} = \frac{5\sqrt{3}}{\sqrt{3 \cdot 5}} = \frac{5\sqrt{3}}{\sqrt{3} \cdot \sqrt{5}} = \frac{5}{\sqrt{5}}
$$
Rationalize by multiplying by $\sqrt{5}$:
$$
\frac{5}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{5\sqrt{5}}{5} = \sqrt{5}
$$
So, the answer is:
$$
\boxed{\sqrt{5}}
$$

---

8. $\frac{8}{\sqrt{2}}$

Multiply the numerator and denominator by $\sqrt{2}$:
$$
\frac{8}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{8\sqrt{2}}{2} = 4\sqrt{2}
$$
So, the answer is:
$$
\boxed{4\sqrt{2}}
$$

---

9. $\frac{8 + \sqrt{12}}{\sqrt{3}}$

Simplify $\sqrt{12}$:
$$
\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}
$$
Rewrite the fraction:
$$
\frac{8 + \sqrt{12}}{\sqrt{3}} = \frac{8 + 2\sqrt{3}}{\sqrt{3}}
$$
Separate into two fractions:
$$
\frac{8 + 2\sqrt{3}}{\sqrt{3}} = \frac{8}{\sqrt{3}} + \frac{2\sqrt{3}}{\sqrt{3}}
$$
Simplify each term:
$$
\frac{8}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{8\sqrt{3}}{3}, \quad \frac{2\sqrt{3}}{\sqrt{3}} = 2
$$
Combine:
$$
\frac{8\sqrt{3}}{3} + 2 = \frac{8\sqrt{3}}{3} + \frac{6}{3} = \frac{8\sqrt{3} + 6}{3}
$$
So, the answer is:
$$
\boxed{\frac{8\sqrt{3} + 6}{3}}
$$

---

10. $\frac{4}{\sqrt{27}}$

Simplify $\sqrt{27}$:
$$
\sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3}
$$
Rewrite the fraction:
$$
\frac{4}{\sqrt{27}} = \frac{4}{3\sqrt{3}}
$$
Rationalize by multiplying by $\sqrt{3}$:
$$
\frac{4}{3\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{3}}{3 \cdot 3} = \frac{4\sqrt{3}}{9}
$$
So, the answer is:
$$
\boxed{\frac{4\sqrt{3}}{9}}
$$

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Final Answer:

$$
\boxed{\frac{\sqrt{2}}{2}, \sqrt{3}, 3\sqrt{2}, \frac{2\sqrt{3}}{3}, 2\sqrt{10}, \frac{7\sqrt{2}}{12}, \sqrt{5}, 4\sqrt{2}, \frac{8\sqrt{3} + 6}{3}, \frac{4\sqrt{3}}{9}}
$$