Math worksheet for rationalizing denominators with various fractions involving square roots.

You're asked to rationalize the denominator of each fraction — that means eliminate any square roots from the denominator by multiplying numerator and denominator by a suitable form of 1 (usually the radical itself or its conjugate). Then simplify if possible.

Let’s go through each one step by step.

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

Multiply numerator and denominator by $\sqrt{2}$:

$$
\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}
$$

✔ Answer: $\boxed{\dfrac{\sqrt{2}}{2}}$

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2. $\frac{6}{\sqrt{2}}$

Multiply numerator and denominator by $\sqrt{2}$:

$$
\frac{6}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{6\sqrt{2}}{2} = 3\sqrt{2}
$$

✔ Answer: $\boxed{3\sqrt{2}}$

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

Multiply numerator and denominator by $\sqrt{3}$:

$$
\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3}
$$

✔ Answer: $\boxed{\dfrac{2\sqrt{3}}{3}}$

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4. $\frac{1}{3\sqrt{12}}$

First, simplify $\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}$, so:

$$
\frac{1}{3 \cdot 2\sqrt{3}} = \frac{1}{6\sqrt{3}}
$$

Now rationalize:

$$
\frac{1}{6\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{6 \cdot 3} = \frac{\sqrt{3}}{18}
$$

✔ Answer: $\boxed{\dfrac{\sqrt{3}}{18}}$

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

Multiply numerator and denominator by $\sqrt{5}$:

$$
\frac{15}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{15\sqrt{5}}{5} = 3\sqrt{5}
$$

✔ Answer: $\boxed{3\sqrt{5}}$

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6. $\frac{12}{\sqrt{6}}$

Multiply numerator and denominator by $\sqrt{6}$:

$$
\frac{12}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{12\sqrt{6}}{6} = 2\sqrt{6}
$$

✔ Answer: $\boxed{2\sqrt{6}}$

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7. $\frac{42}{\sqrt{7}}$

Multiply numerator and denominator by $\sqrt{7}$:

$$
\frac{42}{\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{42\sqrt{7}}{7} = 6\sqrt{7}
$$

✔ Answer: $\boxed{6\sqrt{7}}$

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8. $\frac{8}{3\sqrt{2}}$

Multiply numerator and denominator by $\sqrt{2}$:

$$
\frac{8}{3\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{8\sqrt{2}}{3 \cdot 2} = \frac{8\sqrt{2}}{6} = \frac{4\sqrt{2}}{3}
$$

✔ Answer: $\boxed{\dfrac{4\sqrt{2}}{3}}$

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9. $\frac{2}{\sqrt{11}}$

Multiply numerator and denominator by $\sqrt{11}$:

$$
\frac{2}{\sqrt{11}} \cdot \frac{\sqrt{11}}{\sqrt{11}} = \frac{2\sqrt{11}}{11}
$$

✔ Answer: $\boxed{\dfrac{2\sqrt{11}}{11}}$

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10. $\frac{3}{2\sqrt{7}}$

Multiply numerator and denominator by $\sqrt{7}$:

$$
\frac{3}{2\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{3\sqrt{7}}{2 \cdot 7} = \frac{3\sqrt{7}}{14}
$$

✔ Answer: $\boxed{\dfrac{3\sqrt{7}}{14}}$

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## ✔ Final Answers:

1. $\boxed{\dfrac{\sqrt{2}}{2}}$
2. $\boxed{3\sqrt{2}}$
3. $\boxed{\dfrac{2\sqrt{3}}{3}}$
4. $\boxed{\dfrac{\sqrt{3}}{18}}$
5. $\boxed{3\sqrt{5}}$
6. $\boxed{2\sqrt{6}}$
7. $\boxed{6\sqrt{7}}$
8. $\boxed{\dfrac{4\sqrt{2}}{3}}$
9. $\boxed{\dfrac{2\sqrt{11}}{11}}$
10. $\boxed{\dfrac{3\sqrt{7}}{14}}$

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## 💡 Key Concept:
To rationalize a denominator with a single square root, multiply numerator and denominator by that same square root. This turns the denominator into a rational number (since $\sqrt{a} \cdot \sqrt{a} = a$).

If there’s a coefficient (like $3\sqrt{2}$), treat it as part of the denominator — still multiply by the radical only (not the coefficient) to eliminate the root.

Always simplify the resulting fraction if possible!