Rationalizing denominators involving square roots and algebraic expressions.

It looks like you've uploaded an image containing a list of expressions that need to be rationalized — that is, simplified so that the denominator does not contain any radicals (square roots). These are typical algebra problems involving rationalizing denominators.

Let’s go through each one step by step. The goal is to eliminate radicals from the denominator using conjugates or other algebraic techniques.

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(i) $\frac{1}{3 + \sqrt{2}}$

Step: Multiply numerator and denominator by the conjugate $3 - \sqrt{2}$:

$$
\frac{1}{3 + \sqrt{2}} \cdot \frac{3 - \sqrt{2}}{3 - \sqrt{2}} = \frac{3 - \sqrt{2}}{(3)^2 - (\sqrt{2})^2} = \frac{3 - \sqrt{2}}{9 - 2} = \frac{3 - \sqrt{2}}{7}
$$

✔ Answer: $\boxed{\frac{3 - \sqrt{2}}{7}}$

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(ii) $\frac{1}{\sqrt{6} - \sqrt{5}}$

Multiply by conjugate $\sqrt{6} + \sqrt{5}$:

$$
\frac{1}{\sqrt{6} - \sqrt{5}} \cdot \frac{\sqrt{6} + \sqrt{5}}{\sqrt{6} + \sqrt{5}} = \frac{\sqrt{6} + \sqrt{5}}{(\sqrt{6})^2 - (\sqrt{5})^2} = \frac{\sqrt{6} + \sqrt{5}}{6 - 5} = \sqrt{6} + \sqrt{5}
$$

✔ Answer: $\boxed{\sqrt{6} + \sqrt{5}}$

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(iii) $\frac{16}{\sqrt{41} - 5}$

Multiply by conjugate $\sqrt{41} + 5$:

$$
\frac{16}{\sqrt{41} - 5} \cdot \frac{\sqrt{41} + 5}{\sqrt{41} + 5} = \frac{16(\sqrt{41} + 5)}{(\sqrt{41})^2 - (5)^2} = \frac{16(\sqrt{41} + 5)}{41 - 25} = \frac{16(\sqrt{41} + 5)}{16} = \sqrt{41} + 5
$$

✔ Answer: $\boxed{\sqrt{41} + 5}$

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(iv) $\frac{30}{5\sqrt{3} - 3\sqrt{5}}$

Factor denominator: $= \sqrt{3}(5) - \sqrt{5}(3)$

We can rationalize by multiplying numerator and denominator by the conjugate $5\sqrt{3} + 3\sqrt{5}$:

$$
\frac{30}{5\sqrt{3} - 3\sqrt{5}} \cdot \frac{5\sqrt{3} + 3\sqrt{5}}{5\sqrt{3} + 3\sqrt{5}} = \frac{30(5\sqrt{3} + 3\sqrt{5})}{(5\sqrt{3})^2 - (3\sqrt{5})^2}
$$

Compute denominator:

$$
(25 \cdot 3) - (9 \cdot 5) = 75 - 45 = 30
$$

So:

$$
\frac{30(5\sqrt{3} + 3\sqrt{5})}{30} = 5\sqrt{3} + 3\sqrt{5}
$$

✔ Answer: $\boxed{5\sqrt{3} + 3\sqrt{5}}$

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(v) $\frac{1}{2\sqrt{5} - \sqrt{3}}$

Multiply by conjugate $2\sqrt{5} + \sqrt{3}$:

$$
\frac{1}{2\sqrt{5} - \sqrt{3}} \cdot \frac{2\sqrt{5} + \sqrt{3}}{2\sqrt{5} + \sqrt{3}} = \frac{2\sqrt{5} + \sqrt{3}}{(2\sqrt{5})^2 - (\sqrt{3})^2} = \frac{2\sqrt{5} + \sqrt{3}}{4 \cdot 5 - 3} = \frac{2\sqrt{5} + \sqrt{3}}{20 - 3} = \frac{2\sqrt{5} + \sqrt{3}}{17}
$$

✔ Answer: $\boxed{\frac{2\sqrt{5} + \sqrt{3}}{17}}$

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(vi) $\frac{\sqrt{3} + 1}{2\sqrt{2} - \sqrt{3}}$

Multiply numerator and denominator by conjugate $2\sqrt{2} + \sqrt{3}$:

$$
\frac{\sqrt{3} + 1}{2\sqrt{2} - \sqrt{3}} \cdot \frac{2\sqrt{2} + \sqrt{3}}{2\sqrt{2} + \sqrt{3}} = \frac{(\sqrt{3} + 1)(2\sqrt{2} + \sqrt{3})}{(2\sqrt{2})^2 - (\sqrt{3})^2}
$$

Denominator:

$$
8 - 3 = 5
$$

Numerator:

$$
(\sqrt{3})(2\sqrt{2}) + (\sqrt{3})(\sqrt{3}) + (1)(2\sqrt{2}) + (1)(\sqrt{3}) = 2\sqrt{6} + 3 + 2\sqrt{2} + \sqrt{3}
$$

So:

$$
\frac{2\sqrt{6} + 2\sqrt{2} + \sqrt{3} + 3}{5}
$$

✔ Answer: $\boxed{\frac{2\sqrt{6} + 2\sqrt{2} + \sqrt{3} + 3}{5}}$

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(vii) $\frac{6 - 4\sqrt{2}}{6 + 4\sqrt{2}}$

This is a fraction with radical in both numerator and denominator. Rationalize by multiplying numerator and denominator by the conjugate $6 - 4\sqrt{2}$:

Wait — actually, the denominator is $6 + 4\sqrt{2}$, so conjugate is $6 - 4\sqrt{2}$:

$$
\frac{6 - 4\sqrt{2}}{6 + 4\sqrt{2}} \cdot \frac{6 - 4\sqrt{2}}{6 - 4\sqrt{2}} = \frac{(6 - 4\sqrt{2})^2}{(6)^2 - (4\sqrt{2})^2}
$$

Numerator:

$$
(6)^2 - 2 \cdot 6 \cdot 4\sqrt{2} + (4\sqrt{2})^2 = 36 - 48\sqrt{2} + 16 \cdot 2 = 36 - 48\sqrt{2} + 32 = 68 - 48\sqrt{2}
$$

Denominator:

$$
36 - (16 \cdot 2) = 36 - 32 = 4
$$

So:

$$
\frac{68 - 48\sqrt{2}}{4} = 17 - 12\sqrt{2}
$$

✔ Answer: $\boxed{17 - 12\sqrt{2}}$

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(viii) $\frac{3\sqrt{2} + 1}{2\sqrt{5} - 3}$

Multiply by conjugate $2\sqrt{5} + 3$:

$$
\frac{3\sqrt{2} + 1}{2\sqrt{5} - 3} \cdot \frac{2\sqrt{5} + 3}{2\sqrt{5} + 3} = \frac{(3\sqrt{2} + 1)(2\sqrt{5} + 3)}{(2\sqrt{5})^2 - (3)^2}
$$

Denominator:

$$
4 \cdot 5 - 9 = 20 - 9 = 11
$$

Numerator:

$$
3\sqrt{2} \cdot 2\sqrt{5} = 6\sqrt{10},\quad 3\sqrt{2} \cdot 3 = 9\sqrt{2},\quad 1 \cdot 2\sqrt{5} = 2\sqrt{5},\quad 1 \cdot 3 = 3
$$

So total numerator: $6\sqrt{10} + 9\sqrt{2} + 2\sqrt{5} + 3$

Thus:

$$
\frac{6\sqrt{10} + 9\sqrt{2} + 2\sqrt{5} + 3}{11}
$$

✔ Answer: $\boxed{\frac{6\sqrt{10} + 9\sqrt{2} + 2\sqrt{5} + 3}{11}}$

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(ix) $\frac{b^2}{\sqrt{a^2 + b^2} + a}$

This one is trickier. We want to rationalize the denominator.

Multiply numerator and denominator by the conjugate $\sqrt{a^2 + b^2} - a$:

$$
\frac{b^2}{\sqrt{a^2 + b^2} + a} \cdot \frac{\sqrt{a^2 + b^2} - a}{\sqrt{a^2 + b^2} - a} = \frac{b^2(\sqrt{a^2 + b^2} - a)}{(\sqrt{a^2 + b^2})^2 - a^2}
$$

Denominator:

$$
(a^2 + b^2) - a^2 = b^2
$$

So:

$$
\frac{b^2(\sqrt{a^2 + b^2} - a)}{b^2} = \sqrt{a^2 + b^2} - a
$$

✔ Answer: $\boxed{\sqrt{a^2 + b^2} - a}$

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| (i) | $\frac{3 - \sqrt{2}}{7}$ |
| (ii) | $\sqrt{6} + \sqrt{5}$ |
| (iii) | $\sqrt{41} + 5$ |
| (iv) | $5\sqrt{3} + 3\sqrt{5}$ |
| (v) | $\frac{2\sqrt{5} + \sqrt{3}}{17}$ |
| (vi) | $\frac{2\sqrt{6} + 2\sqrt{2} + \sqrt{3} + 3}{5}$ |
| (vii) | $17 - 12\sqrt{2}$ |
| (viii) | $\frac{6\sqrt{10} + 9\sqrt{2} + 2\sqrt{5} + 3}{11}$ |
| (ix) | $\sqrt{a^2 + b^2} - a$ |

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