Question

s of the square root of a natural number that is not a perfect square, ator and the denomiator by the ___________ number that produces th uare in the denominator.

1. $-\frac{6}{sqrt{2}}$
2. $\frac{2}{sqrt{3}}$
3. $\frac{7}{sqrt{6}}$
4. $\frac{4}{sqrt{6}}$
5. $-\frac{4}{sqrt{7}}$

Explanation:

3) Step1: Rationalize the denominator

Multiply numerator and denominator by $\sqrt{2}$:
$-\frac{6}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = -\frac{6\sqrt{2}}{\sqrt{2 \times 2}}$

3) Step2: Simplify the denominator

$\sqrt{2 \times 2} = \sqrt{4} = 2$, so:
$-\frac{6\sqrt{2}}{2} = -3\sqrt{2}$

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5) Step1: Rationalize the denominator

Multiply numerator and denominator by $\sqrt{3}$:
$\frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{\sqrt{3 \times 3}}$

5) Step2: Simplify the denominator

$\sqrt{3 \times 3} = \sqrt{9} = 3$, so:
$\frac{2\sqrt{3}}{3}$

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7) Step1: Rationalize the denominator

Multiply numerator and denominator by $\sqrt{6}$:
$\frac{7}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{7\sqrt{6}}{\sqrt{6 \times 6}}$

7) Step2: Simplify the denominator

$\sqrt{6 \times 6} = \sqrt{36} = 6$, so:
$\frac{7\sqrt{6}}{6}$

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9) Step1: Rationalize the denominator

Multiply numerator and denominator by $\sqrt{6}$:
$\frac{4}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{4\sqrt{6}}{\sqrt{6 \times 6}}$

9) Step2: Simplify the denominator

$\sqrt{6 \times 6} = \sqrt{36} = 6$, so:
$\frac{4\sqrt{6}}{6} = \frac{2\sqrt{6}}{3}$

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11) Step1: Rationalize the denominator

Multiply numerator and denominator by $\sqrt{7}$:
$-\frac{4}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = -\frac{4\sqrt{7}}{\sqrt{7 \times 7}}$

11) Step2: Simplify the denominator

$\sqrt{7 \times 7} = \sqrt{49} = 7$, so:
$-\frac{4\sqrt{7}}{7}$

Answer:

1. $-3\sqrt{2}$
2. $\frac{2\sqrt{3}}{3}$
3. $\frac{7\sqrt{6}}{6}$
4. $\frac{2\sqrt{6}}{3}$
5. $-\frac{4\sqrt{7}}{7}$