\(\displaystyle \sqrt{81}/\sqrt{6}=9/\sqrt{6}\) We then multiply our fraction by \(\displaystyle \sqrt{6}/\sqrt{6}\) because we cannot leave a radical in the denominator. This gives us \(\displaystyle 9\sqrt{6}/6\). Finally, we can simplify our fraction, dividing out a 3, leaving us with \(\displaystyle 3\sqrt{6}/2\)

Let's combine the two radicals into one radical and simplify.

\(\displaystyle \frac{\sqrt{343x^5}}{\sqrt{49x^3}}=\sqrt{\frac{343x^5}{49x^3}}=\sqrt{7x^2}\) 

Remember, when dividing exponents of same base, just subtract the power.

The final answer is \(\displaystyle x\sqrt{7}\).

If we multiplied top and bottom by \(\displaystyle \sqrt{x}\), we would get nowhere, as this would result: \(\displaystyle \frac{1}{\sqrt{x}}*\frac{\sqrt{x}}{\sqrt{x}}=\frac{\sqrt{x}}{x}\). Instead, let's cross-multiply.

\(\displaystyle \frac{1}{\sqrt{x}}=\frac{4}{1}\)

\(\displaystyle 4\cdot \sqrt{x}=1\cdot 1\)

\(\displaystyle 4\sqrt{x}=1\) 

Then, square both sides to get rid of the radical.

\(\displaystyle (4\sqrt{x})^2=1^2\)

\(\displaystyle 16x=1\) 

Divide both sides by \(\displaystyle 16\).

\(\displaystyle x=\frac{1}{16}\) 

The reason the negative is not an answer is because a negative value in a radical is an imaginary number.

We don't want to have radicals in the denominator. To get rid of radicals, just multiply top and bottom by that radical.

\(\displaystyle \frac{2}{\sqrt{5}}\cdot\frac{\sqrt{5}}{\sqrt{5}}=\frac{2\sqrt{5}}{5}\)

There are two methods we can use to simplify this fraction:

Method 1:

Factor the numerator:

\(\displaystyle \frac{\sqrt{250}}{\sqrt{10}}=\frac{\sqrt{25}\cdot\sqrt{10}}{\sqrt{10}}=\sqrt{25}=5\) 

Remember, we need to factor out perfect squares.

Method 2: 

You can combine the fraction into one big square root.

\(\displaystyle \frac{\sqrt{250}}{\sqrt{10}}=\sqrt{\frac{250}{10}}\) 

Then, you can simplify the fraction.

\(\displaystyle \sqrt{\frac{250}{10}}=\sqrt{25}=5\)

Let's factor the square roots.

\(\displaystyle \frac{\sqrt{21}}{\sqrt{20}}=\frac{\sqrt{3}\cdot\sqrt{7}}{\sqrt{4}\cdot \sqrt{5}}=\frac{\sqrt{3}\cdot \sqrt{7}}{2\cdot \sqrt{5}}\) 

Then, multiply the numerator and the denominator by \(\displaystyle \sqrt{5}\) to get rid of the radical in the denominator.

\(\displaystyle \frac{\sqrt{3}\cdot \sqrt{7}\cdot \sqrt{5}}{2\cdot \sqrt{5}\cdot \sqrt{5}}=\frac{\sqrt{105}}{10}\)

We can definitely eliminate some answer choices. \(\displaystyle 3\) and \(\displaystyle \frac{1}{3}\) don't make sense because we have an irrational number. Next, let's multiply the numerator and denominator of \(\displaystyle \frac{\sqrt{3}}{3}\) by \(\displaystyle \sqrt{3}\). When we simplify radical fractions, we try to eliminate radicals, but here, we are going to go backwards.

\(\displaystyle \frac{\sqrt{3}\cdot \sqrt{3}}{3\cdot \sqrt{3}}=\frac{3}{3\sqrt{3}}=\frac{1}{\sqrt{3}}\)

\(\displaystyle \frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}\), so \(\displaystyle \frac{1}{\sqrt{3}}\) is the answer.

We don't want to have radicals in the denominator. To get rid of radicals, just multiply the numerator and the denominator by that radical.

\(\displaystyle \frac{\sqrt{8}+\sqrt{12}}{\sqrt{6}}\cdot \frac{\sqrt{6}}{\sqrt{6}}=\)\(\displaystyle \frac{\sqrt{48}+\sqrt{72}}{6}\)

Remember to distribute the radical in the numerator when multiplying.

This may be the answer; however, the numerator can be simplified. Let's factor out the squares. 

\(\displaystyle \frac{\sqrt{48}+\sqrt{72}}{6}=\frac{\sqrt{16}\cdot \sqrt{3}+\sqrt{36}\cdot \sqrt{2}}{6}=\frac{4\sqrt{3}+6\sqrt{2}}{6}\) 

Finally, if we factor out a \(\displaystyle 2\), we get:

\(\displaystyle \frac{2}{2}\cdot \frac{2\sqrt{3}+3\sqrt{2}}{3}=\frac{2\sqrt{3}+3\sqrt{2}}{3}\)

Let's get rid of the radicals in the denominator of each individual fraction.

\(\displaystyle \sqrt{\frac{5}{6}}-\sqrt{\frac{7}{8}}=\sqrt{\frac{5\cdot 6}{6\cdot 6}}-\sqrt{\frac{7\cdot 8}{8\cdot 8}}=\frac{\sqrt{30}}{6}-\frac{\sqrt{56}}{8}\) 

Then find the least common denominator of the fractions, which is \(\displaystyle 24\), and multiply them so that they each have a denominator of \(\displaystyle 24\).

\(\displaystyle \frac{4\sqrt{30}}{24}-\frac{3\sqrt{56}}{24}\) 

We can definitely simplify the numerator in the right fraction by factoring out a perfect square of \(\displaystyle 4\).

\(\displaystyle \frac{4\sqrt{30}}{24}-\frac{3\sqrt{56}}{24}=\frac{4\sqrt{30}}{24}-\frac{3\sqrt{4}*\sqrt{14}}{24}=\frac{4\sqrt{30}}{24}-\frac{6\sqrt{14}}{24}\)

Finally, we can factor out a \(\displaystyle 2\):

\(\displaystyle \frac{2}{2}\cdot \frac{2\sqrt{30}-3\sqrt{14}}{12}=\frac{2\sqrt{30}-3\sqrt{14}}{12}\)

That's the final answer.

To get rid of the radical, we need to multiply by the conjugate. The conjugate uses the opposite sign and multiplying by it will let us rationalize the denominator in this problem. The goal is getting an expression of \(\displaystyle a^2-b^2\) in which we are taking the differences of two squares. 

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

This answer is the same as \(\displaystyle -3-2\sqrt{2}\). Remember to distribute the negative sign.