\(\displaystyle (\frac{16}{81})^{1/4}\)

\(\displaystyle =\)\(\displaystyle \frac{16^{1/4}}{81^{1/4}}\)

\(\displaystyle =\)\(\displaystyle \frac{(2\cdot 2\cdot 2\cdot 2)^{1/4}}{(3\cdot 3\cdot 3\cdot 3)^{1/4}}\)

\(\displaystyle =\)\(\displaystyle \frac{2}{3}\)

You can rewrite the equation as \(\displaystyle \sqrt{20x^2}=(x)\sqrt{5} \cdot \sqrt{4}\).

This simplifies to \(\displaystyle 2x\sqrt{5}\).

√75 can be broken down to √25 * √3. Which simplifies to 5√3.

Rewrite what is under the radical in terms of perfect squares:

\(\displaystyle x^{2}=x\cdot x\)

\(\displaystyle x^{4}=x^{2}\cdot x^{2}\)

\(\displaystyle x^{6}=x^{3}\cdot x^{3}\)

Therefore, \(\displaystyle \sqrt{a^{3}b^{4}c^{5}}= \sqrt{a^{2}a^{1}b^{4}c^{4}c^{1}}=ab^{2}c^{2}\sqrt{ac}\).

We know that 25 is a factor of 50. The square root of 25 is 5. That leaves \(\displaystyle \sqrt{2}\) which can not be simplified further.

Multiply by the conjugate and the use the formula for the difference of two squares:

\(\displaystyle \frac{x + \sqrt{3}}{3x + \sqrt{2}}\)

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

\(\displaystyle =\) \(\displaystyle \frac{3x^{2} -x \sqrt{2} + 3x\sqrt{3} - \sqrt{6}}{(3x)^{2} - (\sqrt{2})^{2}}\) 

\(\displaystyle =\) \(\displaystyle \frac{3x^{2} -x \sqrt{2} + 3x\sqrt{3} - \sqrt{6}}{9x^{2} - 2}\)

First find all of the prime factors of \(\displaystyle 468\)

\(\displaystyle 468=6\ast78=6\ast6\ast13=2\ast3\ast2\ast3\ast13\)

So \(\displaystyle \sqrt{468}=\sqrt{2\ast2\ast3\ast3\ast13}=2\ast3\sqrt{13}=6\sqrt{13}\)

\(\displaystyle \sqrt{432}\)

1. We know that \(\displaystyle 432=(144)(3)\), which we can separate under the square root:

\(\displaystyle \sqrt{144\cdot 3}\)

2. 144 can be taken out since it is a perfect square: \(\displaystyle 12\cdot12=144\). This leaves us with:

\(\displaystyle 12\sqrt{3}\)

This cannot be simplified any further.

Write out the common square factors of the number inside the square root.

\(\displaystyle \sqrt{240}= \sqrt{4\times 60}= 2\sqrt{60}\)

Continue to find the common factors for 60.

\(\displaystyle 2\sqrt{60} = 2\sqrt{4\times 15} = 2(2\sqrt{15}) = 4\sqrt{15}\)

Since there are no square factors for \(\displaystyle \sqrt{15}\), the answer is in its simplified form.  It might not have been easy to see that 16 was a common factor of 240.

The answer is: \(\displaystyle 4\sqrt{15}\)

To simplify, we want to find some factors of \(\displaystyle 48\) where at least one of the factors is a perfect square. 

In this case, \(\displaystyle 16\) and \(\displaystyle 3\) are factors of \(\displaystyle 48\), and \(\displaystyle 16\) is a perfect square. 

We can simplify by saying:

\(\displaystyle \sqrt{48} = \sqrt{16\cdot 3} = \sqrt{4^2 \cdot 3} = 4\sqrt{3}\)

We could also recognize that two factors of \(\displaystyle 48\) are \(\displaystyle 4\) and \(\displaystyle 12\). We could approach this way by saying:

\(\displaystyle \sqrt{48}=\sqrt{4 \cdot 12} = \sqrt {2^2 \cdot 12} = 2\sqrt{12}\)

But we wouldn't stop there. That's because \(\displaystyle 12\) can be further factored:

\(\displaystyle 2\sqrt{12} = 2\sqrt{4 \cdot 3} = 2\sqrt{2^2 \cdot 3} = 2\cdot 2 \sqrt{3} = 4\sqrt{3}\)