Evaluate each term.  Write out the factors for each radical and simplify.

\(\displaystyle \sqrt{40} =\sqrt{4\times 10} = \sqrt{4}\cdot \sqrt{10} = 2\sqrt{10}\)

\(\displaystyle \sqrt8 = \sqrt4 \cdot \sqrt2 = 2\sqrt2\)

\(\displaystyle \sqrt{50} = \sqrt{25}\times \sqrt{2} = 5\sqrt{2}\)

Add all the simplified radicals.  Combine like terms.

\(\displaystyle 2\sqrt{10}+ 2\sqrt2+ 5\sqrt2 = 2\sqrt{10}+7\sqrt2\)

The answer is:  \(\displaystyle 2\sqrt{10}+7\sqrt2\)

Simplify all the radicals to their simplest forms. Use the perfect squares as the factors.

\(\displaystyle \sqrt{18} =\sqrt{9\times 2}= \sqrt{9}\cdot \sqrt{2} = 3\sqrt{2}\)

\(\displaystyle \sqrt{50}= \sqrt{25}\cdot \sqrt{2} = 5 \sqrt{2}\)

\(\displaystyle \sqrt{200} = \sqrt{100}\cdot \sqrt{2} = 10\sqrt2\)

Add the like terms together.

\(\displaystyle 3 \sqrt{2}+5 \sqrt{2}+10 \sqrt{2} = 18 \sqrt{2}\)

The answer is:  \(\displaystyle 18 \sqrt{2}\)

Simplify each radical first.

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

\(\displaystyle \sqrt{16y^4}=4y^2\)

Now, subtract those:

\(\displaystyle 2x-4y^2\)

Rewrite the radicals using common factors of perfect squares.

\(\displaystyle \sqrt8+\sqrt{20}+\sqrt{50} = \sqrt{4\times2}+\sqrt{4\times 5}+\sqrt{25\times 2}\)

The equation becomes:

\(\displaystyle \sqrt{4\times2}+\sqrt{4\times 5}+\sqrt{25\times 2}= 2\sqrt2+2\sqrt5 + 5\sqrt2\)

Combine like-terms.

The answer is:  \(\displaystyle 7\sqrt2+2\sqrt5\)

Evaluate each square root by factoring each with factors of perfect squares.

\(\displaystyle \sqrt{27}= \sqrt{9}\cdot \sqrt 3 = 3\sqrt3\)

\(\displaystyle \sqrt{24} = \sqrt{4}\cdot \sqrt 6 = 2\sqrt6\)

\(\displaystyle \sqrt{12} = \sqrt{4} \cdot \sqrt{3} = 2\sqrt3\)

Replace all the terms in the expression.

\(\displaystyle 2\sqrt{27}+4\sqrt{24}+3\sqrt{12} = 2(3\sqrt3)+4(2\sqrt6)+3(2\sqrt{3})\)

\(\displaystyle 6\sqrt3+8\sqrt6+6\sqrt3 = 12\sqrt3+8\sqrt6\) 

The answer is:  \(\displaystyle 12\sqrt3+8\sqrt6\)

Every radical in this expression is simplified except \(\displaystyle \sqrt{12}\).

Simplify by rewriting this radical using factors of perfect squares.

\(\displaystyle \sqrt{12} = \sqrt{4}\cdot \sqrt{3} = 2\sqrt3\)

Replace the term.

\(\displaystyle 2\sqrt3+\sqrt6 +\sqrt{12}=2\sqrt3+\sqrt6 +2\sqrt3\)

Combine like-terms.

The answer is:  \(\displaystyle 4\sqrt3 +\sqrt6\)

Simplify each of the three square root terms separately. Simplify \(\displaystyle \sqrt{72}\) first as follows:

Express radicand 72 as the product of its prime factors:

\(\displaystyle 72 = 2 \cdot 36\)

\(\displaystyle = 2 \cdot 2 \cdot 18\)

\(\displaystyle = 2 \cdot 2 \cdot 2 \cdot 9\)

\(\displaystyle = 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3\)

Look for any prime factors that appear twice; there are two, 2 and 3, so restate the radical as

\(\displaystyle \sqrt{72}\)

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

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

By the Product of Radicals Property, we can restate this as

\(\displaystyle \sqrt{ 2 ^{2} } \cdot \sqrt{ 3 ^{2} } \cdot \sqrt{ 2}\)

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

\(\displaystyle =6 \sqrt{ 2}\)

The second term, \(\displaystyle \sqrt{98}\), can be simplified similarly:

\(\displaystyle 98 = 2 \cdot 49\)

\(\displaystyle = 2 \cdot 7 \cdot 7\)

so

\(\displaystyle \sqrt{98}\)

\(\displaystyle = \sqrt{2 \cdot 7 \cdot 7}\)

\(\displaystyle = \sqrt{ 7 ^{2}\cdot 2 }\)

\(\displaystyle = \sqrt{ 7 ^{2} } \cdot \sqrt{2}\)

\(\displaystyle = 7\sqrt{2}\)

The third term, \(\displaystyle \sqrt{2}\), is already simplified, as 2 is prime. 

Therefore, 

\(\displaystyle \sqrt{72} + \sqrt{98} - \sqrt{2}\) 

can be rewritten, and simplified using distribution:

\(\displaystyle 6 \sqrt{2} + 7 \sqrt{2} -1 \sqrt{2}\)

\(\displaystyle =( 6 + 7 -1 )\sqrt{2}\)

\(\displaystyle = 12 \sqrt{2}\)

\(\displaystyle \small \sqrt3(\sqrt4-\sqrt3)\)

\(\displaystyle \small \sqrt{12}-\sqrt9\)

\(\displaystyle \small \sqrt{12}=\sqrt4 \times \sqrt3=2\sqrt3\)

\(\displaystyle \small \sqrt9=3\)

\(\displaystyle \small \sqrt{12}-\sqrt9=2\sqrt3-3\)

FOIL with difference of squares.  The multiplying cancels the square roots on both terms.  

We can solve this by simplifying the radicals first: \(\displaystyle \sqrt{36} =6\)

Plugging this into the equation gives us: 

\(\displaystyle 6 \times3(6)=108\)