Verified Answer :

When you add any fractions, you need to find a common denominator.  In this case they have no common factors so you just multiply them together.  First multiply the first fraction by \(\frac{x - 2}{x - 2}\) which gets you \(\frac{3x - 6}{ x^{2} - 2x}\).  Then multiply the second fraction by \(\frac{x}{x}\) to get \(\frac{5x}{ x^{2} - 2x}\).  Now you can add them together to get \(\frac{-2x - 6}{ x^{2} - 2x}\).

(b) \(\sqrt{18} = \sqrt{9 * 2} = 3 \sqrt{2}\)
\(\sqrt{72} = \sqrt{36 * 2} = 6 \sqrt{2}\)
now you can pull out root 2 and get \(\sqrt{2}(3 - 1 + 6)\) which equals \(8 \sqrt{2}\)

Eeee12
AWe need a common denominator to be able to subtract the fractions, so by multiplying both the numerator and the denominator of each fraction by the denominator of the other.
\(\frac{x + 2}{x + 2} ( \frac{3}{x}) - \frac{x}{x} ( \frac{5}{x + 2} )\)
Then simplify the fraction:
\(\frac{6 + 3x}{x^{2} + 2x} - \frac{5x}{x^{2} + 2x}\)
Then you can just subtract the nominators from each other, but leave the denominator as it is
\(\frac{(6 + 3x) - 5x}{x^{2} + 2x}\)
to get: 
\(\frac{6 - 2x}{x^{2} + 2x}\)
and then you can simplify to get the final answer:
\(\frac{2(3 - x)}{x(x + 2)}\)

b) you need to have the number inside the root the same to add or subtract:
\(\sqrt{2(9)} - \sqrt{2} + \sqrt{2(36)}\)
Then you can 'take out' the numbers that square root easily:
\(3\sqrt{2} - \sqrt{2} + 6\sqrt{2}\)
now you can just add and subtract using the whole numbers outside the root of 2 ( since 2 is a surd):
\(8\sqrt{2}\)

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