For \(x>0\), which of the following is equivalent to \( \dfrac{1}{\frac{1}{2x}+\frac{1}{x+5}} \) ?

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Let's break the problem into individual parts. We can first deal with the following:

$$ \frac{1}{\colorbox{aqua}{$\frac{1}{2x}+\frac{1}{x+5}$}} $$ $$ \frac{1}{2x}+\frac{1}{x+5}$$

We add these fractions by finding a common denominator.

$$ \frac{x+5}{x+5}\left(\frac{1}{2x}\right)+ \frac{1}{x+5}\left(\frac{2x}{2x}\right) $$ $$ = \frac{x+5}{2x(x+5)}+ \frac{2x}{2x(x+5)} $$ $$ = \frac{x+5+2x}{2x(x+5)} $$ $$ = \frac{3x+5}{2x^2+10x} $$

Replacing this back into the initial expression:

$$ \frac{1}{\colorbox{aqua}{$\frac{1}{2x}+\frac{1}{x+5}$}} =\frac{1}{\colorbox{aqua}{$\frac{3x+5}{2x^2+10x}$}} $$

We can think of this expression as 1 divided by the denominator,

$$ 1 \div \frac{3x+5}{2x^2+10x} $$

Which is the same as:

$$ 1 \times \frac{2x^2+10x}{3x+5} $$ $$ = \boxed{\frac{2x^2+10x}{3x+5}} $$

As always, we can always plug in values to obtain our answer or to check our solution. For example, using \(x=1\),

$$ \frac{1}{\frac{1}{2x}+\frac{1}{x+5}} $$ $$ \frac{1}{\frac{1}{2(1)}+\frac{1}{(1)+5}} $$ $$ =\frac{1}{\frac{1}{2}+\frac{1}{6}}=\frac{1}{\frac{3}{6}+\frac{1}{6}}$$ $$ =\frac{1}{\frac{4}{6} }= \frac{6}{4} $$ $$ = \frac{3}{2} $$

Try plugging in \(x=1\) into the answer chices to verify that the first option is equivalent whereas the others are not.

Note that the method has caveats. There are some values we cannot use for this specific problem, such as \(x=0\), which results in an undefined expression. There is also a chance that multiple options work for certain values of \(x\).

Although approach 1 shows mastery of the math, for testing purposes it is almost always better to use approach 2 to save time. There is also less chance of messing up the algebraic steps to deal with the fractions.