How do you simplify $\dfrac{\sqrt{16}}{\sqrt{4}+\sqrt{2}}$?

Hint: Try to simplify by doing rationalization. This can be done by multiplying $\left( \sqrt{4}-\sqrt{2} \right)$ both in numerator and in denominator. Then do the necessary calculations to obtain the required solution.

Complete step-by-step solution:
Rationalization: It is a method by which we can write a fraction in such a way that the denominator contains only rational numbers.
Considering our question $\dfrac{\sqrt{16}}{\sqrt{4}+\sqrt{2}}$
It can be rationalized by multiplying $\left( \sqrt{4}-\sqrt{2} \right)$ both in numerator and in denominator.
Multiplying $\left( \sqrt{4}-\sqrt{2} \right)$ both in numerator and in denominator, we get
\[\Rightarrow \dfrac{\sqrt{16}\left( \sqrt{4}-\sqrt{2} \right)}{\left( \sqrt{4}+\sqrt{2} \right)\left( \sqrt{4}-\sqrt{2} \right)}\]
For the numerator part \[\sqrt{16}\] is multiplied with \[\sqrt{4}\] and \[\sqrt{2}\] separately, so we get
\[\Rightarrow \sqrt{16}\left( \sqrt{4}-\sqrt{2} \right)=\sqrt{16}\times \sqrt{4}-\sqrt{16}\times \sqrt{2}=4\times 2-4\sqrt{2}=8-4\sqrt{2}\]
For denominator part, as we know $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ so
\[\Rightarrow \left( \sqrt{4}+\sqrt{2} \right)\left( \sqrt{4}-\sqrt{2} \right)={{\left( \sqrt{4} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}=4-2=2\]
Now our expression can be written as
$\Rightarrow \dfrac{8-4\sqrt{2}}{2}$
Taking common ‘2’ form the numerator, we get
$\Rightarrow \dfrac{2\left( 4-2\sqrt{2} \right)}{2}$
Cancelling out ‘2’ both from numerator and denominator we get
$\Rightarrow 4-2\sqrt{2}$
Which can also be written as
$\Rightarrow 2\left( 2-\sqrt{2} \right)$
This is the required solution of the given question.

Note: This particular question can be modified at first as $\dfrac{\sqrt{16}}{\sqrt{4}+\sqrt{2}}=\dfrac{4}{2+\sqrt{2}}$, then can be proceed to rationalization for further simplification. This is the more simple form of the given question. For rationalization the numerator and denominator should be multiplied by the same quantity i.e. $\left( \sqrt{4}-\sqrt{2} \right)$ and for the modified one it should be multiplied by $\left( 2-\sqrt{2} \right)$. Rationalization is done so that we can obtain the denominator in $\left( a+b \right)\left( a-b \right)$ form. Hence in the next step it can be simplified as ${{a}^{2}}-{{b}^{2}}$. For the numerator \[\sqrt{16}\] should be multiplied with \[\sqrt{4}\] and \[\sqrt{2}\] separately and further calculation should be done.