Algebra

Here is my attempt.
To simplify the expression $\dfrac{1}{\sqrt2+\sqrt3}+\dfrac{1}{\sqrt2-\sqrt3}$, we need to rationalize the denominators of the fractions.
 

Let's start by rationalizing the first fraction: $\dfrac{1}{\sqrt2+\sqrt3}\cdot \dfrac{\sqrt2-\sqrt3}{\sqrt2-\sqrt3}= \dfrac{\sqrt2-\sqrt3}{2-3}=-(\sqrt2-\sqrt3)=-\sqrt2+\sqrt3$

 

Next, rationalize the second fraction: $\dfrac{1}{\sqrt2-\sqrt3}\cdot \dfrac{\sqrt2+\sqrt3}{\sqrt2+\sqrt3}= \dfrac{\sqrt2+\sqrt3}{2-3}=-(\sqrt2+\sqrt3)=-\sqrt2-\sqrt3$

 

Now, add the two rationalized fractions together:$-\sqrt2+\sqrt3+(-\sqrt2-\sqrt3)=-\sqrt2+\sqrt3-\sqrt2-\sqrt3=-2\sqrt2$

 

 

Therefore, the simplified expression is $-2\sqrt2$.