(N/A) To solve this,we rationalize each term in the expression.
Step $1$: Rationalize the first term: $\frac{1}{1+\sqrt{2}} = \frac{1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1} = \frac{\sqrt{2}-1}{2-1} = \sqrt{2}-1$.
Step $2$: Rationalize the second term: $\frac{1}{\sqrt{2}+\sqrt{3}} = \frac{1}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{\sqrt{3}-\sqrt{2}}{3-2} = \sqrt{3}-\sqrt{2}$.
Step $3$: Rationalize the third term: $\frac{1}{\sqrt{3}+\sqrt{4}} = \frac{1}{\sqrt{4}+\sqrt{3}} \times \frac{\sqrt{4}-\sqrt{3}}{\sqrt{4}-\sqrt{3}} = \frac{\sqrt{4}-\sqrt{3}}{4-3} = \sqrt{4}-\sqrt{3} = 2-\sqrt{3}$.
Step $4$: Add the results: $(\sqrt{2}-1) + (\sqrt{3}-\sqrt{2}) + (2-\sqrt{3})$.
Step $5$: Simplify the expression: $\sqrt{2} - 1 + \sqrt{3} - \sqrt{2} + 2 - \sqrt{3} = -1 + 2 = 1$.
Thus,the expression equals $1$.