### Instruction: Simplify \[\frac{1}{\dfrac{1}{\sqrt{2}+1} + \dfrac{2}{\sqrt{3}-1}}.\].### Response: I notice that the denominators of the fractions inside the big fraction have radicals, so I wonder if I can rationalize them somehow.
I recall that \[(\sqrt{a}+b)(\sqrt{a}-b) = a-b^2,\] so I multiply the first fraction by \[\frac{\sqrt{2}-1}{\sqrt{2}-1},\] and the second fraction by \[\frac{\sqrt{3}+1}{\sqrt{3}+1}.\]
This gives me \[\frac{1}{\dfrac{\sqrt{2}-1}{2-1^2} + \dfrac{2(\sqrt{3}+1)}{3-1^2}}.\]
I simplify the denominators by performing the subtraction inside the parentheses, and I get \[\frac{1}{\dfrac{\sqrt{2}-1}{1} + \dfrac{2(\sqrt{3}+1)}{2}}.\]
I can now combine the fractions in the denominator by finding a common denominator of 2, and I get \[\frac{1}{\dfrac{2(\sqrt{2}-1)}{2} + \dfrac{2(\sqrt{3}+1)}{2}}.\]
I cancel out the 2's in the numerator and denominator, and I get \[\frac{1}{\sqrt{2}-1 + \sqrt{3}+1}.\]
I simplify the denominator by combining the like terms, and I get \[\frac{1}{\sqrt{2}+\sqrt{3}}.\]
I'm almost done, but I still have a radical in the denominator, so I want to rationalize it by multiplying the fraction by \[\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}.\]
This gives me \[\frac{\sqrt{2}-\sqrt{3}}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})}.\]
I use the same formula as before to expand the denominator, and I get \[\frac{\sqrt{2}-\sqrt{3}}{2-3}.\]
I simplify the denominator by subtracting, and I get \[\frac{\sqrt{2}-\sqrt{3}}{-1}.\]
I can rewrite this as \[-(\sqrt{2}-\sqrt{3}).\]