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Simplify \dfrac{1}{\sqrt2+sqrt3}+\dfrac{1}{\sqrt2-sqrt3}.

 May 5, 2024
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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 \(-2\sqrt2\).

 May 5, 2024

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