<s>[INST] Prove that $\sqrt{k}+\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}$ I try to prove that $\sqrt{k}+\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}$ for all integers k greater or equal to 1 but don't know how to transform this equation full of square roots in order to finish the proof.
 [/INST] First, find a common denominator.
$\sqrt{k}+\frac{1}{\sqrt{k}+\sqrt{k+1}}= \frac{\sqrt{k}(\sqrt{k} + \sqrt{k+1}) + 1}{\sqrt{k} + \sqrt{k+1}}$
Next, distribute.
$\frac{k + \sqrt{k}\sqrt{k+1} + 1}{\sqrt{k} + \sqrt{k+1}}$
Now, we rearrange the terms so we have
$\frac{k + 1 + \sqrt{k}\sqrt{k+1}}{\sqrt{k} + \sqrt{k+1}}$
Now factor out of the numerator a $\sqrt{k+1}$
$\frac{\sqrt{k+1}(\sqrt{k+1} + \sqrt{k})}{\sqrt{k} + \sqrt{k+1}}$
Cancel out common terms, and now we have our answer.
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