Write \(2\) as \(\sqrt{4}\),we'll get \[\frac{1}{2+\sqrt{3}}=\frac{1}{\sqrt{3}+\sqrt{4}}=\frac{4-3}{\sqrt{4}+\sqrt{3}}=\frac{(\sqrt{4})^2-(\sqrt{3})^2}{\sqrt{4}+\sqrt{3}}=\frac{(\sqrt{4}+\sqrt{3})(\sqrt{4}-\sqrt{3})}{\sqrt{4}+\sqrt{3}}=\sqrt{4}-\sqrt{3}=\boxed{2-\sqrt{3}}\] You could generalized to \(\frac{1}{\sqrt{n}+\sqrt{n+1}}\) though
–
Pham Khanh
·
10 months, 3 weeks ago

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TopNewest\(\dfrac{1}{n+\sqrt{n+1}}=\dfrac{n-\sqrt{n+1}}{(n+\sqrt{n+1})(n-\sqrt{n+1})}=\dfrac{n-\sqrt{n+1}}{n^2-n-1}\)

We want denominator to be 1 so,

\(\implies n^2-n-1=1\\ \implies n^2-n-2=0 \\ \implies n^2-2n+n-2=0 \\ \implies n(n-2)+1(n-2)=0 \\ \implies (n+1)(n-2)=0\)

So it only works if \(n=-1,2\) :) – Nihar Mahajan · 11 months, 3 weeks ago

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– Jason Chrysoprase · 11 months, 3 weeks ago

You are smart, thx manLog in to reply

Write \(2\) as \(\sqrt{4}\),we'll get \[\frac{1}{2+\sqrt{3}}=\frac{1}{\sqrt{3}+\sqrt{4}}=\frac{4-3}{\sqrt{4}+\sqrt{3}}=\frac{(\sqrt{4})^2-(\sqrt{3})^2}{\sqrt{4}+\sqrt{3}}=\frac{(\sqrt{4}+\sqrt{3})(\sqrt{4}-\sqrt{3})}{\sqrt{4}+\sqrt{3}}=\sqrt{4}-\sqrt{3}=\boxed{2-\sqrt{3}}\] You could generalized to \(\frac{1}{\sqrt{n}+\sqrt{n+1}}\) though – Pham Khanh · 10 months, 3 weeks ago

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– Jason Chrysoprase · 10 months, 3 weeks ago

Nice ;)Log in to reply

Rationalise the denominator – Rushabh Zambad · 11 months, 3 weeks ago

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