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

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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 \(n=-1,2\) :)

Log in to reply

You are smart, thx man

Log 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

Log in to reply

Nice ;)

Log in to reply

Rationalise the denominator

Log in to reply