[tex] \frac{1}{2+ \sqrt{3} } + \frac{1}{ \sqrt{3} + \sqrt{2} }+ \frac{1}{ \sqrt{2}-1 } [/tex]. Created by mrzapchast. algebra-ru

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September 2021 1 8 Report

[tex] \frac{1}{2+ \sqrt{3} } + \frac{1}{ \sqrt{3} + \sqrt{2} }+ \frac{1}{ \sqrt{2}-1 } [/tex]

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mukus13

Verified answer

$\frac{1}{2+ \sqrt{3} } + \frac{1}{ \sqrt{3}+ \sqrt{2} } + \frac{1}{ \sqrt{2}-1 }=$ $\frac{2- \sqrt{3} }{(2+ \sqrt{3} )(2- \sqrt{3} )} + \frac{ \sqrt{3}- \sqrt{2}}{ (\sqrt{3}+ \sqrt{2})( \sqrt{3}- \sqrt{2})} + \frac{ \sqrt{2}+ 1}{ (\sqrt{2}-1)( \sqrt{2}+1)} =$ $=\frac{2- \sqrt{3} }{2^2- (\sqrt{3})^2} + \frac{ \sqrt{3}- \sqrt{2}}{ (\sqrt{3})^2- (\sqrt{2})^2} + \frac{ \sqrt{2}+ 1}{ (\sqrt{2})^2-1^2} =$ $=\frac{2- \sqrt{3} }{4- 3} + \frac{ \sqrt{3}- \sqrt{2}}{ 3-2} + \frac{ \sqrt{2}+ 1}{ 2-1} =2- \sqrt{3} + \sqrt{3} - \sqrt{2} + \sqrt{2} +1=3$

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aliskairiska2 http://znanija.com/task/21360804

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