Question

calculati suma inverselor numerelor a b c unde a= √3+√2 b= √4+√3 c= √5+√4

Answer 1

   
[tex]\displaystyle \\ \texttt{inversul unui numar }~x ~\texttt{ este }~ \frac{1}{x} \\ \\ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{c} + \frac{1}{b} + \frac{1}{a} =\frac{1}{ \sqrt{5}+ \sqrt{4} } + \frac{1}{\sqrt{4}+ \sqrt{3}} + \frac{1}{\sqrt{3}+ \sqrt{2}} \\ \\ \texttt{Rationalizam numitorii.} [/tex]


[tex]\displaystyl \\ \frac{1}{ \sqrt{5}+ \sqrt{4} } + \frac{1}{\sqrt{4}+ \sqrt{3}} + \frac{1}{\sqrt{3}+ \sqrt{2}} =\\ \\ = \frac{\sqrt{5}- \sqrt{4} }{ (\sqrt{5}+ \sqrt{4})(\sqrt{5}- \sqrt{4})} + \frac{\sqrt{4}- \sqrt{3}}{(\sqrt{4}+ \sqrt{3})(\sqrt{4}- \sqrt{3})} + \frac{\sqrt{3}- \sqrt{2}}{(\sqrt{3}+ \sqrt{2})(\sqrt{3}-\sqrt{2})}= [/tex]


[tex]\displaystyle = \frac{\sqrt{5}- \sqrt{4} }{(\sqrt{5})^2- (\sqrt{4})^2} + \frac{\sqrt{4}- \sqrt{3}}{(\sqrt{4})^2- (\sqrt{3})^2} + \frac{\sqrt{3}- \sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2}= \\ \\ = \frac{\sqrt{5}- \sqrt{4} }{5- 4} + \frac{\sqrt{4}- \sqrt{3}}{4- 3} + \frac{\sqrt{3}- \sqrt{2}}{3-2}= \\ \\ = \frac{\sqrt{5}- \sqrt{4} }{1} + \frac{\sqrt{4}- \sqrt{3}}{1} + \frac{\sqrt{3}- \sqrt{2}}{1}= [/tex]


$= \sqrt{5}- \underline{\sqrt{4}} + \underline{\sqrt{4}}- \underline{\underline{\sqrt{3}}} +\underline{\underline{\sqrt{3}}}- \sqrt{2} = \boxed{\sqrt{5} - \sqrt{2} }$