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Sa se demonstreze: 1/radicaldin2+1  +1/radicaldin3+radicaldin2 +... +1/radicaldin100+radicaldin99 =9

Răspuns :

[tex] ^{\sqrt{n+1}-\sqrt{n})}\frac{1}{ \sqrt{n+1}+ \sqrt{n} } =\sqrt{n+1}-\sqrt{n}\\ \sqrt{2} -1+ \sqrt{3} - \sqrt{2}+...+ \sqrt{100} - \sqrt{99} =-1+ \sqrt{100} =-1+10=9 [/tex]