Question

Calculați partea întreaga din nr
$\frac{1}{ \sqrt{1} + \sqrt{2} } + \frac{1}{ \sqrt{2} + \sqrt{3} } + ... + \frac{1}{ \sqrt{98} + \sqrt{99} }$
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Answer 1

 

$\displaystyle\bf\\\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{97}+\sqrt{98}}+\frac{1}{\sqrt{98}+\sqrt{99}}=\\\\\\\text{Schimbam intre ei termenii de la numitor. (Adunarea este comutativa.)}\\\\\\=\frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+...+\frac{1}{\sqrt{98}+\sqrt{97}}+\frac{1}{\sqrt{99}+\sqrt{98}}=$

.

$\displaystyle\bf\\\text{Rationalizam numitorii:}\\\\\\=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{98}-\sqrt{97}}{98-97}+\frac{\sqrt{99}-\sqrt{98}}{99-98}=\\\\\\=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{98}-\sqrt{97}+\sqrt{99}-\sqrt{98}=$

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$\displaystyle\bf\\ \text{Se reduc termenii asemenea.}\\\\=-\sqrt{1}+\sqrt{99}=\\\\=\sqrt{99}-1\\\\\sqrt{81}<\sqrt{99}<\sqrt{100}\\\\9<\sqrt{99}<10\\\\\implies~\Big[\sqrt{99}\Big]=9~~~daaar~~~\Big[\sqrt{99}-1\Big]=9-1=\boxed{\boxed{\bf8}}$