Đáp án:
Giải thích các bước giải:
$M= \frac{1}{\sqrt[]{1}+\sqrt[]{2}}+\frac{1}{\sqrt[]{2}+\sqrt[]{3}}+\frac{1}{\sqrt[]{3}+\sqrt[]{4}}+\dots +\frac{1}{\sqrt[]{99}+\sqrt[]{100}}$
Xét $\frac{1}{\sqrt[]{n}+\sqrt[]{n+1}}$
$\frac{1}{\sqrt[]{n}+\sqrt[]{n+1}}= \frac{1.(\sqrt[]{n}-\sqrt[]{n+1})}{(\sqrt[]{n}+\sqrt[]{n+1}).(\sqrt[]{n}-\sqrt[]{n+1})} = \frac{\sqrt[]{n}-\sqrt[]{n+1}}{n-(n+1)} = \frac{\sqrt[]{n}-\sqrt[]{n+1}}{-1} = -(\sqrt[]{n}-\sqrt[]{n+1} ) = \sqrt[]{n+1} - \sqrt[]{n}$
thay $n = 1;2;3;...;99$ ta được
$\frac{1}{\sqrt[]{1}+\sqrt[]{2}} = \sqrt[]{2} - \sqrt[]{1}$
$\frac{1}{\sqrt[]{2}+\sqrt[]{3}} = \sqrt[]{3} - \sqrt[]{2}$
$\frac{1}{\sqrt[]{3}+\sqrt[]{4}} = \sqrt[]{4} - \sqrt[]{3}$
$...$
$\frac{1}{\sqrt[]{99}+\sqrt[]{100}} = \sqrt[]{100} - \sqrt[]{99}$
$M= \frac{1}{\sqrt[]{1}+\sqrt[]{2}}+\frac{1}{\sqrt[]{2}+\sqrt[]{3}}+\frac{1}{\sqrt[]{3}+\sqrt[]{4}}+\dots +\frac{1}{\sqrt[]{99}+\sqrt[]{100}}$
$⇒ M = (\sqrt[]{2} - \sqrt[]{1}) + (\sqrt[]{3} - \sqrt[]{2}) + (\sqrt[]{4} - \sqrt[]{3}) + ... + (\sqrt[]{100} - \sqrt[]{99})$
$⇒M = \sqrt[]{2} - \sqrt[]{1} + \sqrt[]{3} - \sqrt[]{2} + \sqrt[]{4} - \sqrt[]{3} + ... + \sqrt[]{100} - \sqrt[]{99}$
$⇒ M = \sqrt[]{100} - \sqrt[]{1} $
$⇒ M = 10 - 1 $
$⇒ M = 9 $
Vậy $ M = 9 $
