$ \text{tính giá trị biểu thức } $ $M = \frac{1}{\sqrt[]{1}+\sqrt[]{2}}+\frac{1}{\sqrt[]{2}+\sqrt[]{3}}+\frac{1}{\sqrt[]{3}+\sqrt[]{4}}+\dots +\frac

$ \text{tính giá trị biểu thức } $
$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}} $

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  1. 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 $ 

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  2. 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 $ 

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