$ \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

0 bình luận về “$ \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. 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 $ 

   Trả lời

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 $ 

   Trả lời