e) $\frac{1}{1+\sqrt[]{2} }$ + $\frac{1 }{\sqrt[]{2}+ \sqrt[]{3} }$ +....+$\frac{1}{\sqrt[]{99}+\sqrt[]{100} }$
Ta có : $\frac{1}{1+\sqrt[]{2} }$ + $\frac{1 }{\sqrt[]{2}+ \sqrt[]{3} }$ +....+$\frac{1}{\sqrt[]{99}+\sqrt[]{100} }$
= $\sqrt[]{2}$ -$\sqrt[]{1}$ +$\sqrt[]{3}$ -$\sqrt[]{2}$ +...+$\sqrt[]{100}$ -$\sqrt[]{99}$
= $\sqrt[]{100}$ -$\sqrt[]{1}$ =10-1=9
Giải thích cách làm trên: Ta có Dạng Tổng Quát sau:
∵∴ $\frac{1 }{\sqrt[]{k}+\sqrt[]{k+1} }$ =$\frac{(\sqrt[]{k}+1-\sqrt[]{k})}{\sqrt[]{k+1}-\sqrt[]{k}).(\sqrt[]{k+1}+\sqrt[]{k})}$ =$\frac{(\sqrt[]{k+1}-\sqrt[]{k})}{k+1-k}$ = $\sqrt[]{k+1}$ -$\sqrt[]{k}$ (k>0)
_______________________________∞∞Δ∞∞____________________________∞∞Δ∞∞_____
