a) Biến đổi vế phải, ta có:
$\dfrac{1}{\sqrt[]{n}}-\dfrac{1}{\sqrt[]{n+1}}$
$=\dfrac{\sqrt[]{n+1}-\sqrt[]{n}}{\sqrt[]{n}(\sqrt[]{n+1})}$
$=\dfrac{(\sqrt[]{n+1}-\sqrt[]{n})(\sqrt[]{n+1}+\sqrt[]{n})}{\sqrt[]{n}(\sqrt[]{n+1})(\sqrt[]{n+1}+\sqrt[]{n})}$
$=\dfrac{n+1-n}{(n+1)\sqrt[]{n}+n\sqrt[]{n+1}}$
$=\dfrac{1}{(n+1)\sqrt[]{n}+n\sqrt[]{n+1}}$
$=VT$ (điều phải chứng minh)
b) $S=\dfrac{1}{2\sqrt[]{1}+1\sqrt[]{2}}+\dfrac{1}{3\sqrt[]{2}+2\sqrt[]{3}}+...+\dfrac{1}{400\sqrt[]{399}+399\sqrt[]{400}}$
$=\dfrac{1}{1}-\dfrac{1}{\sqrt[]{2}}+\dfrac{1}{\sqrt[]{2}}-\dfrac{1}{\sqrt[]{3}}+...+\dfrac{1}{\sqrt[]{399}}-\dfrac{1}{\sqrt[]{400}}$
$=1-\dfrac{1}{\sqrt[]{400}}$
$=1-\dfrac{1}{20}$
$=\dfrac{19}{20}$
