Đáp án:
$B = 3$
Giải thích các bước giải:
Ta có:
$\ \ \dfrac{1}{\sqrt{n} + \sqrt{n+3}}$
$=\dfrac{\sqrt{n+3} -\sqrt{n}}{\left(\sqrt{n} + \sqrt{n+3}\right)\left(\sqrt{n+3} -\sqrt{n}\right)}$
$=\dfrac{\sqrt{n+3} -\sqrt n}{3}$
Áp dụng:
$\quad B =\dfrac{1}{\sqrt1 +\sqrt4} + \dfrac{1}{\sqrt4 +\sqrt7}+\cdots + \dfrac{1}{\sqrt{97}+\sqrt{100}}$
$\to B = \dfrac{\sqrt4 -\sqrt1}{3} + \dfrac{\sqrt7-\sqrt4}{3}+\cdots + \dfrac{\sqrt{100}-\sqrt{97}}{3}$
$\to B = \dfrac{\sqrt4 - \sqrt1 + \sqrt7 -\sqrt4 +\cdots +\sqrt{100} - \sqrt{97}}{3}$
$\to B =\dfrac{\sqrt{100} -\sqrt1}{3}$
$\to B = \dfrac{9}{3}$
$\to B = 3$
