Đáp án:

$\dfrac{24}{25}$

Giải thích các bước giải:

\(\begin{array}{l}
\dfrac{1}{2\sqrt1 + 1\sqrt2} + \dfrac{1}{3\sqrt2 + 2\sqrt3} + \dfrac{1}{3\sqrt4 + 4\sqrt3}+\cdots + \dfrac{1}{625\sqrt{624} + 624\sqrt{625}}\\
= \dfrac{1}{\sqrt{1.2}\left(\sqrt1 + \sqrt2\right)} + \dfrac{1}{\sqrt{2.3}\left(\sqrt2 + \sqrt3\right)} +  \dfrac{1}{\sqrt{3.4}\left(\sqrt3 + \sqrt4\right)} +\cdots +  \dfrac{1}{\sqrt{624.625}\left(\sqrt{624}+ \sqrt{625}\right)} \\
= \dfrac{\sqrt2 - \sqrt 1}{\sqrt{1.2}(2 - 1)} + \dfrac{\sqrt3 - \sqrt2}{\sqrt{2.3}(3- 2)} + \dfrac{\sqrt4 - \sqrt3}{\sqrt{3.4}(4 - 3)} + \cdots + \dfrac{\sqrt{625} - \sqrt{624}}{\sqrt{624.625}(625-624)}\\
= \dfrac{\sqrt2 - \sqrt 1}{\sqrt{1.2}} + \dfrac{\sqrt3 - \sqrt2}{\sqrt{2.3}} + \dfrac{\sqrt4 - \sqrt3}{\sqrt{3.4}} + \cdots + \dfrac{\sqrt{625} - \sqrt{624}}{\sqrt{624.625}}\\
= 1 - \dfrac{1}{\sqrt2} + \dfrac{1}{\sqrt2} - \dfrac{1}{\sqrt3} + \dfrac{1}{\sqrt3} - \dfrac{1}{\sqrt4} + \cdots + \dfrac{1}{\sqrt{624}} - \dfrac{1}{\sqrt{625}}\\
= 1 - \dfrac{1}{\sqrt{625}}\\
= 1 - \dfrac{1}{25}\\
= \dfrac{24}{25}
\end{array}\)

Đáp án:

$\dfrac{24}{25}$

Giải thích các bước giải:

Công thức tổng quát với số nguyên dương $k$ ta có:

$\dfrac{1}{(k+1)\sqrt{k}+k\sqrt{k+1}}=\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}$

Chứng minh:

$\dfrac{1}{(k+1)\sqrt{k}+k\sqrt{k+1}}\\=\dfrac{1}{\sqrt{k+1}\sqrt{k}(\sqrt{k+1}+\sqrt{k})}\\=\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k+1}\sqrt{k}}\\=\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}$

Áp dụng ta có:

$\dfrac{1}{2\sqrt{1}+1\sqrt{2}}=\dfrac{1}{(1+1)\sqrt{1} +1\sqrt{1+1}}=1-\dfrac{1}{\sqrt{2}}\\\dfrac{1}{3\sqrt{2}+2\sqrt{3}}=\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\\...\\\dfrac{1}{625\sqrt{624}+624\sqrt{625}}=\dfrac{1}{\sqrt{624}}-\dfrac{1}{\sqrt{625}}\\\to \dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{625\sqrt{624}+624\sqrt{625}}=1-\dfrac{1}{\sqrt{625}}=1-\dfrac{1}{25}=\dfrac{24}{25}$