Ta có công thức: \(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{\left(n+1\right)-n}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{1}{\sqrt{n\left(n+1\right)^2}+\sqrt{n^2\left(n+1\right)^2}}=\frac{1}{n\cdot\sqrt{n+1}+\sqrt{n}\cdot\left(n+1\right)}\)
Áp dụng vào biểu thức, ta có:
\(\frac{1}{2\sqrt1+1\sqrt2}+\frac{1}{3\sqrt2+2\sqrt3}+\cdots+\frac{1}{25\sqrt{24}+24\sqrt{25}}\)
\(=\frac{1}{\sqrt1}-\frac{1}{\sqrt2}+\frac{1}{\sqrt2}-\frac{1}{\sqrt3}+\cdots+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)
\(=1-\frac15=\frac45\)
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