1: \(\dfrac{2}{3}\): \(\dfrac{3}{4}\) :\(\dfrac{4}{5}\) :.....:\(\dfrac{2024}{2025}\)
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5 tháng 6 2017
\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{[\left(n+1\right)\sqrt{n}-n\sqrt{n+1}].[\left(n+1\right)\sqrt{n}+n\sqrt{n+1}]}\)
=\(\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{\sqrt{n}}{n}-\dfrac{\sqrt{n+1}}{n+1}\)
=\(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Áp dụng ta có S=\(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-...+\dfrac{1}{\sqrt{2024}}-\dfrac{1}{\sqrt{2025}}=1-\dfrac{1}{\sqrt{2025}}=1-\dfrac{1}{45}=\dfrac{44}{45}\)
15 tháng 10 2018
Ta có công thức tổng quát:
\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}\left(n+1-n\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Vậy \(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{2025\sqrt{2024}+2024\sqrt{2025}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{2024}}-\dfrac{1}{\sqrt{2025}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2025}}=1-\dfrac{1}{45}=\dfrac{44}{45}\)
\(1:\dfrac{2}{3}:\dfrac{3}{4}:\dfrac{4}{5}:...:\dfrac{2024}{2025}\)
= \(1\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{2025}{2024}=\dfrac{2025}{2}\)