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\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{[\left(n+1\right)\sqrt{n}-n\sqrt{n+1}].[\left(n+1\right)\sqrt{n}+n\sqrt{n+1}]}\)
=\(\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{\sqrt{n}}{n}-\dfrac{\sqrt{n+1}}{n+1}\)
=\(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Áp dụng ta có S=\(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-...+\dfrac{1}{\sqrt{2024}}-\dfrac{1}{\sqrt{2025}}=1-\dfrac{1}{\sqrt{2025}}=1-\dfrac{1}{45}=\dfrac{44}{45}\)
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) Ta thấy:
\(4=1+3=1+\sqrt{9}\)
\(1+2\sqrt{2}=1+\sqrt{2^2\cdot2}=1+\sqrt{8}\)
Mà: \(\sqrt{8}< \sqrt{9}\)
\(\Rightarrow1+\sqrt{8}< 1+\sqrt{9}\)
\(\Rightarrow\dfrac{1}{1+\sqrt{8}}>\dfrac{1}{1+\sqrt{9}}\)
\(\Rightarrow\dfrac{1}{1+2\sqrt{2}}>\dfrac{1}{4}\)
2) Ta thấy:
\(2018< 2024\)
\(\Rightarrow\sqrt{2018}< \sqrt{2024}\) (1)
\(2025< 2026\)
\(\Rightarrow\sqrt{2025}< \sqrt{2026}\) (2)
Từ (1) và (2) ta có:
\(\sqrt{2018}+\sqrt{2025}< \sqrt{2024}+\sqrt{2026}\)
ta có : \(\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+\dfrac{1}{\sqrt{4}+\sqrt{5}}+...+\dfrac{1}{\sqrt{2025}+\sqrt{2026}}\)
\(=\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}+\dfrac{\sqrt{4}-\sqrt{3}}{\left(\sqrt{3}+\sqrt{4}\right)\left(\sqrt{4}-\sqrt{3}\right)}+...+\dfrac{\left(\sqrt{2026}-\sqrt{2025}\right)}{\left(\sqrt{2026}+\sqrt{2025}\right)\left(\sqrt{2026}-\sqrt{2025}\right)}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\sqrt{5}-\sqrt{4}+...+\sqrt{2026}-\sqrt{2025}\)
\(=-\sqrt{2}+\sqrt{2026}\)
\(A>\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{4}+\sqrt{5}}+\dfrac{1}{\sqrt{6}+\sqrt{7}}+...+\dfrac{1}{\sqrt{2024}+\sqrt{2025}}\)
\(\Rightarrow2A>\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+\dfrac{1}{\sqrt{4}+\sqrt{5}}+...+\dfrac{1}{\sqrt{2024}+\sqrt{2025}}\)
\(\Rightarrow2A>\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2025}-\sqrt{2024}\)
\(\Rightarrow2A>\sqrt{2025}-\sqrt{1}=44\)
\(\Rightarrow A>22\) (đpcm)
\(x=\sqrt{\dfrac{2\sqrt{3}+2-6\sqrt{3}}{2\sqrt{3}\left(2\sqrt{3}+2\right)}}=\sqrt{\dfrac{2-4\sqrt{3}}{2\sqrt{3}\left(2\sqrt{3}+2\right)}}\) ko tồn tại vì 2-4căn 3<0
4: Ta có: \(\dfrac{6}{1-\sqrt{3}}-\dfrac{3\sqrt{3}+3}{\sqrt{3}+1}\)
\(=-3-3\sqrt{3}-3\)
\(=-6-3\sqrt{3}\)
5: Ta có: \(\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}\)
\(=-\sqrt{2}-\sqrt{2}\)
\(=-2\sqrt{2}\)
Ko biết