\(Rútgọn\):
\(a,\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{48}+\sqrt{49}}\)
\(b,\frac{1}{\sqrt{3}-\sqrt{5}}+\frac{1}{\sqrt{3}+\sqrt{5}}\)
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NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 6 2020
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{\sqrt{n}}{n}-\frac{\sqrt{n+1}}{n+1}\)
\(\Rightarrow A=\frac{1}{1}-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{3}+...+\frac{\sqrt{48}}{48}-\frac{\sqrt{49}}{49}\)
\(=1-\frac{\sqrt{49}}{49}=1-\frac{7}{49}=1-\frac{1}{7}=\frac{6}{7}\)
22 tháng 10 2021
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+2\sqrt{12}}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-2\sqrt{75}}}}\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)
\(C=\sqrt{4+5}\)
\(C=3\)
Y
13 tháng 6 2019
2.+ \(\left(2n+1\right)^2=4n^2+4n+1>4n^2+4n\)
\(\Rightarrow2n+1>\sqrt{4n\left(n+1\right)}=2\sqrt{n\left(n+1\right)}\)
+ \(\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(2n+1\right)\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(A< \frac{1}{2}\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{48}}-\frac{1}{\sqrt{49}}\right)\)
\(\Rightarrow A< \frac{1}{2}\)
Y
13 tháng 6 2019
1. + \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\left(n+1\right)-n}{\left(n+1\right)\sqrt{n}}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(n+1\right)\sqrt{n}}\)
\(< \frac{\left(\sqrt{n+1}-\sqrt{n}\right)\cdot2\sqrt{n+1}}{\sqrt{n}\left(n+1\right)}=2\cdot\frac{n+1-\sqrt{n\left(n+1\right)}}{\left(n+1\right)\sqrt{n}}=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(A< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
\(\Rightarrow A< 2\)
Bài 2 tạm thời chưa nghĩ ra :))
a) \(\frac{1}{1+\sqrt{2}}\)+\(\frac{1}{\sqrt{2}+\sqrt{3}}\)+\(\frac{1}{\sqrt{3}+\sqrt{4}}\)+...+\(\frac{1}{\sqrt{48}+\sqrt{49}}\)
=\(\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right).\left(\sqrt{2}-1\right)}\)+\(\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right).\left(\sqrt{3}-\sqrt{2}\right)}+...+\)\(\frac{\sqrt{49}-\sqrt{48}}{\left(\sqrt{49}+\sqrt{48}\right).\left(\sqrt{49}-\sqrt{48}\right)}\)
=\(\frac{\sqrt{2}-1}{2-1}\)+\(\frac{\sqrt{3}-\sqrt{2}}{3-2}\)+...+\(\frac{\sqrt{49}-\sqrt{48}}{49-48}\)
=\(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\)\(\sqrt{49}-\sqrt{48}\)
=\(\sqrt{49}-1\)=\(7-1\)= \(6\)