Đặt A=\(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+.......+\frac{1}{\sqrt{100}}\)
\(=\frac{2}{2\sqrt{1}}+\frac{2}{2\sqrt{2}}+\frac{2}{2\sqrt{3}}+.......+\frac{2}{2\sqrt{100}}\)
\(< 2\left(\sqrt{1}-\sqrt{0}+\sqrt{2}-\sqrt{1}+.........+\sqrt{100}-\sqrt{99}\right)\)
\(=2.\sqrt{100}=20\)
\(\Rightarrow A< 20\left(đpcm\right)\)
1-a,\(A=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
b,\(B=\frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+.....\frac{1}{\sqrt{99}+\sqrt{100}}=9\) =9
Bạn nào chứng minh giúp mình với
Mình biết ơn nhiều lắm ạ!
Chứng minh \(\frac{\sqrt{n}}{\sqrt{n+1}}+1< \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Mình cần gấp lol
Tính P = \(\frac{4+\sqrt{3}}{\sqrt{1}+\sqrt{3}}+\frac{8+\sqrt{15}}{\sqrt{3}+\sqrt{5}}+...+\frac{2n+\sqrt{n^2-1}}{\sqrt{n-1}+\sqrt{n+1}}+...+\frac{240+\sqrt{14399}}{\sqrt{119}+\sqrt{121}}\)
CM \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{n}}>\sqrt{n}\)\(\sqrt{n}\)
CM \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{n}}>\sqrt{n}\)
CMR : với mọi số tự nhiên n > 1, ta có :
a) \(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+n}< \frac{3}{4}\)
b) \(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}>2\left(\sqrt{n+1}-1\right)\)
\(\left(1+\frac{1}{\sqrt{1}}\right)\left(1+\frac{1}{\sqrt{2}}\right)\left(1+\frac{1}{\sqrt{3}}\right)...\left(1+\frac{1}{\sqrt{n}}\right)\)
Biểu thức này có công thức ntn ạ
Rút gọn các biểu thức,
a> A= \(\frac{1}{\sqrt{1}+\sqrt{2}}\) + \(\frac{1}{\sqrt{2}+\sqrt{3}}\)+ \(\frac{1}{\sqrt{3}+\sqrt{4}}\)+ ......... + \(\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
b> B= \(\frac{1}{\sqrt{1}-\sqrt{2}}\)- \(\frac{1}{\sqrt{2}-\sqrt{3}}\)- \(\frac{1}{\sqrt{3}-\sqrt{4}}\)- .......... - \(\frac{1}{\sqrt{24}-\sqrt{25}}\)
