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\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
có \(\frac{1}{2\cdot3}< \frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3\cdot4}< \frac{1}{3^2}< \frac{1}{2\cdot3}\)
\(\frac{1}{4\cdot5}< \frac{1}{4^2}< \frac{1}{3\cdot4}\)
...
\(\frac{1}{9\cdot10}< \frac{1}{9^2}< \frac{1}{8\cdot9}\)
\(\Rightarrow\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}>A>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}>A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow1-\frac{1}{9}>A>\frac{1}{2}-\frac{1}{10}\)
\(\Rightarrow\frac{8}{9}>A>\frac{2}{5}\)
Bạn ơi, sai rồi, mình k nhầm
làm sao mà \(\frac{1}{2^2}< \frac{1}{1.2}\)được
Ta có : \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{2015.2015}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(=1-\frac{1}{2015}=\frac{2014}{2015}< 1\)
=> A < 1 (đpcm)
\(a,\frac{5}{16}:0,125-\left(2\frac{1}{4}-0,6\right)\cdot\frac{10}{11}\)
\(=\frac{5}{16}:\frac{1}{8}-\left(\frac{9}{4}-\frac{3}{5}\right)\cdot\frac{10}{11}\)
\(=\frac{5}{2}-\frac{33}{20}\cdot\frac{10}{11}\)
\(=\frac{5}{2}-\frac{3}{2}\)
\(=1\)
\(\frac{5}{16}:0,125-\left(2\frac{1}{4}-0,6\right).\frac{10}{11}\)
\(\Rightarrow\frac{5}{16}:\frac{125}{1000}-\left(\frac{9}{4}-\frac{6}{10}\right).\frac{10}{11}\)
\(\Rightarrow\frac{5}{16}:\frac{1}{8}-\left(\frac{9}{4}-\frac{3}{5}\right).\frac{10}{11}\)
\(\Rightarrow\frac{5}{16}.\frac{8}{1}-\left(\frac{45}{20}-\frac{12}{20}\right).\frac{10}{11}\)
\(\Rightarrow\frac{5}{2}-\frac{33}{20}.\frac{10}{11}\)
\(\Rightarrow\frac{5}{2}-\frac{3.1}{2.1}\)
\(\Rightarrow\frac{5}{2}-\frac{3}{2}=\frac{2}{2}=1\)
\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}\)
\(\Rightarrow A>\frac{1}{70}+\frac{1}{70}+\frac{1}{70}+...+\frac{1}{70}\)(60 số hạng)
\(\Rightarrow A>\frac{60}{70}>\frac{60}{80}=\frac{3}{4}\)
Vậy \(A>\frac{3}{4}\left(đpcm\right)\)
\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}\)
\(\Rightarrow A>\frac{1}{70}+...+\frac{1}{70}\)(60 số hạng)
\(\Rightarrow A>\frac{60}{70}>\frac{60}{60}=\frac{3}{4}\)