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Ta có : \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)
\(.\) \(.\)
\(.\)
\(.\) \(.\)
\(.\) \(.\)
\(\frac{1}{2013^2}< \frac{1}{2012\cdot2013}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.........+\frac{1}{2013^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.....+\frac{1}{2012\cdot2013}\)
Mà \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.....+\frac{1}{2012\cdot2013}=1-\frac{1}{2013}< 1\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{2013^2}< 1\)
Nhớ k cho mình nhé!
Chúc các bạn học tốt!
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{19}{20}\)
\(=\frac{1.2.3.....19}{2.3.4.....20}\)
\(=\frac{1}{20}\)
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{20}\right)\)
\(B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{18}{19}.\frac{19}{20}\)
\(B=\frac{1}{20}\)
Hok tốt
Ta có: \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)
\(=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}\right)\)
\(< \frac{1}{2^2}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=\frac{1}{2^2}\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{1}{2^2}\left(2-\frac{1}{7}\right)=\frac{1}{2}-\frac{1}{28}< \frac{1}{2}\)
Vậy \(A< \frac{1}{2}\).
Ta có:
\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\)\(\frac{1}{19}\)
\(B=\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{15}\right)+\left(\frac{1}{16}+...+\frac{1}{19}\right)\)
\(\Rightarrow B>\left(\frac{1}{15}+\frac{1}{15}+\frac{1}{15}+...+\frac{1}{15}\right)+\left(\frac{1}{20}+...+\frac{1}{20}\right)\)
\(B>\frac{4}{5}+\frac{1}{5}\)
\(B>1\)\(\left(đpcm\right)\)
\(=\frac{\frac{6}{5}:\left(\frac{6}{5}-\frac{5}{4}\right)}{\frac{8}{25}+\frac{2}{25}}+\frac{\frac{2027}{25}:\frac{9}{4}}{\left(\frac{38}{7}-\frac{9}{4}\right):\frac{267}{56}}\)
\(=\frac{\frac{6}{5}:\left(\frac{-1}{20}\right)}{\frac{2}{5}}+\frac{\frac{8180}{225}}{\frac{89}{28}:\frac{167}{56}}\)
\(=\frac{-12}{5}:\frac{2}{5}+\frac{8180}{225}:\frac{178}{167}\)
\(=-1+...\)ra số to vcl
Đề sai à ???
Điều kiện a \(\ne\) 0, a \(\ne\) -1
Xét vế phải:
\(\frac{1}{a+1}+\frac{1}{a\left(a+1\right)}\)
= \(\frac{a\left(a+1\right)+\left(a+1\right)}{\left(a+1\right)a\left(a+1\right)}\)
= \(\frac{\left(a+1\right)\left(a+1\right)}{a\left(a+1\right)\left(a+1\right)}\)
= \(\frac{1}{a}\)(đpcm)
ta có \(\frac{1}{a+1}\)+ \(\frac{1}{a\left(a+1\right)}\)= \(\frac{a}{a.\left(a+1\right)}\)+ \(\frac{1}{a.\left(a+1\right)}\)( chỗ này ta có đc là nhờ bước quy đồng ) = \(\frac{a+1}{a.\left(a+1\right)}\)= \(\frac{1}{a}\)( còn chỗ này thì ta có nhờ rút gọn )
^_^ chúc bn học tốt ...........^_^
\(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{100}\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A< 1-\frac{1}{10}=\frac{9}{10}\)
\(=>A>\frac{65}{132}\)
