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\(B=\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.305.308}\)
\(\Rightarrow\frac{6}{5}B=\frac{6}{5.8.11}+\frac{6}{8.11.14}+...+\frac{6}{302.305.308}\)
\(=\frac{11-5}{5.8.11}+\frac{14-8}{8.11.14}+...+\frac{308-302}{302.305.308}\)
\(=\frac{1}{5.8}-\frac{1}{8.11}+\frac{1}{8.11}-\frac{1}{8.11}+...+\frac{1}{302.305}-\frac{1}{305.308}\)
\(=\frac{1}{5.8}-\frac{1}{305.308}< \frac{1}{5.8}\)
Ta gọi biểu thức đó là D
\(D=\frac{5}{2}\left[\frac{1}{5.8}-\frac{1}{8.11}+....+\frac{1}{302.305}-\frac{1}{305.308}\right]\)
\(D=\frac{5}{2}.\left[\frac{1}{5.8}-\frac{1}{305.308}\right]\)
\(D=\frac{4695}{75152}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{48}-\frac{1}{49}\)
\(\Rightarrow1-A-\frac{1}{50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-...-\frac{1}{48}+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow\frac{49}{50}-A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{48}+\frac{1}{49}+\frac{1}{50}\)
\(-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{50}\right)\)
\(\Rightarrow\frac{49}{50}-A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{25}\)
\(\Rightarrow\frac{49}{50}-A=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
\(\Rightarrow A=\frac{49}{50}-\left(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+\frac{1}{29}+...+\frac{1}{50}\right)\)
Ta có :
\(\frac{1}{26}< \frac{1}{25};\frac{1}{27}< \frac{1}{25};\frac{1}{28}< \frac{1}{25};\frac{1}{29}< \frac{1}{25};\frac{1}{30}< \frac{1}{25};\)
\(\frac{1}{31}< \frac{1}{30};\frac{1}{32}< \frac{1}{30};..;\frac{1}{39}< \frac{1}{30};\frac{1}{40}< \frac{1}{30};\)
\(\frac{1}{41}< \frac{1}{40};\frac{1}{42}< \frac{1}{40};...;\frac{1}{49}< \frac{1}{40};\frac{1}{50}< \frac{1}{40}\)
\(\Rightarrow\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}< 5.\frac{1}{25}+10.\frac{1}{30}+10.\frac{1}{40}\)
\(\Rightarrow\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{47}{60}< \frac{48}{60}=\frac{4}{5}\)
\(\Rightarrow A=\frac{49}{50}-\left(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+\frac{1}{29}+...+\frac{1}{50}\right)>\frac{49}{50}-\frac{4}{5}=\frac{9}{50}>\frac{10}{50}=\frac{1}{5}\)
\(\Rightarrow A>\frac{1}{5}\)( đpcm )
a, ta xét:
\(\frac{1}{2}< \frac{2}{3}\)
\(\frac{3}{4}< \frac{4}{5}\)
\(\frac{5}{6}< \frac{6}{7}\)
.....
\(\frac{99}{100}< \frac{100}{101}\)
=>\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{99}{100}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{100}{101}\)
hay:A<B(đpcm)
b,\(A.B=\frac{1}{2}.\frac{3}{4}.....\frac{99}{100}.\frac{2}{3}.\frac{4}{5}.....\frac{100}{101}\)
\(=\frac{1.2.3....100}{2.3.4....101}=\frac{1}{101}\)
c,vì A<B (theo phần a)
=>A.A<B.A
Mà B.A=\(\frac{1}{101}\)
=>A2<101
Mà A2=\(\left(\frac{1}{2}.\frac{3}{4}.....\frac{99}{100}\right)^2\)
=>\(\left(\frac{1}{2}.\frac{3}{4}.....\frac{99}{100}\right)^2\)<\(\frac{1}{101}\)<\(\frac{1}{100}=\frac{1}{10^2}\)
=>\(\left(\frac{1}{2}.\frac{3}{4}.....\frac{99}{100}\right)^2\)<\(\frac{1}{10^2}\)
=>\(\frac{1}{2}.\frac{3}{4}....\frac{99}{100}< \frac{1}{10}\)
Hay A<\(\frac{1}{10}\)
Lời giải:
\(B=\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.205.308}\)
\(\Rightarrow \frac{6}{5}B=\frac{6}{5.8.11}+\frac{6}{8.11.14}+...+\frac{6}{302.305.308}\)
\(=\frac{11-5}{5.8.11}+\frac{14-8}{8.11.14}+...+\frac{308-302}{302.305.308}\)
\(=\frac{1}{5.8}-\frac{1}{8.11}+\frac{1}{8.11}-\frac{1}{11.14}+...+\frac{1}{302.305}-\frac{1}{305.308}\)
\(=\frac{1}{5.8}-\frac{1}{305.308}< \frac{1}{5.8}\)
\(\Rightarrow B< \frac{1}{40}.\frac{5}{6}\Leftrightarrow B< \frac{1}{48}\)