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\(2B=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{23}-\frac{1}{25}\right)\)
\(2B=2\left(\frac{1}{3}-\frac{1}{25}\right)\)
\(2B=2\times\frac{22}{75}\)
\(B=\frac{44}{75}\)
\(=4\left(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{53.55}\right)\)
\(=4\left(\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+...+\frac{1}{53}-\frac{1}{55}\right)\)
\(=4\left(\frac{1}{5}-\frac{1}{55}\right)\)
\(=4.\frac{2}{11}\)
\(=\frac{8}{11}\)
\(\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{2013.2015}=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2013.2015}\right)=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=2.\left(\frac{2015}{2015}-\frac{1}{2015}\right)\)
\(=2.\frac{2014}{2015}\)
\(=\frac{4028}{2015}\)
Ta có :
\(A=2.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+............+\frac{2}{53.55}\right)\)
\(\Rightarrow A=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+..............+\frac{1}{53}-\frac{1}{55}\right)\)
\(\Rightarrow A=2.\left(\frac{1}{5}-\frac{1}{55}\right)=2.\frac{2}{11}=\frac{4}{11}\)
k nha bạn !!!
4/3.5+4/5.7+4/7.9+4/9.11
=4.(1/3.5+1/5.7+1/7.9+1/9.11)
=4.1/2.(2/3.5+2/5.7+2/7.9+2/9.11)
=2.(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11)
=2.(1/3-1/11)
=2.8/33
=16/33
4/3.5+4/5.7+4/7.9+4/9.11
=4.2/2.3.5+4.2/2.5.7+4.2/2.7.9+4.2/2.9.11
=4/2.2/3.5+4/2.2/5.7+4/2.2/7.9+4/2.2/9.11
=4/2.(2/3.5+2/5.7+2/7.9+2/9.11)
=4/2.(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11)
=2.(1/3-1/11)
=2.8/33
=16/33
S=\(\frac{2}{5.7}\)+\(\frac{2}{7.9}\)+....+\(\frac{2}{93.95}\)+\(\frac{3}{95.98}\)+\(\frac{4}{98.102}\)+\(\frac{5}{102.107}\)+\(\frac{2012}{107.2119}\)
S=\(\frac{1}{5}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{9}\)+....+\(\frac{1}{93}\)-\(\frac{1}{95}\)+\(\frac{1}{95}\)-\(\frac{1}{98}\)+\(\frac{1}{98}\)-\(\frac{1}{102}\)+\(\frac{1}{102}\)-\(\frac{1}{107}\)+\(\frac{1}{107}\)-\(\frac{1}{2119}\)
S=\(\frac{1}{5}\)-\(\frac{1}{2119}\)
S=\(\frac{2114}{10595}\)
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{99}\right)\)
\(=\left(\frac{2}{2}+\frac{1}{2}\right)\left(\frac{3}{3}+\frac{1}{3}\right)\left(\frac{4}{4}+\frac{1}{4}\right).....\left(\frac{99}{99}+\frac{1}{99}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)
\(=\frac{3.4.5....100}{2.3.4....99}=\frac{100}{2}=50\)
\(\frac{x}{3.5}+\frac{x}{5.7}+\frac{x}{7.9}+...+\frac{x}{13.15}=\frac{4}{45}\)
\(\Leftrightarrow\frac{x}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{13.15}\right)=\frac{4}{45}\)
\(\Leftrightarrow\frac{x}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{4}{45}\)
\(\Leftrightarrow\frac{x}{2}.\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{4}{45}\)
\(\Leftrightarrow\frac{x}{2}.\frac{4}{15}=\frac{4}{45}\)
\(\Leftrightarrow\frac{x}{2}=\frac{4}{45}:\frac{4}{15}\)
\(\Leftrightarrow\frac{x}{2}=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}.2\)
\(\Leftrightarrow x=\frac{2}{3}\)
Vậy x = \(\frac{2}{3}\)
_Chúc bạn học tốt_
`A=4/[5.7]+4/[7.9]+4/[9.11]+...+4/[21.23]+4/[23.25]`
`A=2.(2/[5.7]+2/[7.9]+....+2/[23.25])`
`A=2.(1/5-1/7+1/7-1/9+....+1/23-1/25)`
`A=2.(1/5-1/25)`
`A=2. 4/25`
`A=8/25`