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`2/(3.5)+2/(5.7)+....+2/(2015.2017)`
`=1/3-1/5+1/5-1/7+....+1/2016-1/2017`
`=1/3-1/2017=2014/6051`
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2015.2017}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)
\(=\dfrac{1}{3}-\dfrac{1}{2017}\)
\(=\dfrac{2017}{6051}-\dfrac{3}{6051}=\dfrac{2014}{6051}\)
\(M=1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9999}\\ =\dfrac{3}{3}+\dfrac{3}{15}+\dfrac{3}{35}+...+\dfrac{3}{9999}\\ =\dfrac{3}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\\ =\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =\dfrac{3}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{3}{2}\cdot\dfrac{100}{101}=\dfrac{150}{101}\)
A bn lướt xuống dưới mà xem cách làm
nhưng của bn là cho 3 ra ngoài nha
\(=3.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{47.49}\right)\)
\(=3.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{47}-\dfrac{1}{49}\right)\)
\(=3.\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)
\(=3.\dfrac{46}{147}\)
\(=\dfrac{46}{49}\)
\(\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{47.49}\)
=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{47}-\dfrac{1}{49}\right)\)
=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)
=\(\dfrac{3}{2}.\dfrac{46}{147}\)
=\(\dfrac{23}{49}\)
\(\dfrac{x+3}{15}=\dfrac{1}{3}\left(1\right)\\ \Leftrightarrow x+3=\dfrac{1}{3}\cdot15\\ \Leftrightarrow x+3=5\\ \Rightarrow x=5-3\\ \Rightarrow x=2\)
Vậy tập nghiệm phương trình (1) là \(\left\{2\right\}\)
\(M=\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}=\frac{3\left(\frac{1}{5}+\frac{1}{7}-\frac{3}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{3}{4}\) \(\frac{3}{4}\) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=2-\frac{2}{101}=\frac{200}{101}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(B=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{101}\right)\)
\(B=2.\frac{100}{101}=\frac{200}{101}\)
\(S=\dfrac{7}{3.5}+\dfrac{7}{5.7}+\dfrac{7}{7.9}+...+\dfrac{7}{2015.2017}\)
\(\dfrac{2}{7}S=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2015.2017}\)
\(\dfrac{2}{7}S=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)
\(\dfrac{2}{7}S=\dfrac{1}{3}-\dfrac{1}{2017}\)
\(\dfrac{2}{7}S=\dfrac{2014}{6051}\)
\(S=\dfrac{4028}{42357}\)
\(S=\dfrac{7}{3.5}+\dfrac{7}{5.7}+\dfrac{7}{7.9}+...+\dfrac{7}{2015.2107}\)
\(S=\dfrac{7}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2015.2017}\right)\)
\(S=\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}...+\dfrac{1}{2015}-\dfrac{1}{2017}\right)\)
\(S=\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{2017}\right)\)
\(S=\dfrac{7}{2}.\dfrac{2014}{6051}\)
\(S=\dfrac{4028}{42357}\)
\(B=\dfrac{2^{24}\cdot3^5-2^{24}\cdot3^4}{2^{24}\cdot3^5}+1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{301}-\dfrac{1}{303}\)
\(=\dfrac{2^{24}\cdot3^4\left(3-1\right)}{2^{24}\cdot3^5}+\dfrac{302}{303}\)
\(=\dfrac{2}{3}+\dfrac{302}{303}=\dfrac{202+302}{303}=\dfrac{504}{303}\)
=168/101
\(P=\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{2015.2017}\)
\(P=3\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2015.2017}\right)\)
\(P=3.\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\right)\)
\(P=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{2017}\right)\)
\(P=\dfrac{3}{2}.\dfrac{2014}{6051}\)
\(P=\dfrac{1007}{2017}\)
1007/2017