Question

\(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{17\times19}\)=

Answer 1

Đặt \(A=\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{17x19}\)

=>\(2xA=2x\left(\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{17x19}\right)\)

=>\(2xA=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{17x19}\)

=>\(2xA=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}\)

=>\(2xA=1-\frac{1}{19}=\frac{18}{19}\)

=>\(A=\frac{18}{19}:2=\frac{9}{19}\)

Answer 2

(\(\frac{1}{1}-\frac{1}{3}\left(\right)+\left(\right)\frac{1}{3}-\frac{1}{5}\left(\right)+\left(\right)\frac{1}{5}-\frac{1}{7}\left(\right)+....+\left(\right)\frac{1}{17}-\frac{1}{19}\left(\right)\)\(\frac{1}{19}\)

\(\frac{1}{1}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+....+\left(\frac{1}{17}-\frac{1}{17}\right)-\frac{1}{19}\)

\(\frac{1}{1}-\frac{1}{19}=\frac{18}{19}\)

Answer 3

=2x(\(\frac{1}{1\times3}+\)\(\frac{1}{3\times5}+\)\(\frac{1}{5\times7}+\)..........+\(\frac{1}{17\times19}\))

=\(\frac{2}{1\times3}+\)\(\frac{2}{3\times5}+\)\(\frac{2}{5\times7}+\)............+\(\frac{2}{17\times19}\)

=\(\frac{1}{1}-\)\(\frac{1}{3}+\)\(\frac{1}{3}-\)\(\frac{1}{5}+\)\(\frac{1}{5}-\frac{1}{7}\)\(+\)..........\(+\)\(\frac{1}{17}-\frac{1}{19}\)

=\(\frac{1}{1}-\frac{1}{19}\)

=\(\frac{19}{19}-\frac{1}{19}\)

=\(\frac{18}{19}\)

Answer 4

\(A=\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+...+\frac{1}{17x19}\)

   \(=\frac{1}{2}\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{17x19}\right)\)

    \(=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{17}-\frac{1}{19}\right)\)

  \(=\frac{1}{2}\left(1-\frac{1}{19}\right)=\frac{1}{2}\frac{18}{19}=\frac{9}{19}\)