1.3.5+3.5.(

\(A=9\left(\dfrac{4}{1\cdot3\cdot5}+\dfrac{4}{3\cdot5\cdot7}+...+\dfrac{4}{25\cdot27\cdot29}\right)\)

\(=9\left(\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+...+\dfrac{1}{25\cdot27}-\dfrac{1}{27\cdot29}\right)\)

\(=9\left(\dfrac{1}{3}-\dfrac{1}{783}\right)=\dfrac{260}{87}\)

đáp án là 277.1937

277.1937

\(\frac{4}{1x3x5}+\frac{4}{3x5x7}+...+\frac{4}{9x11x13}\)

\(=\frac{1}{1x3}-\frac{1}{3x5}+\frac{1}{3x5}-...+\frac{1}{9x11}-\frac{1}{11x13}\)

\(=\frac{1}{3}-\frac{1}{143}\)

\(=\frac{140}{429}\)

4/1x3x5+4/3x5x7+4/5x7x9+4/7x9x11+4/9x11x13

=1/1x3-1/3x5+1/3x5-1/5x7+1/5x7-1/7x9+1/7x9-1/9x11+1/9x11-1/11x13

=1/1x3-1/11x13

=1/3-1/143=140/429

\(\dfrac{4}{1\times3\times5}+\dfrac{4}{3\times5\times7}+\dfrac{4}{5\times7\times9}+\dfrac{4}{7\times9\times11}\)

=\(\dfrac{5-1}{1\times3\times5}+\dfrac{7-3}{3\times5\times7}+\dfrac{9-5}{5\times7\times9}+\dfrac{11-7}{7\times9\times11}\)

=\(\dfrac{1}{1\times3}-\dfrac{1}{3\times5}+\dfrac{1}{3\times5}-\dfrac{1}{5\times7}+...+\dfrac{1}{9\times11}-\dfrac{1}{11\times13}\)

=\(\dfrac{1}{3}-\dfrac{1}{143}=\dfrac{140}{429}\)

giúp mình với ạ

\(\dfrac{3}{1\cdot3\cdot5}+\dfrac{3}{3\cdot5\cdot7}+\dfrac{3}{5\cdot7\cdot9}+\cdot\cdot\cdot+\dfrac{3}{37\cdot39\cdot41}\)

\(=\dfrac{3}{4}\left(\dfrac{4}{1\cdot3\cdot5}+\dfrac{4}{3\cdot5\cdot7}+\dfrac{4}{5\cdot7\cdot9}+\cdot\cdot\cdot+\dfrac{4}{37\cdot39\cdot41}\right)\)

\(=\dfrac{3}{4}\left(\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+\dfrac{1}{5\cdot7}-\dfrac{1}{7\cdot9}+\cdot\cdot\cdot+\dfrac{1}{37\cdot39}-\dfrac{1}{39\cdot41}\right)\)

\(=\dfrac{3}{4}\left(\dfrac{1}{3}-\dfrac{1}{39\cdot41}\right)\)

\(=\dfrac{3}{4}\cdot\dfrac{532}{1599}=\dfrac{133}{533}\)

#Ayumu