Question

A = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)

Answer 1

\(A=\frac{1}{2}.\left(\frac{1}{3.5}+\frac{1}{5.7}+...\frac{1}{37.39}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{39}\right)=\frac{1}{2}.\frac{12}{39}=\frac{6}{39}\)

Answer 2

Ta đặt nhân tử chung nha :

  \(A=\frac{1}{2}\left(\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{37.39}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{39}\right)\)

\(=\frac{1}{2}.\frac{12}{39}\)

\(=\frac{6}{39}\)

Answer 3

=\(\frac{5-3}{3\cdot5}\)+ \(\frac{7-5}{5\cdot7}\)+ \(\frac{9-7}{7\cdot9}\)+...+ \(\frac{39-37}{37\cdot39}\)

= \(\frac{5}{3\cdot5}\)- \(\frac{3}{3\cdot5}\)+ \(\frac{7}{5\cdot7}\)- \(\frac{5}{5\cdot7}\)+ \(\frac{9}{7\cdot9}\)- \(\frac{7}{7\cdot9}\)+...+ \(\frac{39}{37\cdot39}\)- \(\frac{37}{37\cdot39}\)

= \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+ \(\frac{1}{7}\)- \(\frac{1}{9}\)+...+ \(\frac{1}{37}\)- \(\frac{1}{39}\)

= \(\frac{1}{3}\)- \(\frac{1}{39}\)

=\(\frac{4}{13}\)