\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\)

\(A=1-\frac{1}{51}\)

\(A=\frac{50}{51}\)

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)

\(2A=3\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\right)\)

\(2A=3\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(2A=3\left(1-\frac{1}{51}\right)\)

\(2A=3.\frac{50}{51}\)

\(2A=\frac{50}{17}\Rightarrow A=\frac{25}{17}\)'

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)

\(A=\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)

\(A=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(A=\frac{3}{2}.\left(1-\frac{1}{51}\right)\)

\(A=\frac{3}{2}.\frac{50}{51}=\frac{25}{17}\)

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)

\(A=3.\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{50}\right)\)

\(A=\frac{3}{2}\left(1-\frac{1}{50}\right)\)

\(A=\frac{3}{2}.\frac{49}{50}\)

\(A=\frac{147}{100}\)

A=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{49.51}\)

\(\frac{1}{2}\).A=\(\frac{3}{1.3.2}+\frac{3}{3.5.2}+\frac{3}{5.7.2}+....+\frac{3}{49.51.2}\)

\(\frac{1}{2}.A\)=\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{51}\)

\(\frac{1}{2}.A\)=\(1-\frac{1}{51}\)

\(\frac{1}{2}.A\)=\(\frac{50}{51}\)

A=\(\frac{50}{51}.2\)

A=\(\frac{100}{51}\)

A=3/1.3+3/3.5+3/5.7+............+3/49.51

A=3/1-3/3=3/3-3/5+3/5-3/7+...............+3/49-3/51

A=1-1/3+1/3-1/5+1/5-1/7+.....................+1/39-1/51

A=1-1/51

A=50/51

A\(=3\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...\frac{1}{49.51}\right) \)

    \(=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...\frac{2}{49.51}\right)\)

  \(=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

     =\(\frac{3}{2}\left(1-\frac{1}{51}\right)\) 

    \(=\frac{3}{2}.\frac{50}{51}\)   

  \(=\frac{25}{17}\)

ko tin cu an thu nguyen cai cum dau vao may tinh cer :v

\(\frac{1}{1.3}+\left|\frac{-1}{3.5}\right|+\left|\frac{-1}{5.7}\right|+...+\left|\frac{-1}{49.51}\right|=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{51}\right)=\frac{1}{2}\left(1-\frac{1}{51}\right)=\frac{1}{2}.\frac{50}{51}=\frac{25}{51}\)

    \(\frac{1}{1.3}+\left|-\frac{1}{3.5}\right|+\left|-\frac{1}{5.7}\right|+...+\left|-\frac{1}{49.51}\right|\)

\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49\cdot51}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{51}\right)\)

\(=\frac{1}{2}.\frac{50}{51}\)

\(=\frac{25}{51}\)

a, Ta có \(A=\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{49.51}\)

\(=\frac{3}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{49.51}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)\)

\(=\frac{1}{2}-\frac{3}{102}=\frac{48}{102}=\frac{24}{51}\)

b,Ta có \(\frac{1}{2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}\)

\(=\frac{2-1}{2}+\frac{4-2}{2.4}+\frac{7-4}{4.7}+\frac{11-7}{7.11}+\frac{16-11}{11.16}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\)

\(=\frac{15}{16}\)

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

!!!!!!!!!!!!!!!!!!!!!

!!!!!!!!!!!!!!1111

\(a)\) \(A=\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{49.51}\)

\(A=3\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{49.50}\right)\)

\(2A=3\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{49.50}\right)\)

\(2A-A=3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(A=3\left(\frac{1}{3}-\frac{1}{50}\right)\)

\(A=1-\frac{3}{50}\)

\(A=\frac{47}{50}\)

Vậy \(A=\frac{47}{50}\)

\(b)\) \(B=\frac{1}{2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}\)

\(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\)

\(B=1-\frac{1}{16}\)

\(B=\frac{15}{16}\)

Vậy \(B=\frac{15}{16}\)

Chúc bạn học tốt ~

=1/2.(2/1.3 + 2/3.5 + 2/5.7 +.....+ 2/49.51)

=1/2.(1-1/3+1/3-1/5+1/5-1/7+.....+1/49-1/51)

=1/2.(1-1/51)

=1/2.50/51

=25/51

=1/2.(2/1.3 + 2/3.5 + 2/5.7 +.....+ 2/49.51)

=1/2.(1-1/3+1/3-1/5+1/5-1/7+.....+1/49-1/51)

=1/2.(1-1/51)

=1/2.50/51

=25/51

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)

\(=\frac{1}{2.}\left(1-\frac{1}{1}+\frac{1}{3}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{51}\right)\)

\(=\frac{1}{2}.\frac{50}{51}\)

\(=\frac{25}{51}\)