A=1/1-1/2+1/2-1/3+1/3-1/4+....+1/49-1/50

A=1/1-1/50

A=49/50

Vay A=49/50

B=1/3-1/5+1/5-1/7....+1/37-1/39

B=1/3-1/39

b=36/117

B=4/13

39%.21%=18% ; 1000/125.125/1000

Câu C sai đề rùi!

Đề đúng : C =1/1.2 + 1/2.3 +......+ 1/49.50

C = 1 - 1/2 + 1/2 - 1/3 +......+1/49 - 1/50

C = 1 - 1/50

C = 49/50

D = 2/3.5 + 2/5.7 +.......+ 2/37.39

D = 1/3 - 1/5 + 1/5 - 1/7 +.......+ 1/37 - 1/39

D = 1/3 - 1/39

D = 12/39

\(B=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{37.39}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{37}-\frac{1}{39}=\frac{1}{3}-\frac{1}{39}=\frac{12}{39}=\frac{4}{13}\)

Okok

Tham khảo :Câu hỏi của hoàng quỳnh dương - Toán lớp 7 - Học toán với OnlineMath

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

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

   \(=\frac{1}{3}-\frac{1}{39}\)

   \(=\frac{4}{13}\)

Study well ! >_<

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

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

\(B=\frac{1}{3}-\frac{1}{39}=\frac{13}{39}-\frac{1}{39}=\frac{12}{39}\)

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

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

\(\Rightarrow B=\frac{1}{3}-\frac{1}{39}=\frac{12}{39}\)

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

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

\(=\frac{1}{3}-\frac{1}{39}\)

\(=\frac{13}{39}-\frac{1}{39}\)

\(=\frac{12}{39}=\frac{4}{13}\)

ta có A=1/3-1/5+1/5-1/7+1/7-1/9+....+1/37-1/39

          =1/3-1/39

          =12/39

A = 2/3.5 + 2/5.7 + 2/7.9 + ... + 2/37.39

=1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/37 - 1/39

=1/3 - 1/39

=12/39

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)

\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(A=\dfrac{1}{1}-\dfrac{1}{50}=\dfrac{49}{50}\)

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(A=\dfrac{1}{1}-\dfrac{1}{50}=\dfrac{50}{50}-\dfrac{1}{50}=\dfrac{49}{50}\)

A = \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)

= \(\left(1-\dfrac{1}{2}\right)\)+\(\left(\dfrac{1}{2}-\dfrac{1}{3}\right)\)+...+\(\left(\dfrac{1}{49}-\dfrac{1}{50}\right)\)

= \(\left(1-\dfrac{1}{50}\right)\) = \(\dfrac{49}{50}\)