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

\(=\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+...+\frac{2}{37}-\frac{2}{39}\)

\(=\frac{2}{3}-\frac{2}{39}\)

\(=\frac{8}{13}\)

Ta có:

\(\frac{2}{3.5}=\frac{1}{3}-\frac{1}{5}\)

\(\frac{2}{5.7}=\frac{1}{5}-\frac{1}{7}\)

\(\frac{2}{7.9}=\frac{1}{7}-\frac{1}{9}\)

\(......................................\)

\(\frac{2}{37.39}=\frac{1}{37}-\frac{1}{39}\)

nên \(C=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)

\(C=\frac{1}{3}-\frac{1}{39}=\frac{4}{13}\)

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

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

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

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

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

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

\(2A=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\right).2\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

\(2A=1-\frac{1}{99}\)
\(2A=\frac{98}{99}\)

\(A=\frac{98}{99}:2\)

\(A=\frac{49}{99}\)

a) \(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{24\cdot25}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{25}\)
\(\Leftrightarrow\frac{4}{25}\)

a/ =1/5-1/6+1/6-1/7+1/7-1/8+...+1/24-1/25=1/5-1/25=4/25

Mấy câu còn lại thì từ từ!

b/ \(2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\right)=2\left(1-\frac{1}{101}\right)=2.\frac{100}{101}=\frac{200}{101}\)

Ai thấy đúng thì tik nha Đa tạ!!!!!!!! Chúc bn năm ms zui zẻ!!!!!!!!!!

a, \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)

\(=\frac{1}{5}-\frac{1}{25}\)

\(=\frac{4}{25}\)

b, \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

Gọi biểu thức trên là A 

\(C=\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{37\cdot39}\)

\(2C=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{37\cdot39}\)

\(2C=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\)

\(2C=\dfrac{1}{3}-\dfrac{1}{39}\)

\(2C=\dfrac{4}{13}\)

\(C=\dfrac{2}{13}\)

\(C=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\)

\(C=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\right)\)

\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\right)\)

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

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

\(C=\frac{4}{26}=\frac{2}{13}\)

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

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

Ta đặt nhân tử chung nha :

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

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

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

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

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

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

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

= \(\frac{1}{3}\)- \(\frac{1}{39}\)

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

a, Ta có:

\(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{0,6-\frac{3}{9}+\frac{3}{11}}+\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{14}}{-1-\frac{3}{7}+\frac{3}{28}}=\frac{2\left(0,2-\frac{1}{9}+\frac{1}{11}\right)}{3\left(0,2-\frac{1}{9}+\frac{1}{11}\right)}+\frac{2\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{28}\right)}{-3\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{28}\right)}=\frac{2}{3}+\frac{-2}{3}=0\)

k đúng cho mình nha. Thanks!!!

a, bày cho mình cách viết bằng phân số đi , mình trình bày cách làm cho. k đúng cho mình nha.

\(=\frac{2}{3}\times\frac{0,1-\frac{1}{9}+\frac{1}{11}}{0,1-\frac{1}{9}+\frac{1}{11}}+\frac{-\frac{2}{3}\times\left(-1-\frac{3}{7}+\frac{3}{28}\right)}{-1-\frac{3}{7}+\frac{3}{28}}\) 

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

=\(-\frac{5}{6}\)