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

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

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

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

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

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

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

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

Ta đặt nhân tử chung nha :

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

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

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

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

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

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

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

= \(\frac{1}{3}\)- \(\frac{1}{39}\)

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

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{87.89}\)

= \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{87}-\frac{1}{89}\right)\)

= \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{89}\right)\)

= \(\frac{1}{2}.\frac{86}{267}=\frac{43}{267}\)

~~~
Đáp số to quá, tớ không chắc là mình đúng đâu.

#Sunrise

=1/3-1/5+1/5-1/7+1/7-1/9+.....+1/87-1/89

=1/3-1/89

=86/267

trả lừoi

43/267

nha

hok tốt

http://olm.vn/hoi-dap/question/772291.html

sau 3 phút có kết quả tùy bạn

A=\(\frac{37}{78}\)

B=\(\frac{9}{38}\)

\(S=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\)

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

\(=\frac{1}{2}\left(1-\frac{1}{99}\right)\)

\(=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)

S=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+......+\frac{1}{95.97}+\frac{1}{97.99}\)

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

S=\(\frac{1}{2}.\left(1-\frac{1}{99}\right)\)

S=\(\frac{1}{2}.\frac{98}{99}\)

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

S = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

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

= \(\frac{1}{2}\). (\(1-\frac{1}{99}\))

= \(\frac{1}{2}\). \(\frac{98}{99}\) = \(\frac{49}{99}\)

=1/3-1/5+1/5-1/7+1/7-1/9+....+1/97-1/99

= 1/3 -1/99

=32/99

tích cho mình nha

=1/3-1/5+1/7-1/7+1/9-1/9+...+1/97-1/99

=1/3-1/99

=32/99

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

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

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

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

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

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