=>A=\(\frac{7}{2}\)(\(\frac{1}{1}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+...+\(\frac{1}{99}\)-\(\frac{1}{101}\))

=>A=\(\frac{7}{2}\)(1-\(\frac{1}{101}\))

=>A=\(\frac{350}{101}\)

7/2 ( \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{99}-\frac{1}{101}\))

7/2 ( 1 - 1/101 ) 

7/2 x 100/101

=350/101 

Rút gọn :

\(\frac{-63}{81}=\frac{-7\cdot9}{9\cdot9}=\frac{-7}{9}\)

\(\frac{-5\cdot6}{9\cdot35}=\frac{-5\cdot2\cdot3}{3\cdot3\cdot5\cdot7}=\frac{-2}{3\cdot7}=\frac{-2}{21}\)

\(\frac{7\cdot2+8}{2\cdot14\cdot5}=\frac{14+8}{2\cdot2\cdot7\cdot5}=\frac{22}{2\cdot2\cdot7\cdot5}=\frac{2\cdot11}{2\cdot2\cdot7\cdot5}=\frac{11}{2\cdot7\cdot5}=\frac{11}{70}\)

\(\frac{-63}{81}=\frac{-7}{9}\)

\(\frac{-5.6}{9.35}=\frac{-5.2.3}{3.3..5.7}=\frac{-2}{3.7}=\frac{-2}{21}\)

\(\frac{7.2+8}{2.14.5}=\frac{22}{2.2.7.5}=\frac{2.11}{2.2.7.5}=\frac{11}{70}\)

a=1/3x5+1/5x7+...+1/2003x2005

a=1x2/3x5x2+1x2/5x7x2+...+1x2/2003x2005x2

a=1/2(2/3x5+2/5x7+...+2/2003x2005)

a=1/2x(1/3-1/5+1/5-1/7+...+1/2003-1/2005)

a=1/2x(1/3-1/2005)

a=1/2x2002/6015

a=1001/6015

A = 1/3.5 + 1/5.7 + 1/7.9 + .... + 1/2003.2005 

2A = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + .... + 1/2003 - 1/2005

2A = 1/3 - 1/2005 = 2002/6015 

=>A = 1001/6015

gọi biểu thức đó là A

\(A=\frac{1}{3.5}+\frac{1}{5.7}+.......+\frac{1}{2009.2011}\)

\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{2009}-\frac{1}{2011}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2011}\right)\)

\(A=\frac{1}{2}.\left(\frac{2008}{6033}\right)\)

\(A=\frac{1004}{6033}\)

mink nghĩ vậy bạn ạ

C.mơn bạn nha ! ^_^

Cho e hỏi cái này. Ở câu 1 ý, cuối đề là \(-\frac{1}{7}\) sao xuống dưới phải đổi thành -1 thế ạ ? E chưa hiểu lắm :<

ờ thì kiểu :
\(\frac{1}{7}.\frac{1}{3}+\frac{1}{7}.\frac{1}{2}-\frac{1}{7}=\frac{1}{7}.\frac{1}{3}+\frac{1}{7}.\frac{1}{2}+\frac{1}{7}.\left(-1\right)=\frac{1}{7}.\left(...\right)\)

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{96}\)

\(2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\right)=2.\frac{15}{96}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right).\left(2x+3\right)}=\frac{5}{16}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{5}{16}\)

\(\frac{1}{3}-\frac{1}{2x+3}=\frac{5}{16}\)

\(\frac{1}{2x+3}=\frac{1}{3}-\frac{5}{16}\)

\(\frac{1}{2x+3}=\frac{1}{48}\)

=> 2x + 3 = 48

=> 2x = 48 - 3

=> 2x = 45

=> x = 45/2

a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)

\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\)

\(=\frac{1}{2}-\frac{1}{3n+2}=\frac{3n+2}{2\cdot\left(3n+2\right)}-\frac{2}{2\cdot\left(3n+2\right)}\)

\(=\frac{3n+2-2}{6n+4}=\frac{3n}{6n+4}=VP\)

chết phần a quên nhân vs 1/3

\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)

\(=2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

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

\(=1-\frac{1}{100}\Rightarrowđpcm\)

Ta có :

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}< 1\)\(\left(đpcm\right)\)