\(A=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)=\dfrac{1}{2}\left(\dfrac{100}{101}\right)=\dfrac{50}{101}\)

\(A=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)=\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{50}{101}\)

= 1/2 . (1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ...+ 1/99 + 1/101)

= 1/2 . (1/1 - 1/101)

= 1/2 . 100/101

= 50/101

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

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

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

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

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

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

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

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

\(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{\left(2x-1\right)\cdot\left(2x+1\right)}=\dfrac{49}{99}\)

\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2x-1\right)\cdot\left(2x+1\right)}=\dfrac{98}{99}\)

\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}=\dfrac{98}{99}\)

\(1-\dfrac{1}{2x+1}=\dfrac{98}{99}\)

\(\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\)

\(\dfrac{2x}{2x+1}=\dfrac{98}{99}\)

=> 2x=98

=> x=49

\(B=1.3+3.5+5.7+.....+95.97+97.99\)

\(\frac{2}{B}=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{95.97}+\frac{2}{97.99}\)

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

\(\frac{2}{B}=\frac{1}{1}-\frac{1}{99}=\frac{90}{99}=\frac{30}{33}\)

\(B=\frac{2}{\frac{30}{33}}=\frac{2.33}{30}=\frac{33}{15}\)

\(6A=1.3.6+3.5.6+5.7.6+...+97.99.6\)

= \(1.3\left(5+1\right)+3.5\left(7-1\right)+5.7\left(9-3\right)+...97.99\left(101-95\right)\)

= \(.3.5+1.3+3.5.7-1.3.5+5.7.9-3.5.7+...+97.99.101-97.97.99\)

= 3 + 97 .99 . 101 

= \(\frac{1+97.33.101}{2}\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}\right)=\dfrac{1}{2}\cdot\dfrac{2008}{2009}=\dfrac{1004}{2009}\)

  A=1.3+3.5+5.7+...+95.97+97.99

6A=1.3.6+3.5.6+5.7.6+...+95.97.96+97.99.96

    =1.3.(5+1)+3.5.(7-1)+...+95.97.(99-93)+97.99.(101-95)

    =1.1.3+1.3.5-1.3.5+3.5.7-....-95.97.99+97.99.101

    =3.97.99.101

=>A=\(\frac{3+97.99.101}{6}=\frac{1+97.33.101}{2}\)\(=161651\)

Chép sai đề bài

the minh do lai ban

D=1+3-5-7+9+11....-397-399

\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + ..... + \(\dfrac{2}{95.97}\)

= 1 - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + .... + \(\dfrac{1}{95}\) - \(\dfrac{1}{97}\)

= \(1-\dfrac{1}{97}\) 

= \(\dfrac{96}{97}\)

\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{95\times97}\)

\(=\dfrac{2}{3}\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{95\times97}\right)\)

\(=\dfrac{2}{3}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)

\(=\dfrac{2}{3}\left(1-\dfrac{1}{97}\right)\)\(=\dfrac{2}{3}\times\dfrac{96}{97}\)\(=\dfrac{64}{97}\)

=3.(3/1.3+3/3.5+3/5.7+...+3/95.97+3/97.99)

=3(1-1/3+1/3-1/5+1/5-1/7+...+1/95-1/97+1/97-1/99)

=3[(1-1/99)+(1/5-1/5)+(1/7-1/7)+...+(1/97-1/97)]

=3(1-1/99)=3(99/99-1/99)=3.98/99=1.98/33=98/33

Neu la 3 ma ko phai la 3^2 thi sao : Tinh gium minh nha .