\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

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

\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

~ Hok tốt ~

\(\)

Viết thành 2 . (1/3.5 + 1/5.7 + 1/7.9 + ...+ 1/97.99

Tui hk bít nữa

 vipboyss5: \(\frac{32}{99}\)chứ ko phải 33

1/3.5+1/5.7+1/7.9+...+1/97.99

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

=1/3-1/99

=33/99-1/99

=32/99

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

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

M=1/3-1/99

M=32/99

\(\frac{32}{99}\)

A = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/95 - 1/97 + 1/97 - 1/99

A = 1/3 - 1/99

A = 32/99

BẠN TICK CHO MÌNH NHA !

\(B=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)

   \(=2.(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99})\)

   \(=2.(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99})\)

   \(=2.(\dfrac{1}{3}-\dfrac{1}{99})\)

   \(=2.\dfrac{1}{297}\)

   =\(\dfrac{2}{297}\)

a.  

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

\(M=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

b.

\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right]\)

\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{199}\right]=\frac{291}{995}\)

mk đầu tiên nha bạn

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

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

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

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

ta có A=1-1/3+1/2-1/5+..................1/95-1/97+1/97-1/99

        A=1-1/99

        A=98/99

Cho A =2/1.3+2/3.5+2/5.7+2/7.9+....2/97.99 

A=1-1/3+1/3-1/5+1/5-1/7+..........+1/97-1/98

A=1-1/98

A=98/99

A = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/95 - 1/97 + 1/97 - 1/99

A = 1/3 - 1/99

A = 32/99

BẠN TICK CHO MÌNH NHA

32/99

A = \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{95.97}+\dfrac{2}{97.99}\)

A = \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{99}\)

A = \(\dfrac{1}{3}-\dfrac{1}{99}\)

A = \(\dfrac{32}{99}\)

Chúc bạn học tốt !

2/3.5+ 2 /5.7+ 2/7.9+...+ 2/97.99
=1/3 - 1/5 + 1/5 - 1 /7 +.... + 1/97 - 1/99
=1/3 - 1/99
=32/99

m=/3.5+2/5.7+2/7.9+.....+2/97.99

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

m=1/3-1/99

=32/99