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

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

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

Tự tính

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

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

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

32/99

k với nghe bạn

và chúc chueeuf nay thi tốt

Giải:

M=\(\dfrac{3}{3.5}+\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{95.97}+\dfrac{3}{97.99}\) 

M=\(\dfrac{3}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{95.97}+\dfrac{2}{97.99}\right)\) 

M=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{99}\right)\) 

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

M=\(\dfrac{3}{2}.\dfrac{32}{99}\) 

M=\(\dfrac{16}{33}\) 

Chúc bạn học tốt!

 M= 3/3.5 + 3/5.7 + 3/7.9 +.......+ 3/95.97 + 3/97.99

=3.1/2 ( 2/3.5+...+2/97.99)

=3.1/2(1/3- 1/5+...+1/97+1/99)

=3.1/2(1/3- 1/99)

=(3/2).(32/99)

=96/891

n/xét

3/3.5=(3/3-3/5).1/2

3/5.7=(3/5-3/7).1/2

...

3/97.99=(3/97-3/99).1/2

vậy M=(3/3-3/5).1/2+(3/5-3/7).1/2+...+(3/97-3/99).1/2

⇒M=1/2.(3/3-3/7+3/5-3/7+...+3/97-3/99)

      =1/2.(3/3-3/99)

     =1/2.32/33

   M =16/33

VẬY M=16/33

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

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

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

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

~ Hok tốt ~

\(\)

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

Tui hk bít nữa

=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 .

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