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

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

\(\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+....+\frac{3}{49.51}\)

= 3. \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{49}-\frac{1}{51}\right)\)

=\(\frac{3}{2}\left(\frac{1}{3}-\frac{1}{51}\right)\)

=\(\frac{3}{2}.\frac{16}{51}\)

=\(\frac{8}{17}\)

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\(A=\dfrac{3}{3.5} + \dfrac{3}{5.7} + ... + \dfrac{3}{97.99}\)

\(\Rightarrow A=\dfrac{3}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\right)\)

\(\Rightarrow A=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(\Rightarrow A=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)

\(\Rightarrow A=\dfrac{3}{2}.\dfrac{32}{99}\)

\(\Rightarrow A=\dfrac{16}{33}\)

Vậy \(A=\dfrac{16}{33}\)

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

= \(\dfrac{3}{2}.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\right)\)

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

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

= \(\dfrac{3}{2}.\dfrac{32}{99}\)

= \(\dfrac{3.32}{2.99}\)= \(\dfrac{3.2.3.6}{2.11.3.3}\)= \(\dfrac{6}{11}\)

cho mình hỏi là tại sao lại đặt \(\dfrac{3}{2}\) ra ngoài ?

Đặt A = \(\dfrac{3}{3.5}+\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{47.49}\)

2A = \(\dfrac{3.2}{3.5}+\dfrac{3.2}{5.7}+\dfrac{3.2}{7.9}+...+\dfrac{3.2}{47.49}\)

2A = 3\(\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{47.49}\right)\)

2A = 3 \(\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{47}-\dfrac{1}{49}\right)\)

2A = 3 \(\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)

2A = 3 . \(\dfrac{46}{147}\)

2A = \(\dfrac{46}{49}\)

=> A = \(\dfrac{46}{49}\) : 2

=> A = \(\dfrac{23}{49}\)

B=\(\dfrac{3}{3.5}.\dfrac{3}{5.7}.....\dfrac{3}{47.49}\)

B=\(\dfrac{3}{2}.\left(\dfrac{2}{3.5}.\dfrac{2}{5.7}.....\dfrac{2}{47.49}\right)\)

B=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{47}-\dfrac{1}{49}\right)\)

B=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)

B=\(\dfrac{3}{2}.\dfrac{46}{147}\)

B=\(\dfrac{23}{49}\)

a) Ta có: \(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{2499}{2500}\)

\(=\dfrac{1\cdot3}{2^2}\cdot\dfrac{2\cdot4}{3^2}\cdot\dfrac{3\cdot5}{4^2}\cdot...\cdot\dfrac{49\cdot51}{50^2}\)

\(=\dfrac{1}{50}\cdot\dfrac{51}{2}=\dfrac{51}{100}\)

b) Ta có: \(B=\dfrac{3}{3\cdot5}+\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+...+\dfrac{3}{47\cdot49}\)

\(=\dfrac{3}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{47\cdot49}\right)\)

\(=\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{46}{147}=\dfrac{138}{294}=\dfrac{23}{49}\)