\(Q=\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{47.49}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{47}-\frac{1}{49}\)

\(=\frac{1}{3}-\frac{1}{49}\)

\(=\frac{46}{147}\)

Vậy \(Q=\frac{46}{147}\)

Ta có : \(\frac{2}{3}Q=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{47.49}\)

\(\Rightarrow\frac{2}{3}Q=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}\)

\(\Rightarrow\frac{2}{3}Q=\frac{1}{3}-\frac{1}{49}=\frac{49}{147}-\frac{3}{147}=\frac{46}{147}\)

\(\Rightarrow Q=\frac{46}{147}\div\frac{2}{3}=\frac{138}{294}=\frac{23}{49}\)

Vậy ...

\(C=\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{3}{47\cdot49}\)

\(\Rightarrow\frac{2}{3}C=\frac{2}{3}\cdot\left(\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{3}{47\cdot49}\right)\)

\(\Rightarrow\frac{2}{3}C=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{47\cdot49}\)

\(\Rightarrow\frac{2}{3}C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{47}-\frac{1}{49}\)

\(\Rightarrow\frac{2}{3}C=\frac{1}{3}-\frac{1}{49}\)

\(\Rightarrow\frac{2}{3}C=\frac{46}{147}\)

\(\Rightarrow C=\frac{46}{147}:\frac{2}{3}\)

\(\Rightarrow C=\frac{23}{49}\)

3/3.5+3/5.7+3/7.9+.....+3/47.49

=1-1/5+1/5-1/7+...+1/47-1/49

=1-1/49

=48/49

C = 3/2 . (2/3.5 + 2/5.7 + 2/7.9 + ... + 2/47.49)

C = 3/2 . (1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/47 - 1/49)

C = 3/2 . (1/3 - 1/49) = 3/2 . 46/147 = 23/49

Chúc bn HỌC TỐT nhá!!!!

= 3/3 - 3/5 + 3/5 - 3/7 + ... + 3/47 - 3/49

= 3/3 - 3/49

= 46/49

= lưu ý : phép tính đầu là gach bỏ các số là : 3/5 + 3/5 - 5/7 + 5/7 - ... + 3/47

Chúc bạn học giỏi

1/1-1/3+1/3-1/5+1/5-1/7+...... +1/47-1/49

 3/1.3+3/3.5+3/5.7+......+3/47.49

=1/1-1/3+1/3-1/5+1/5-1/7+........+1/47-1/49

=1/1-1/49

=49/49-1/49

=48/49

= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 +......+ 1/47 - 1/49

= 1 - 1/49

= 48/49 nha!

*** Ai k mk mk k lại !!***

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

\(=3.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{47.49}\right)\)

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

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

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

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

\(\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{47.49}\)

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

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

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

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

=3/2.(1/3-1/5+1/5-1/7+...+1/47-1/49)

=3/2.(1/3-1/49)

=3/2.46/147=23/49

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

Đặt \(A=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)

\(\Rightarrow A=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)=\frac{3}{2}.\frac{56}{305}=\frac{84}{305}\)