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

9 tháng 5 2018

thanks

25 tháng 2 2017

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

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

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

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

Vậy \(Q=\frac{46}{147}\)

25 tháng 2 2017

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 ...

1 tháng 4 2018

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

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

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

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

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

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

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

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

1 tháng 4 2018

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

10 tháng 5 2018

= 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

20 tháng 4 2016

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

20 tháng 4 2016

 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

3 tháng 4 2017

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

11 tháng 5 2018

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

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

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

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

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

11 tháng 5 2018

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

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

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

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

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

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!

9 tháng 5 2021

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

M=3/2.(2/3.5 +2/5.7 +2/7.9 +...+2/95.97 +2/97.99)

M=3/2.(1/3 -1/5 +1/5-1/7 +1/7-1/9+...+1/95-1/97+1/97-1/99)

M=3/2.(1/3-1/99)

M=3/2.32/99

M=16/33

11 tháng 5 2015

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

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

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