Em xem lại đề câu B nhé\(B=\dfrac{3}{2}+\dfrac{3}{6}+\dfrac{3}{12}+\dfrac{3}{20}+...+\dfrac{3}{\left(n-1\right).n}\\ =3.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{\left(n-1\right).n}\right)\\ =3.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\right)=3.\left(1-\dfrac{1}{n}\right)=3.\dfrac{n-1}{n}=3-\dfrac{3}{n}.\)

\(C=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{30.32}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{30}-\dfrac{1}{32}\\ =1-\dfrac{1}{32}=\dfrac{31}{32}.\)

\(D=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{n+1}-\dfrac{1}{n+3}\right)\\ =\dfrac{1}{2}.\left(1-\dfrac{1}{n+3}\right)=\dfrac{1}{2}.\dfrac{n+2}{n+3}.\)

a. \(\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{3-1}{3}=\dfrac{2}{3}\); \(\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5-3}{15}=\dfrac{2}{15}\)

b. Ta có \(VP=\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{2}{3}\) mà \(VP=\dfrac{2}{3}\) \(\Rightarrow VT=VP\)

Ta có \(VP=\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{2}{15}\) mà \(VP=\dfrac{2}{3.5}=\dfrac{2}{15}\) \(\Rightarrow VT=VP\)

c. \(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{97.99}+\dfrac{2}{99.101}\)

\(=2\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{97.99}+\dfrac{1}{99.101}\right)\)

\(=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=2\left(1-\dfrac{1}{101}\right)\) \(=\dfrac{200}{101}\)

a: \(\dfrac{1}{1}-\dfrac{1}{3}=1-\dfrac{1}{3}=\dfrac{2}{3}\)

\(\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{2}{15}\)

b: \(\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{3}{3}-\dfrac{1}{3}=\dfrac{2}{3}\)

\(\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5}{15}-\dfrac{3}{15}=\dfrac{2}{15}\)

c: Ta có: \(A=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\)

\(=\dfrac{100}{101}\)

\(\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{2015.2017}\)

\(=\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2015.2017}\right)\)

\(=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(=\frac{3}{2}.\left(1-\frac{1}{2017}\right)\)

\(=\frac{3}{2}.\frac{2016}{2017}\)

\(=\frac{3024}{2017}\)

_Chúc bạn học tốt_

(X-21.23):11+39=40

\(\frac{3}{1.3}\)+  \(\frac{3}{3.5}\)+ ... + \(\frac{3}{2015.2017}\) = (3 - 1) : 2 + (1 - \(\frac{3}{5}\)) : 2 + ... + (\(\frac{3}{2015}\)- \(\frac{3}{2017}\)) : 2

                                                                   = (3 - 1 + 1 -\(\frac{3}{5}\)+ ... + \(\frac{3}{2015}\) - \(\frac{3}{2017}\)) : 2

                                                                   = (3 - \(\frac{3}{2017}\)) : 2 

                                                                   = \(\frac{6048}{2017}\) : 2

                                                                   = \(\frac{3024}{2017}\)

D = \(\frac{3}{1.3}\) + \(\frac{3}{3.5}\)+.......+ \(\frac{3}{49.51}\)

D = \(\frac{3.2}{1.3.2}\)+ \(\frac{3.2}{3.5.2}\)+ .....+ \(\frac{3.2}{49.51.2}\)

D = \(\frac{3}{2}\)( \(\frac{2}{1.3}\)+ \(\frac{2}{3.5}\)+...+ \(\frac{2}{49.51}\))

D = \(\frac{3}{2}\)( \(\frac{1}{1}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)-\(\frac{1}{5}\)+ .....+ \(\frac{1}{49}\)- \(\frac{1}{51}\))

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

D = \(\frac{3}{2}\) x  \(\frac{50}{51}\)

D = \(\frac{25}{17}\)

Vậy D = \(\frac{25}{17}\)

**** xcho mình nha bn !!!! 

tại sao lạ là 3/2 hả bạn

                           \(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)

                 \(=\frac{2}{3}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)

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

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

                  \(=\frac{2}{3}.\frac{50}{51}=\frac{20}{51}\)

              Ủng hộ mk nha !!! ^_^

25/17 mới đúng

\(\frac{3}{1.3}\)+ \(\frac{3}{3.5}\)+ \(\frac{3}{5.7}\)+...+ \(\frac{3}{49.51}\)

= \(\frac{3}{2}\)( \(\frac{2}{1.3}\)+ \(\frac{2}{3.5}\)+ \(\frac{2}{5.7}\)+...+ \(\frac{2}{49.51}\))

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

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

= \(\frac{3}{2}\). \(\frac{50}{51}\)

= \(\frac{25}{17}\).

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

= 1- 1/99

= 98/99

Mãi yêu chàng trai Song Tử:bạn làm vầy là ngu roày

3/1.3+3/3.5+3/5.7+....+3/97.99=98/99

3/1.3 + 3/3.5 + 3/5.7 + ....... + 3/49.51

= 3 x ( 1/1.3 + 1/3.5 + 1/5.7 + .... + 1/49.51 )

= 3 x ( 1 - 1/51 )

= 3 x      50/51

=       150/151

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)

\(A=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)

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

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

\(A=\frac{3}{2}.\frac{50}{51}=\frac{25}{17}\)

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 !!***