Đặt \(A=\)\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{143}\)

\(=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{11.13}\)

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\)

\(2A=\frac{1}{3}-\frac{1}{13}=\frac{10}{39}\)

\(A=\frac{5}{39}\)

Câu còn lại cx dựa như vậy nhé bn ! 

Chúc bn hc tốt <3

câu c hình như sai đề hả bn

A=1/1.3 + 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11 + 1/11.13 + 1/13.15

A=1/1 - 1/3 +1/3 - 1/5 +1/5 -1/7+......+1/13 - 1/15

A=1 - 1/15

A=1/14

bạn ơi: 

1/195 mới bằng 1/13.15 mà!!

1/3+1/15+1/35+1/63+1/99+1/143+1/180=1/1.3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15

=1/2.(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+.....+1/13-1/15)

=1/2.(1-1/15)

=1/2.14/15

=7/15

chúc học tốt

\(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{25}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)

\(=\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)

\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...+\dfrac{1}{11}-\dfrac{1}{13}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)

\(=\left(1-\dfrac{1}{3}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)

\(=\dfrac{2}{3}\cdot\dfrac{1}{2}+\dfrac{1}{25}\)

\(=\dfrac{1}{3}+\dfrac{1}{25}\)

\(=\dfrac{28}{75}\)

cám ơi bạn

cảm ơn bạn rất nhiều

a)\(A=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{195}=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{2}{15}\)

b)\(M=1+3+3^2+...+3^{25}=\frac{3^{26}-1}{3-1}=\frac{3^{26}-1}{2}

bạn đọc lại đề bài b) đi

Giải: 
Đặt A = 1/3+1/15+1/35+1/63+1/99+1/143+1/195 
2A= 2/(1.3) + 2/(3.5) + 2/(5.7) + 2/(7.9)+2/(9.11) + 2/(11.13)+2/(13.15) 
2A=1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9... 
2A=1/1-1/15=14/15 
Vậy A=14/15 : 2 = 7/15

Nhấn đúng mk nha             Tran Dan 

  \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+..+\frac{1}{143}+\frac{1}{195}\)

=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{13.15}\)

= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+..+\frac{1}{13}-\frac{1}{15}\)

= \(1-\frac{1}{15}=\frac{14}{15}\)

tick đúng nha 

A = \(\frac{1}{1\times3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{13\times15}=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{13\times15}\right)\)

A = \(\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{1}{2}\times\left(1-\frac{1}{15}\right)=\frac{7}{15}\)

a=78/35

b=22/12

c=1/1

d=40202090/4040090

e=1,24025667172...

f=871,82

ko biết đúng ko [0_0'] hihi

Gọi dãy là A ta có :

A = 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11 + 1/11.13

A = 1/2 . ( 2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + 2/11.13 )

A = 1/2 . ( 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13 )

A = 1/2 . ( 1/3 - 1/13 )

A = 1/2 . 10/39

A = 5/39

\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)

=\(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)

=\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

=\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)\)

=\(\frac{1}{2}.\frac{10}{39}\)

=\(\frac{5}{39}\)

ai có thể giải theo cách cấp 1 ko

Hỏi đi hỏi lại vậy.          

1/15+1/35+1/63+1/99+1/143

=1/2x(1/15+1/35+1/63+1/99+1/143)

=1/2x(2/3x5+2/5x7+2/7x9+2/9x11+2/11x13)

=1/2x(1/3-1/5+1/7-1/9+1/9-1/11+1/11-1/13)

=1/2x(1/3-1/13)

=1/2x10/39

=5/39