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

B = 1/4 + 1/15 + 1/35 + 1/63 + 1/99 + 1/143 + 1/195

= 1/4 + 1/(3.5) + 1/(5.7) + 1/(7.9) + 1/(9.11) + 1/(11.13) + 1/(13.15)

= 1/4 + 1/2.(1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13 + 1/13 - 1/15)

= 1/4 + 1/2.(1/3 - 1/15)

= 1/4 + 1/2 . 4/15

= 1/4 + 2/15

= 23/60

1/2 + 1/6 + 1/12 + 1/20 + 1/30

= 1/1*2 + 1/2*3 + 1/3*4 + 1/ 4*5 + 1/5*6

= 1/1 - 1/2 + 1/2 - 1/3 + 1/4 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6

=1/1 - 1/6

= 6/6 - 1/6

= 5/6

\(tíchnha\) Trương Thị Hoàng Hà

=1/1.2 + 1/2.3 + 1/3.4 .......+1/5.6

=1-1/2 + 1/2-1/3+....+1/5-1/6

=1-1/6

=5/6

= 1/1*2 + 1/2*3 +1/2*4 + 1/4*5 + 1/5*6

=1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 -1/6

=1/1 - 1/6

=6/6 - 1/6

=5/6

ko ghi đề:

\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{15}\right)\)

\(=\frac{1}{2}.\left(\frac{5}{15}-\frac{1}{15}\right)\)

\(=\frac{1}{2}.\frac{4}{15}\)

\(=\frac{4}{30}=\frac{2}{15}\)

cố gắng lên nha bn ,đây cũng ko phải là một bà toán khó đối với hs lớp 6 đâu nha !!!!

                                TRY MY BEST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

\(\frac{7}{15}\)nha ban

bạn có thể trình bày cách làm cho mình ko

nhâ1 máy tính thôi bạn

mik nha

Đặt \(A=1\frac{7}{15}-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}-\frac{1}{143}-\frac{1}{195}\)

\(\Rightarrow A=\frac{22}{15}-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\right)\)

Đặt \(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)

\(\Rightarrow B=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\)

\(\Rightarrow2B=2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\right)\)

\(\Rightarrow2B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\)

\(\Rightarrow2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)

\(\Rightarrow2B=1-\frac{1}{15}\)

\(\Rightarrow2B=\frac{14}{15}\)

\(\Rightarrow B=\frac{14}{15}:2\Rightarrow B=\frac{7}{15}\)

\(\Rightarrow A=\frac{22}{15}-\frac{7}{15}\Rightarrow A=\frac{15}{15}=1\)

đáp án là 59​/15

   mình chắc chắn

=22/15- 1/1.3 - 1/3.5 - 1/5.7 -.........- 1/11.13 - 1/13.15

=22/15 - (1/1.3+1/3.5+....+1/13.17)

=22/15 - 1/2(2/1.3+2/3.5.........+2/13.17)

=22/15 - 1/2(1-1/3+1/3-1/4+.............+1/13-1/17)

=22/15 - 1/2(1-1/17)

=22/15-8/17

=254/255

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

\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)

\(=\frac{1}{2}\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\right)\)

\(=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)

\(=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{1}{2}\cdot\frac{4}{15}=\frac{2}{15}\)

Dấu \(.\)là dấu nhân 

\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)

\(=\frac{1}{2}.\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}+\frac{2}{195}\right)\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)\)

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

\(=\frac{1}{2}.\left(1-\frac{1}{15}\right)\)

\(=\frac{1}{2}.\frac{14}{15}\)

\(=\frac{7}{15}\)

~ Ủng hộ nhé 

Đặt \(A=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)

\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)

Suy ra ; \(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)

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

\(=1-\frac{1}{15}=\frac{14}{15}\)

=> A = \(\frac{14}{15}:2=\frac{14}{15}.\frac{1}{2}=\frac{7}{15}\)

Gọi dãy trên là A

\(\Leftrightarrow A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{13\cdot15}\)

\(\Leftrightarrow2A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{13\cdot15}\)

\(\Leftrightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\)

\(\Leftrightarrow2A=1-\frac{1}{15}\)

\(\Leftrightarrow2A=\frac{14}{15}\)

\(\Leftrightarrow A=\frac{7}{15}\)