\(B=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}\)

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

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

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{1}{2}\cdot\frac{10}{39}=\frac{5}{39}\)

\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{1.13}\)

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

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

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{1}{2}.\frac{10}{39}=\frac{5}{39}\)

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

\(\Rightarrow2B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\)

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

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

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

\(\Rightarrow B=\frac{10}{39}:2=\frac{5}{39}\)

2S=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)

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

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

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

a. Ta có:A= 1/1.3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15

A=1/2(1/1.3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15)

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

A=2(1-1/15)

A=1/2.14/15

A=7/15

phần b nè

pt \(\Rightarrow90-6ab=3a\)\(\Leftrightarrow3a\left(b+2\right)=90\)vì b>0 \(\Leftrightarrow a=\frac{30}{b+2}\)mà a,b \(\inℕ^∗\)

\(\Rightarrow\)b+2\(\inƯ\left(30\right)\)MÀb\(\inℕ^∗\)\(b+2\in\left\{3;5;6;10;15;30\right\}\)khi đó tìm đc b \(\rightarrow\)thau vào tìm a . nhớ thử lại vào pt ban đầu nhé 

k cho mk nha mn ^.^

S=1/3 + 1/15 + 1/35 + 1/63 + 1/99

=>S=1/1*3+1/3*5+1/5*7+1/7*9+1/9*11

Nhân cả hai vế với 2 ta được:

=>2S=1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11

=>2S=1-1/11

=>2S=11/11-1/11

=>2S=10/11

=>S=10/11 : 2

=>S=5/11

  Vậy S= 5/11

\(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)

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

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

\(S=\frac{1}{2}\left(1-\frac{1}{11}\right)\)

\(S=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)

cảm ơn bạn nha

A=1/1*3+1/3*5+1/5*7+.....+1/99*101

A=1/3*(1-1/3+1/3-1/5+1/5-1/7+.......+1/99-1/101)

A=1/3*(1-1/101)

A=1/3*100/101

A=300/301

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(A=\frac{1}{1}-\frac{1}{101}=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)

\(2B=\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{9.11}\)

\(2B=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\)

\(2B=\frac{1}{1}-\frac{1}{11}=\frac{10}{11}\)

\(B=\frac{10}{11}:2=\frac{10}{11}.\frac{1}{2}=\frac{5}{11}\)

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

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

\(=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)

B = 1/3 + 1/15 + 1/35 + 1/63 + 1/99

B = 1/2 × (2/1×3 + 2/3×5 + 2/5×7 + 2/7×9 + 2/9×11)

B = 1/2 × (1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11)

B = 1/2 × (1 - 1/11)

B = 1/2 × 10/11

B = 5/11

mình không biết nữa bằng bao nhiêu ấy nhỉ .......? .......? Sory ^.^

1/3 + 13/15 + 33/35 + 61/63 + 97/99

= 45/11 ( mình không tiện giải, để khi khác giải sau)

Chúc bạn may mắn!

= 45/11

mik làm biếng ghi lâu lắm bạn ạ !!!

k mik nhaaaaaaaaaaaaaaaaaaaaaa

\(B=\frac{1}{5.6}+\frac{1}{10.9}+\frac{1}{15.12}+...+\frac{1}{3350.2013}\)

\(B=\frac{1}{5.3}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{670.671}\right)\)

\(B=\frac{1}{15}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{670}-\frac{1}{671}\right)\)

\(B=\frac{1}{15}.\left(1-\frac{1}{671}\right)\)

\(B=\frac{1}{15}.\frac{670}{671}=\frac{134}{2013}\)

Nguyễn Huy Thắngsoyeon_Tiểubàng giảiSilver bulletLê Nguyên HạoPhương AnVõ Đông Anh Tuấnsoyeon_Tiểubàng giảiLê Thị Linh ChiNguyễn Huy Tú

đề sai ak 2 con S 

B=2/1.3 + 2/3.5 + 2/5.7 +...+ 2/299.301

B=1-1/3+1/3-1/5+1/5-1/7+...+1/299-1/301=1-1/301=300/301

\(Ta có: \frac{2}{3}=\frac{1}{1}-\frac{1}{3}\);

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

\(\frac{2}{35}=\frac{1}{5}-\frac{1}{7}\) ; ... ; \(\frac{2}{89999}=\frac{1}{299}-\frac{1}{301}\).

=> B= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{299}-\frac{1}{301}\)

=> B=\(\frac{1}{1}-\frac{1}{301}\)

=> B=\(\frac{300}{301}\)

A = 1/5 + 1/13 + 1/14 + 1/15 + 1/60 + 1/61 + 1/62 + 1/63

Ta có : A = 1/5 + 1/13 + 1/14 + 1/15 + 1/60 + 1/61 + 1/62 + 1/63 < 1/5 + 1/12 + 1/12 + 1/12 + 1/60 + 1/60 + 1/60 

               = A < 1/5 + 1/4 + 1/20 

               = A < 1/2

Vậy A < 1/12