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

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

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

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

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

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

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

\(=\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+...+\frac{1}{13x15}\)

\(=\frac{1}{2}x\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+...+\frac{2}{13x15}\right)\)

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

\(=\frac{1}{2}x\left(1-\frac{1}{15}\right)=\frac{1}{2}x\frac{14}{15}=\frac{7}{15}\)

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

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

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

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

=\(\frac{14}{15}:2\)

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

mk giải đúng ko vậy , sai thì nói ra mk còn sửa nha

\(C=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{49.51}\)

\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{49}-\frac{1}{51}\right)\)

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

\(C=\frac{1}{2}.\frac{16}{51}=\frac{8}{51}\)

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

\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{8}{51}\)

C=\(\frac{1}{3.5}\)+\(\frac{1}{5.7}\)+\(\frac{1}{7.9}\)+......+\(\frac{1}{49.51}\)

C=(\(\frac{2}{3.5}\)+\(\frac{2}{5.7}\)+\(\frac{2}{7.9}\)+.......+\(\frac{2}{49.51}\)) :2

C=(\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{9}\)+....+\(\frac{1}{49}\)-\(\frac{1}{51}\)) :2

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

C=\(\frac{16}{51}\):2=\(\frac{8}{51}\)

1/3 + 1/15 + 1/35+ 1/63 +...... + 1/195

=  1/3 + 1/3x5 + 1/5 x7 + 1/7x9 + ....+1/13x15

= 1/3+1/3-1/5+1/5-1/7+1/7-1/9+....+1/13-1/15 ( vì +- nên rút gọn )

= 1/3+1/3-1/15

=3/5

=1/1.3+1/3.5+1/5.7+...+1/13.15

=1/2.2(1/1.3+1/3.5+1/5.7+...+1/13.15)

=1/2(2/1.3+2/3.5+2/5.7+...+2/13.15)

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

=1/2[(1-1/15)+(1/3-1/3)+(1/5-1/5)+...+(1/13-1/15)]

=1/2[(1-1/15)+0+...+0=1/2(1-1/15)=1/2.14/15=14/30=7/15

1/3 + 1/15 + 1/35+ 1/63 +...... + 1/195

=  1/3 + 1/3x5 + 1/5 x7 + 1/7x9 + ....+1/13x15

= 1/3+1/3-1/5+1/5-1/7+1/7-1/9+....+1/13-1/15 ( vì +- nên rút gọn )

= 1/3+1/3-1/15

=3/5

A=1/3.5+1/5.7+1/7.9+...+1/99.101

2A= 2/3.5+2/5.7+2/7.9+...+2/99.101

2A= 1/3-1/5+1/5-1/7-1/7+1/7-1/9+...+1/99-1/101

2A=1/3-1/101=98/303

A=(98/303)/2=49/303

A=97

   666

A=1/3.5 + 1/5.7 + 1/7.9 +...+ 1/99.101

=1/2.[(1/3-1/5) + (1/5-1/7) + ... + 1/99-1/101)]

=1/2.(1/3-1/101)

=49/303

\(A=1/3.5+1/5.7+1/7.9+…+1/99.101\)

A.2=2/3.5+2/5.7+2/7.9+…+2/99.101

A.2=1/3-1/5+1/5-1/7+1/7-1/9+...+1/99-1/101

Vậy

A.2=1/3-1/101=98/303

A=98/303/2=49/303

Đúng không

A = 1/15 + 1/35 + 1/63 + 1/99 + ... + 1/9999

   = 1/3x5 + 1/5x7 + 1/7x9 + 1/9x11 + ... + 1/99x101

A x 2 = 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11 + ... + 2/99x101

         = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + ... + 1/99 - 1/101

         = 1/3 - 1/101 = 98/303

Vậy A = 98/303 : 2 = 49/303

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

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

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

\(A=\frac{1}{2}\left(\frac{2}{3.5}+...+\frac{2}{11.13}\right)\)

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

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

\(A=\frac{1}{2}\cdot\frac{10}{39}=\frac{5}{39}\)

P/s: Có thể tính sai :<

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

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

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

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

Sửa lại : 

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

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

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

\(=\frac{1}{2}\cdot(\frac{1}{3}-\frac{1}{13})=\frac{1}{2}\cdot\frac{10}{39}=\frac{1\cdot10}{2\cdot39}=\frac{1\cdot5}{1\cdot39}=\frac{5}{39}\)

Hoàng Thiên làm đúng rồi bạn mk nhầm rồi nhé 

siêu tốc

\(2000+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)=2000+\frac{50}{3.101}\)

Ta có: \(2000+\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)

=\(2000+\left(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+...+\frac{1}{99x101}\right)\)

Đặt A=\(\left(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+...+\frac{1}{99x101}\right)\)

=> 2xA =\(\left(\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+...+\frac{2}{99x101}\right)\)

2xA = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)

2xA = \(\frac{1}{3}-\frac{1}{101}\)

2xA = \(\frac{98}{303}\)

A = \(\frac{98}{606}=\frac{49}{303}\)

=> \(2000+\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)=2000+\frac{49}{303}=\frac{606049}{303}\)

Bạn xem rút gọn được thì rút nhé

2000+49/303=........................