B = 1/3*5 + 1/5*7 + 1/7*9 + 1/9*11 + 1/11*13

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

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

   = 1/2 * ( 1/3 - 1/11)

   = 1/2 * 8/33

   = 4/33

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

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

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

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

\(B=\frac{1}{2}.\frac{12}{39}\)

\(B=\frac{2}{13}\)

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

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

\(\Rightarrow\)\(2B=\frac{1}{3}-\frac{1}{13}\)\(=\frac{13}{39}-\frac{3}{39}=\frac{10}{39}\)\(\Rightarrow B=\frac{10}{\frac{39}{2}}=\frac{10}{39\cdot2}=\frac{5}{39}\)

bấm vào chữ 0 đúng sẽ ra đáp án 

A = 98/303

Cơn Mưa Tình Yêu

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

A     = 1/(3x5) + 1/(5x7) + 1/(7x9) + 1/(9x11) + ... + 1/(99 x 101)

Ax2 = 2/(3x5) + 2/(5x7) + 2/(7x9) + 2/(9x11) + ... + 2/(99 x 101)

Ax2 = 1/3 – 1/5 + 1/5 – 1/7 + 1/7 – 1/9 + 1/9 – 1/11 + ...+ 1/99 – 1/101

Ax2 = 1/3 – 1/101 = 98/303

A    =  98/303 : 2

A    =  49/303

bai toan nay don

49/303

duyệt đi

\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\)

\(\Rightarrow\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)

\(\Rightarrow\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{3}-\frac{1}{101}\right)\)

\(\Rightarrow\frac{1}{2}.\frac{98}{303}\)

\(\Rightarrow\frac{49}{303}\)

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

= 1/(3x5) + 1/(5x7) + 1/(7x9) + ... + 1/(99x101)

= (1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ...+ 1/99 - 1/101) : 2

= (1/3 - 1/101) : 2 

= 98/303 : 2

= 49/303

\(y=\frac{1}{x^2+\sqrt{x}}\)

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

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

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

=1/2.(1/3-1/5+1/5-1/7+1/6-1/9+1/9-1/11+1/11-1/13=1/2.(1/3-1/13)=1/2.10/39=5/39

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

A=

49/303,xin lỗi bạn mk làm biếng viết lời giải nếu cần nói mk nha

\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+........+\frac{1}{9999}\)

\(2A=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+.......+\frac{2}{9999}\)

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+......+\frac{2}{99.101}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.......+\frac{1}{99}-\frac{1}{101}\)

\(2A=\frac{1}{3}-\frac{1}{101}\)

\(2A=\frac{98}{303}\)

\(A=\frac{98}{303}:2=\frac{98}{303}.\frac{1}{2}=\frac{49}{303}\)

49/303

tick minh di ban

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

= 1/(3x5) + 1/(5x7) + 1/(7x9) + ... + 1/(99x101)

= (1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ...+ 1/99 - 1/101) : 2

= (1/3 - 1/101) : 2 

= 98/303 : 2

= 49/303

ĐS: 49/303

Tick nha 

A = 1/15 + 1/35 + 1/63 + 1/99 +........1/9999
A = 1/3x5 + 1/5x7 + 1/7x9 + .......+ 1/99x101
A = 1/2 x (1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 +..... + 1/99 - 1/101)
A = 1/2 X ( 1/3 - 1/99)
A = 1/2 x ( 1/3 - 1/101)
A = 1/2 ( 98/303)
A = 49/303

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

đặt A = 1/15 + 1/35 + 1/63 + 1/99 + 1/143

A       = 1/3X5 + 1/5X7 + 1/7X9 + 1/9X11 + 1/11X13

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

A x 2 = 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11 + 2/11x13

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

A x 2 = 1/3 - 1/13

A x 2 = 13/39 - 3/39

A x 2 = 10/39

A      =10/39 : 2

A       = 5/39

1/15 + 1/35 + 1/63 + 1/99 + 1/143   = 5/39 nhé bạn !

A=1/3*5+1/5*7+1/7*9+1/9*11+1/11*13

2A=2/3*5+2/5*7+2/7*9+2/9*11+2/11*13=2/3-2/5+2/5-2/7+2/7+2/9-2/9+2/11-2/11+2/13=2/3+2/13=26/39+6/39        =32/39

=>A=16/39

Ta có:

\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{2915}+\frac{1}{3135}\)

Coi \(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{53.55}+\frac{1}{55.57}\)

\(\Rightarrow2A=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{53.55}+\frac{1}{55.57}\right)\)

\(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{53.55}+\frac{2}{55.57}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{53}-\frac{1}{55}+\frac{1}{55}-\frac{1}{57}\)

\(=\frac{1}{3}-\frac{1}{57}\)

\(=\frac{19}{57}-\frac{1}{57}=\frac{18}{570}=\frac{6}{19}\)

\(\Rightarrow A=\frac{6}{19}:2=\frac{3}{19}\)

Vậy tổng trên bằng \(\frac{3}{19}\)

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

\(2A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}=\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{13}\)

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

\(\Rightarrow A=\dfrac{10}{39}.\dfrac{1}{2}=\dfrac{10}{78}=\dfrac{5}{39}\)