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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{101}\)

\(A=\left(\frac{1}{3}-\frac{1}{101}\right):2\)= 49/303

\(A=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+...+\frac{1}{99\times101}\)

\(=\frac{1}{2}\times\left(\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}+...+\frac{2}{99\times101}\right)\)

\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\right)\)

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

\(=\frac{1}{2}\times\frac{98}{303}=\frac{49}{303}\)

Vậy A = 49/303.

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

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

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

\(A=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)

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

a)Ta có:
​A= 1/15+1/35+1/63+1/99+1/143
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
2A= 1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13
Đơn giản đi ta được:
2A= 1/3-1/13
2A= 10/39
A= 5/39
Vậy A= 5/39   

5/39 **** mik nha Xuân

5

39

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

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

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

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

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

49/303 nha bạn

Kb với mình rồi mình giải kĩ cho

@@@@@###

=1/3*5+1/5*7+1/7*9+...+1/99*101

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

=1/3-1/101

=98/303

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

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

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

\(a=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-....-\frac{1}{101}\right)\)

\(a=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)=\frac{49}{303}\)