49/303

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

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

\(A=\frac{49}{303}\)

ta có:

A=\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+....+\frac{1}{999}\)

A=\(\frac{1}{3\text{x}5}+\frac{1}{5\text{x}7}+\frac{1}{7\text{x}9}+\frac{1}{9\text{x}11}+...+\frac{1}{?\text{x}?}\)

A=

Thử các mẫu số ta thấy

Các mẫu số có dạng x(x + 2)

Số 999 khác dạng x(x + 2)

Bạn xem lại đề 

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

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

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

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

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

A = \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{99.101}\)

   = \(\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{99.101}\right):2\)\(=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right):2\)

   = \(\left(\frac{1}{3}-\frac{1}{101}\right):2=\frac{101-3}{303}:2=\frac{98}{303}:2=\frac{49}{303}\)

Dấu chấm trong bài là dấu nhân nha !

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

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

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

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

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

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

\(A=\frac{49}{303}\)

Ủng hộ mk nha !!! ^_^

bấm vào chữ 0 đúng sẽ ra đáp án 

A = 98/303

Cơn Mưa Tình Yêu

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

a=lộn