\(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+...+\frac{9998}{9999}\)

\(=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+\left(1-\frac{1}{35}\right)+\left(1-\frac{1}{63}\right)+...+\left(1-\frac{1}{9999}\right)\)

\(=\left(1-\frac{1}{1\cdot3}\right)+\left(1-\frac{1}{3\cdot5}\right)+\left(1-\frac{1}{5\cdot7}\right)+\left(1-\frac{1}{7\cdot9}\right)+...+\left(1-\frac{1}{99\cdot101}\right)\)

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

Có tất cả : (101 - 3) : 2 + 1 = 50 chữ số 1 => (1 + 1 + 1 + .... + 1) = 1 x 50 = 50 

\(\Rightarrow50-\frac{1}{2}\cdot\left(1-\frac{1}{101}\right)\)

\(=50-\frac{1}{2}\cdot\frac{100}{101}=50-\frac{100}{101}=\frac{4950}{101}\)

Vậy \(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+...+\frac{9998}{9999}=\frac{4950}{101}\)

cái này tính cái gì thế

ko hiểu

\(E=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+...+\left(1-\frac{1}{9999}\right)\)

    \(=\left(1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{9999}\right)\)

     \(=50-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)\)

     \(=50-\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

     \(=50-\frac{1}{2}.\left(1-\frac{1}{101}\right)=50-\frac{1}{2}.\frac{100}{101}=50-\frac{50}{101}=\frac{5000}{101}\)

Cảm ơn bạn nha !!!  

\(=3,571\)

\(A=\dfrac{2}{3}+\dfrac{14}{15}+\dfrac{34}{35}+...+\dfrac{9998}{9999}\\ =\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{15}\right)+\left(1-\dfrac{1}{35}\right)+...+\left(1-\dfrac{1}{9999}\right)\\ =\left(1+1+1+...+1\right)\left(\text{có 50 số 1}\right)-\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{9999}\right)\\ =50\cdot1-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{99\cdot101}\right)\\ =50-\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =50-\left(1-\dfrac{1}{101}\right)\\ =50-1+\dfrac{1}{101}\\ =49+\dfrac{1}{101}\\ =\dfrac{4949+1}{101}\\ =\dfrac{4950}{101}\)

\(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}\)

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

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

\(=5+\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\)

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

\(=5+\frac{5}{11}=\frac{60}{11}\)

A = \(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}\)

A = ( 1 - 1/3 ) + ( 1 - 1/15 ) + ( 1 - 1/35 ) + ( 1 - 1/63 ) + ( 1 - 1/99 )

A = ( 1 + 1 + 1 + 1 + 1 ) - ( 1/3 + 1/15 + 1/35 + 1/63 + 1/99 )

A = 5 - \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\)

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

A = 5 - ( 1 - 1/11 ) 

A = 5 - 10/11

A = 45/11

Dấu \(.\)là dấu nhân 

\(A=\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}\)

\(\Rightarrow A=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+\left(1-\frac{1}{35}\right)+\left(1-\frac{1}{63}\right)+\left(1-\frac{1}{99}\right)\)

\(\Rightarrow A=\left(1+1+1+1+1\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)

\(\Rightarrow A=5-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\)

\(\Rightarrow A=5-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

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

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

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

\(\Rightarrow A=5-\frac{5}{11}\)

\(\Rightarrow A=\frac{55}{11}-\frac{5}{11}\)

\(\Rightarrow A=\frac{50}{11}\)

~ Ủng hộ nhé 

\(\frac{32}{9}\)

2/3+14/15+34/35+62/63=32/9