Question

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}+\dfrac{2}{99.101}\)

giúp mình với ạ, giải từng bước nhé ạ

 

Answer 1

\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{101}\right)=2\cdot\dfrac{98}{303}=\dfrac{196}{303}\)

Answer 2

= 2/3 . 2/5 + 2/5 . 2/7 + ... + 2/99 . 2/101

= 2/3 - 2/5 + 2/5 - 2/7 + ... + 2/99 - 2/101

= 2/3 - 2/101

= 196/303

Answer 3

2/3 - 2/5 + 2/5 - 2/7 + 2/7 - 2/9 + .... + 2/97 - 2/99 + 2/99 - 2/101

 = 2/3 - 2/101

= 196/303

Answer 4

\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ =\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)