Question

Thực hiện phép tính hợp lí nếu có thể:

\(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right)\)

Answer 1

\(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right)\)

\(=\) \(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\)

\(=\) \(\dfrac{1}{4}-\dfrac{1}{49}\)

\(=\) \(\dfrac{49}{196}-\dfrac{4}{196}\)

\(=\) \(\dfrac{45}{196}\)

Answer 2

Ta có : \(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right)\)

= 5.\(\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)\)

= 5. \(\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\)

=5. \(\dfrac{45}{196}\)

=\(\dfrac{225}{196}\)

Answer 3

\(\dfrac{1}{4.9}+\dfrac{1}{9.14}+...+\dfrac{1}{44.49}=\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+...+\dfrac{5}{44.49}\right)=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right)=\dfrac{1}{5}.\dfrac{45}{196}=\dfrac{9}{196}\)