Câu hỏi của Nguyễn Tấn Đạt

Question

Tính $A=\dfrac{1}{3}.\dfrac{1}{7}+\dfrac{1}{7}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{15}+...+\dfrac{1}{95}.\dfrac{1}{99}$

Nguyễn Huy Tú · Accepted Answer

$A=\dfrac{1}{3}.\dfrac{1}{7}+\dfrac{1}{7}.\dfrac{1}{11}+...+\dfrac{1}{95}.\dfrac{1}{99}$

$=\dfrac{1}{3.7}+\dfrac{1}{7.11}+...+\dfrac{1}{95.99}$

$=\dfrac{1}{4}\left(\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{95.99}\right)$

$=\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{99}\right)$

$=\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{99}\right)$

$=\dfrac{8}{99}$

Vậy $A=\dfrac{8}{99}$Câu trả lời: $A=\dfrac{1}{3}.\dfrac{1}{7}+\dfrac{1}{7}.\dfrac{1}{11}+...+\dfrac{1}{95}.\dfrac{1}{99}$

$=\dfrac{1}{3.7}+\dfrac{1}{7.11}+...+\dfrac{1}{95.99}$

$=\dfrac{1}{4}\left(\dfrac{4}{3.7}+\dfrac{4}{7.11}+...+\dfrac{4}{95.99}\right)$

$=\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{99}\right)$

$=\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{99}\right)$

$=\dfrac{8}{99}$

Vậy $A=\dfrac{8}{99}$

Nguyễn Tấn Đạt · Answer

Nguyễn Huy Tú kinh :vCâu trả lời: Nguyễn Huy Tú kinh :v