\(P=\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{6}\right)+...+\left(1-\dfrac{1}{1225}\right)+\left(1-\dfrac{1}{1275}\right)\\ \Rightarrow\dfrac{P}{2}=\left(\dfrac{1}{2}-\dfrac{1}{6}\right)+\left(\dfrac{1}{2}-\dfrac{1}{12}\right)+...+\left(\dfrac{1}{2}-\dfrac{1}{2550}\right)\\ =\left(\dfrac{1}{2}-\dfrac{1}{2\cdot3}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3\cdot4}\right)+...+\left(\dfrac{1}{2}-\dfrac{1}{50\cdot51}\right)\\ =\dfrac{1}{2}\cdot49-\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}\right)\\ =\dfrac{49}{2}-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{50}-\dfrac{1}{51}\right)\\ =\dfrac{49}{2}-\dfrac{1}{2}+\dfrac{1}{51}=\dfrac{1225}{51}\\ \Rightarrow P=\dfrac{2450}{51}\)

Bạn tham khảo câu trả lời tại đây nhé:

Câu hỏi của Trần Thị Phương - Toán lớp 6 - Học toán với OnlineMath

Học tốt mùa covid nha !!!!

cíu mik cần gấp

thế này mà là lớp 1 ! ảo thật đấy 

một vé báo cáo không phải bài toán lớp 1

\(B=\left(\frac{1}{3}-1\right).\left(\frac{1}{6}-1\right).\left(\frac{1}{10}-1\right).......\left(\frac{1}{1225}-1\right)\left(\frac{1}{1275}-1\right)\)

\(B=\frac{-2}{3}.\frac{-5}{6}.\frac{-9}{10}......\frac{-1224}{1225}.\frac{-1274}{1275}\)

\(B=\frac{-4}{6}.\frac{-10}{12}.\frac{-18}{20}......\frac{-2448}{2450}.\frac{-2548}{2550}\)

\(B=\frac{-4}{2.3}.\frac{-10}{3.4}.\frac{-18}{4.5}.....\frac{-2448}{49.50}.\frac{-2548}{50.51}\)

\(\Rightarrow\)B có : ( 50 - 2 ) : 1 + 1 = 49 ( số hạng )

\(\Rightarrow B=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}........\frac{2448}{49.50}.\frac{2548}{50.51}.\left(-1\right)\)

     \(B=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.........\frac{48.51}{49.50}.\frac{49.52}{50.51}.\left(-1\right)\)

     \(B=\frac{\left(1.2.3...48.49\right).\left(4.5.6......51.52\right)}{\left(2.3.4......49.50\right).\left(3.4.5.....50.51\right)}.\left(-1\right)\)

     \(B=\frac{52}{50.3}.\left(-1\right)\)

      \(B=\frac{26}{75}.\left(-1\right)\)

Vậy \(B=\frac{-26}{75}\)