5) a) Sửa: \(A=\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\right):\left(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\)
\(A=\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\right):\left(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\right):\left[\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{100}\right)\right]\)
\(A=\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\right):\left[\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\cdot2\right]\)
\(A=\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\right):\left[\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\cdot2\right]\)
\(A=\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\right):\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\right]\)
\(A=\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\right):\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{50}-1-\dfrac{1}{2}-...-\dfrac{1}{50}\right)+\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\right)\right]\)
\(A=\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\right):\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\right)\)
\(A=1\)
5) b) \(B=\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{6}\right)\cdot\left(1-\dfrac{1}{10}\right)\cdot...\cdot\left(1-\dfrac{1}{780}\right)\)
\(B=\dfrac{2}{3}\cdot\dfrac{5}{6}\cdot\dfrac{9}{10}\cdot...\cdot\dfrac{779}{780}\)
\(B=\dfrac{2\cdot2}{3\cdot2}\cdot\dfrac{5\cdot2}{6\cdot2}\cdot\dfrac{9\cdot2}{10\cdot2}\cdot...\cdot\dfrac{779\cdot2}{780\cdot2}\)
\(B=\dfrac{4}{6}\cdot\dfrac{10}{12}\cdot\dfrac{18}{20}\cdot...\cdot\dfrac{1558}{1560}\)
\(B=\dfrac{1\cdot4}{2\cdot3}\cdot\dfrac{2\cdot5}{3\cdot4}\cdot\dfrac{3\cdot6}{4\cdot5}\cdot...\cdot\dfrac{38\cdot41}{39\cdot40}\)
\(B=\dfrac{1\cdot4\cdot2\cdot5\cdot3\cdot6\cdot...\cdot39\cdot41}{2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot39\cdot40}\)
\(B=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot38\right)\cdot\left(4\cdot5\cdot6\cdot...\cdot41\right)}{\left(2\cdot3\cdot4\cdot...\cdot39\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot40\right)}\)
\(B=\dfrac{1\cdot41}{39\cdot3}\)
\(B=\dfrac{41}{117}\)
(đề bài: tìm x biết)
