Câu hỏi của Băng Thị Thủy Tiên

\(b.75\%-1\dfrac{2}{3}+0,5:\dfrac{5}{12}\)\(=\dfrac{3}{4}-\dfrac{5}{3}+\dfrac{6}{5}\)\(=\dfrac{17}{60}\)\(d.\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+....+\dfrac{1}{399}+\dfrac{1}{483}\)\(=\left(\dfrac{2}{3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+.....+\dfrac{2}{21\cdot19}+\dfrac{2}{21\cdot23}\right)\cdot\dfrac{1}{2}\)\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{23}\right)\cdot\dfrac{1}{2}\)\(=\left(1-\dfrac{1}{23}\right)\cdot\dfrac{1}{2}\)\(=\dfrac{22}{23}\cdot\dfrac{1}{2}\)\(=\dfrac{11}{23}\) Câu trả lời: \(b.75\%-1\dfrac{2}{3}+0,5:\dfrac{5}{12}\)\(=\dfrac{3}{4}-\dfrac{5}{3}+\dfrac{6}{5}\)\(=\dfrac{17}{60}\)\(d.\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+....+\dfrac{1}{399}+\dfrac{1}{483}\)\(=\left(\dfrac{2}{3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+.....+\dfrac{2}{21\cdot19}+\dfrac{2}{21\cdot23}\right)\cdot\dfrac{1}{2}\)\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{23}\right)\cdot\dfrac{1}{2}\)\(=\left(1-\dfrac{1}{23}\right)\cdot\dfrac{1}{2}\)\(=\dfrac{22}{23}\cdot\dfrac{1}{2}\)\(=\dfrac{11}{23}\)