oi y b bai 5 giong het de thi toan cua truong minh sang nay
đây nè!
\(\dfrac{13}{14}\)+\(\dfrac{13}{35}\)+\(\dfrac{13}{65}\)+\(\dfrac{13}{104}\)+...+\(\dfrac{13}{1274}\)- X + 2\(\dfrac{1}{5}\)=0
\(\dfrac{13}{14}\)+\(\dfrac{13}{35}\)+\(\dfrac{13}{65}\)+\(\dfrac{13}{104}\)+...+\(\dfrac{13}{1274}\)- X = 0 - 2\(\dfrac{1}{5}\) \(\dfrac{13}{14}\)+\(\dfrac{13}{35}\)+\(\dfrac{13}{65}\)+\(\dfrac{13}{104}\)+...+\(\dfrac{13}{1274}\)- X = -2\(\dfrac{1}{5}\) Đặt A= \(\dfrac{13}{14}\)+\(\dfrac{13}{35}\)+\(\dfrac{13}{65}\)+\(\dfrac{13}{104}\)+...+\(\dfrac{13}{1274}\) = 13 (\(\dfrac{1}{14}\)+\(\dfrac{1}{35}\)+\(\dfrac{1}{65}\)+\(\dfrac{1}{104}\)+...+\(\dfrac{1}{1274}\)) = 13 (\(\dfrac{2}{28}\)+\(\dfrac{2}{70}\)+\(\dfrac{2}{130}\)+\(\dfrac{2}{208}\)+...+\(\dfrac{2}{2548}\)) =13 ( \(\dfrac{2}{4.7}\)+\(\dfrac{2}{7.10}\)+\(\dfrac{2}{10.13}\)+\(\dfrac{2}{13.16}\)+...+\(\dfrac{2}{49.52}\)) = 13.\(\dfrac{2}{3}\) ( \(\dfrac{3}{4.7}\)+\(\dfrac{3}{7.10}\)+\(\dfrac{3}{10.13}\)+\(\dfrac{3}{13.16}\)+...+\(\dfrac{3}{49.52}\)) = \(\dfrac{13}{3}\) (\(\dfrac{1}{4}\)_\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{10}\)+\(\dfrac{1}{10}\)-\(\dfrac{1}{13}\)+\(\dfrac{1}{13}\)-\(\dfrac{1}{16}\)+...+\(\dfrac{1}{49}\)-\(\dfrac{1}{52}\)) = \(\dfrac{13}{3}\) ( \(\dfrac{1}{4}\)-\(\dfrac{1}{52}\)) =\(\dfrac{13}{3}\).\(\dfrac{3}{13}\) = 1 Thay A vào phép tính ban đầu ta được: 1 - X = -2\(\dfrac{1}{5}\) X = 1- (-2\(\dfrac{1}{5}\)) X = 1+ 2\(\dfrac{1}{5}\) X = 3\(\dfrac{1}{5}\) Vậy X = 3\(\dfrac{1}{5}\) Xong rồi đó! Mà bài 2 khoanh câu D chứ!