\(=\dfrac{1}{99.97}-\left(\dfrac{1}{2}.\dfrac{2}{97.95}+\dfrac{1}{2}.\dfrac{2}{95.93}+...+\dfrac{1}{2}.\dfrac{2}{5.3}+\dfrac{1}{2}.\dfrac{2}{3.1}\right)\\ =\dfrac{1}{99.97}-\dfrac{1}{2}.\left(\dfrac{97-95}{97.95}+\dfrac{95-93}{95.93}+....+\dfrac{5-3}{5.3}+\dfrac{3-1}{3.1}\right)\\ =\dfrac{1}{99.97}-\dfrac{1}{2}.\left(\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{93}-\dfrac{1}{95}+...+\dfrac{1}{3}-\dfrac{1}{5}+1-\dfrac{1}{3}\right)\)
\(=\dfrac{1}{99.97}-\dfrac{1}{2}.\left[\left(\dfrac{1}{95}-\dfrac{1}{95}\right)+....+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{97}\right)\right]\\ =\dfrac{1}{99.97}-\dfrac{1}{2}.\left(1-\dfrac{1}{97}\right)\\ =\dfrac{1}{99.97}-\dfrac{1}{2}.\left(\dfrac{97-1}{97}\right)\\ =\dfrac{1}{99.97}-\dfrac{1}{2}.\dfrac{96}{97}\\ =\dfrac{1}{99.97}-\dfrac{48}{97}\\ =\dfrac{1-48.99}{99.97}\\ =\dfrac{1-4752}{99.97}\\ =\dfrac{-4751}{99.97}=\dfrac{-4751}{9603}\)
các bạn giúp mik giải đề cương vs nha . mong sự giúp đỡ của các bạn . bài nào bạn biết làm cx đ ko cần làm hết 
