Ta có:
\(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
\(=\dfrac{1}{99.97}-\left(\dfrac{1}{97.95}+\dfrac{1}{95.93}+...+\dfrac{1}{5.3}+\dfrac{1}{3.1}\right)\)
\(=\dfrac{1}{99.97}=\dfrac{1}{2}\left(\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{93}-\dfrac{1}{95}+...+\dfrac{1}{3}-\dfrac{1}{5}+1-\dfrac{1}{3}\right)\)
\(=\dfrac{1}{99.97}-\dfrac{1}{2}\left(1-\dfrac{1}{97}\right)\)
\(=\dfrac{1}{99.97}-\dfrac{1}{2}.\dfrac{96}{97}\)
\(=\dfrac{1}{9603}-\dfrac{48}{97}\)
\(=\dfrac{-4751}{9603}\)
Vậy \(\dfrac{1}{99.97}-\dfrac{1}{97.95}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}=\dfrac{-4751}{9603}\)
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mk đg cần nhưng còn nhìu bài để làm cho nên giúp mk vs nha Thanks trước
