\(B=\dfrac{1}{99.97}-\dfrac{1}{97.95}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
\(B=\dfrac{1}{97.99}-\left(\dfrac{1}{95.97}+...+\dfrac{1}{3.5}+\dfrac{1}{1.3}\right)\)
Đặt \(C=\dfrac{1}{95.97}+...+\dfrac{1}{3.5}+\dfrac{1}{1.3}\)
\(C=\dfrac{1}{95.97}+...+\dfrac{1}{3.5}+\dfrac{1}{1.3}\)
\(C=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{95.97}\)
\(C=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{95.97}\right):2\)
\(2C=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{95.97}\)
\(2C=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}
+...+\dfrac{1}{95}-\dfrac{1}{97}\)
\(2C=\dfrac{1}{1}-\dfrac{1}{97}\)
\(2C=\dfrac{96}{97}\)
\(C=\dfrac{96}{97}:2=\dfrac{48}{97}\)
Thay C vào ta được:
\(B=\dfrac{1}{97.99}-\dfrac{48}{97}\)
\(99B=\dfrac{99}{97.99}-\dfrac{48.99}{97}\)
\(99B=\dfrac{1}{97}-\dfrac{4752}{97}\)
\(99B=-\dfrac{4751}{97}\)
\(B=-\dfrac{4751}{97}:99=-\dfrac{4751}{9603}\)