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- Hướng dẫn giải

$\begin{array}{l}B=\frac{3}{99.96}-\frac{3}{96.93}-\frac{3}{93.90}-\mathrm{...}-\frac{3}{7.4}-\frac{3}{4}\\ =\frac{3}{99.96}-\left(\frac{3}{96.93}+\frac{3}{96.93}+\mathrm{...}+\frac{3}{7.4}+\frac{3}{4.1}\right)\\ =\frac{3}{99.96}-\left(\frac{3}{1.4}+\frac{3}{4.7}+\mathrm{...}+\frac{3}{90.93}+\frac{3}{93.96}\right)\\ =\frac{3}{99.96}-\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\mathrm{...}+\frac{1}{90}-\frac{1}{93}+\frac{1}{93}-\frac{1}{96}\right)\\ =\left(\frac{1}{96}-\frac{1}{99}\right)-\left(1-\frac{1}{96}\right)\\ =\frac{1}{96}-\frac{1}{99}-1+\frac{1}{96}\\ =\frac{1}{48}-1-\frac{1}{99}\\ =-\frac{47}{48}-\frac{1}{99}=\frac{-1567}{1584}\end{array}$