Câu trả lời:
\(A=\frac{\frac{98}{2}+1+\frac{97}{3}+1+.....+\frac{2}{98}+1+\frac{1}{99}+1+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{99}+\frac{1}{100}}=\frac{\frac{100}{2}+\frac{100}{3}+........+\frac{100}{98}+\frac{100}{99}+\frac{100}{100}}{\frac{1}{2}+\frac{1}{3}+......+\frac{1}{99}+\frac{1}{100}}\)
\(=\frac{100\left(\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{100}\right)}{\left(\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{100}\right)}=100\)