tính riêng:
\(\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\)
=\(\left(\frac{100}{99}-1\right)+\left(\frac{100}{98}-1\right)+\left(\frac{100}{97}-1\right)+...+\left(\frac{100}{2}-1\right)+99\)
=\(100.\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2}\right)+99-98\)
=\(100.\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+...+\frac{1}{2}\right)\)
vậy \(\left(\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}\right):\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)=100\)
chúc bạn học tốt ^^