Đáp án: $\dfrac{A}B=\dfrac1{100}$
Giải thích các bước giải:
Ta có:
$B=\dfrac{99}1+\dfrac{98}{2}+\dfrac{97}{3}+....+\dfrac1{99}$
$\to B=99+\dfrac{98}{2}+\dfrac{97}{3}+....+\dfrac1{99}$
$\to B=(1+\dfrac{98}{2})+(1+\dfrac{97}{3})+....+(1+\dfrac1{99})+1$
$\to B=\dfrac{2+98}{2}+\dfrac{3+97}{3}+....+\dfrac{99+1}{99}+1$
$\to B=\dfrac{100}{2}+\dfrac{100}{3}+....+\dfrac{100}{99}+\dfrac{100}{100}$
$\to B=100(\dfrac12+\dfrac13+...+\dfrac1{99}+\dfrac1{100})$
$\to B=100A$
$\to \dfrac{A}B=\dfrac1{100}$
