$\text{a, A=(1-2)+(3-4)+(5-6)+...+(99-100)+101=$\underbrace{-1-1-1-...-1}_{50 chữ số 1}$+101=-50+101=51}$
$\text{b, $\frac{101+100+99+..+1}{(101-100)+(99-98)+...+(3-2)+1}$=$\frac{101.102:2}{\underbrace{1+1+1+...+1}_{51 chữ số 1}}$=$\dfrac{5151}{51}$=101 }$