$\dfrac{1}{n(n+1)}=\dfrac{n+1-n}{n(n+1)}=\dfrac{n+1}{n(n+1)}-\dfrac{n}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}\\ a)A=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\cdots+\dfrac{1}{99.100}\\ =\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\cdots+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{2}-\dfrac{1}{100}\\ =\dfrac{49}{100}\\ b)B=\dfrac{1}{30}+\dfrac{1}{42}+\cdots+\dfrac{1}{132}\\ =\dfrac{1}{5.6}+\dfrac{1}{6.7}+\cdots+\dfrac{1}{11.12}\\ =\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\cdots+\dfrac{1}{11}-\dfrac{1}{12}\\ =\dfrac{1}{5}-\dfrac{1}{12}\\ =\dfrac{7}{60}$