Giải thích các bước giải:
Ta có :
$B=\dfrac{144}{3}+\dfrac{144}{15}+\dfrac{144}{35}+..+\dfrac{144}{399}$
$\to B=72(\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+..+\dfrac{21-19}{19.21})$
$\to B=72(\dfrac11-\dfrac13+\dfrac13-\dfrac15+\dfrac15-\dfrac17+..+\dfrac{1}{19}-\dfrac{1}{21})$
$\to B=72(1-\dfrac{1}{21})=\dfrac{480}{7}$
Lại có :
$A=\dfrac{27}{4}+\dfrac{72}{9}+\dfrac{135}{16}+\dfrac{216}{25}+..+\dfrac{891}{100}$
$\to A=9(\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+\dfrac{24}{25}+..+\dfrac{99}{100})$
$\to A=9(1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+1-\dfrac{1}{4^2}+1-\dfrac{1}{5^2}+..+1-\dfrac{1}{10^2})$
$\to A=9(9-(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+..+\dfrac{1}{10^2}))$
$\to A=9(9-(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+..+\dfrac{1}{10.10}))$
$\to A>9(9-(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+..+\dfrac{1}{9.10}))$
$\to A>9(9-(\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+..+\dfrac{10-9}{9.10}))$
$\to A>9(9-(\dfrac11-\dfrac12+\dfrac12-\dfrac13+..+\dfrac19-\dfrac{1}{10}))$
$\to A>9(9-(1-\dfrac{1}{10}))$
$\to A>\dfrac{729}{10}>\dfrac{480}{7}=B$
$\to A>B$
