B4
$A = \dfrac{1}{2.3} + \dfrac{1}{3.4} + \dfrac{1}{4.5} + ... + \dfrac{1}{19.20}$
$=\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{19} - \dfrac{1}{20}$
$= \dfrac{1}{2} - \dfrac{1}{20}$
$ = \dfrac{9}{20}$\
Câu B sai đề
$B = \dfrac{1}{7} + \dfrac{1}{91} + \dfrac{1}{247} + \dfrac{1}{475} + \dfrac{1}{775} + \dfrac{1}{1147}$
$= \dfrac{1}{1.7} + \dfrac{1}{7.13} + \dfrac{1}{13.19} + ... + \dfrac{1}{25. 31} + \dfrac{1}{31. 37}$
$=\dfrac{1}{6}\bigg(\dfrac{9}{1.7} + \dfrac{9}{7.13} + \dfrac{9}{13.19} + ... + \dfrac{9}{25. 31} + \dfrac{9}{31. 37}\bigg)$
$=\dfrac{1}{6}\bigg(\dfrac{1}{1} - \dfrac{1}{37}\bigg)$
$= \dfrac{1}{6} . \dfrac{36}{37}$
$= \dfrac{6}{37}$
B5
$M = \dfrac{1}{11} + \dfrac{1}{12} + \dfrac{1}{13} +... + \dfrac{1}{19} < \dfrac{9}{11} < 1$
Mà mỗi số hạng của $M$ lớn hơn 0 => $M > 0$
Do đó: $0 < M < 1 $ ⇒ $M$ không khải số nguyên
$N = \dfrac{2}{6} + \dfrac{2}{8} + \dfrac{2}{10} + \dfrac{2}{7} + \dfrac{2}{9} + \dfrac{2}{11}$
$\dfrac{2}{6} + \dfrac{2}{8} + \dfrac{2}{10} + \dfrac{2}{7} + \dfrac{2}{9} + \dfrac{2}{11} > \dfrac{2.6}{12} = 1$
$\dfrac{2}{6} + \dfrac{2}{8} + \dfrac{2}{10} + \dfrac{2}{7} + \dfrac{2}{9} + \dfrac{2}{11} < \dfrac{2.6}{6} =2$
⇒ $1< \dfrac{2}{6} + \dfrac{2}{8} + \dfrac{2}{10} + \dfrac{2}{7} + \dfrac{2}{9} + \dfrac{2}{11} < 2$
⇒ $N$ không phải số nguyên
B6
$A = \dfrac{7}{3.4} - \dfrac{9}{4.5} + \dfrac{11}{5.6} - \dfrac{13}{6.7} + \dfrac{15}{7.8} - \dfrac{17}{3.9} + \dfrac{19}{9.10}$
$= \dfrac{3+4}{3.4} - \dfrac{4+5}{4.5} + \dfrac{5+6}{5.6} - \dfrac{6+7}{6.7} + \dfrac{7+8}{7.8} - \dfrac{8+9}{8.9} + \dfrac{9+10}{9.10}$
$=\dfrac{3}{3.4} + \dfrac{4}{3.4} - \dfrac{4}{4.5} - \dfrac{5}{4.5}+ \dfrac{5}{5.6} + ... - \dfrac{8}{8.9} - \dfrac{9}{8.9} + \dfrac{9}{9.10} + \dfrac{10}{9.10}$
$=\dfrac{1}{4} + \dfrac{1}{3} - \dfrac{1}{5} - \dfrac{1}{4} + ... + \dfrac{1}{10} + \dfrac{1}{9}$
$= \dfrac{1}{3} + \dfrac{1}{10}$
$= \dfrac{13}{30}$
