\(\begin{array}{l}
\text{Bài 1:}\\
A = \dfrac{1}{1\times 2} + \dfrac{1}{2 \times 3}+\dfrac{1}{3\times 4} + \cdots + \dfrac{1}{999\times 1000}\\
\quad =\dfrac{2 - 1}{1\times 2} + \dfrac{3-2}{2 \times 3}+\dfrac{4-3}{3\times 4} + \cdots + \dfrac{1000-999}{999\times 1000}\\
\quad = \dfrac{2}{1\times 2} - \dfrac{1}{1\times 2} + \dfrac{3}{2\times 3} - \dfrac{2}{2\times 3} + \cdots + \dfrac{1000}{999\times 1000} - \dfrac{999}{999\times 1000}\\
\quad = 1 - \dfrac12 + \dfrac12 - \dfrac13 + \cdots + \dfrac{1}{999} - \dfrac{1}{1000}\\
\quad = 1 - \dfrac{1}{1000}\\
\quad = \dfrac{1000}{1000} - \dfrac{1}{1000}\\
\quad = \dfrac{999}{1000}\\
\text{Bài 2:}\\
B = \dfrac{3}{1\times 4} + \dfrac{3}{4\times 7} + \dfrac{3}{7\times 10} +\cdots + \dfrac{3}{181\times 184}\\
\quad = \dfrac{4-1}{1\times 4} + \dfrac{7-4}{4\times 7} + \dfrac{10-7}{7\times 10} +\cdots + \dfrac{184-181}{181\times 184}\\
\quad = \dfrac{4}{1\times 4} -\dfrac{1}{1\times 4}+ \dfrac{4}{4\times 7} - \dfrac{4}{4\times 7}+\cdots + \dfrac{184}{181\times 184}-\dfrac{181}{181\times 184}\\
\quad = 1 - \dfrac{1}{4} + \dfrac14 - \dfrac17 + \cdots + \dfrac{1}{181} - \dfrac{1}{184}\\
\quad = 1 - \dfrac{1}{184}\\
\quad = \dfrac{184}{184} - \dfrac{1}{184}\\
\quad = \dfrac{183}{184}\\
\text{Bài 3:}\\
C = \dfrac{4}{1\times 2} + \dfrac{4}{2 \times 3}+\dfrac{4}{3\times 4} + \cdots + \dfrac{4}{99\times 100}\\
\quad = 4\times \left(\dfrac{1}{1\times 2} + \dfrac{1}{2 \times 3}+\dfrac{1}{3\times 4} + \cdots + \dfrac{1}{99\times 100}\right)\\
\quad =4\times \left(\dfrac{2 - 1}{1\times 2} + \dfrac{3-2}{2 \times 3}+\dfrac{4-3}{3\times 4} + \cdots + \dfrac{100-99}{99\times 100}\right)\\
\quad = 4\times \left(1 -\dfrac12 + \dfrac12 - \dfrac13 + \cdots + \dfrac{1}{99} - \dfrac{1}{100}\right)\\
\quad = 4\times \left(1 - \dfrac{1}{100}\right)\\
\quad = 4 \times \left(\dfrac{100}{100} - \dfrac{1}{100}\right)\\
\quad = 4\times \dfrac{99}{100}\\
\quad = \dfrac{99}{25}\\
\text{Bài 4:}\\
D = \dfrac{7}{1\times 3} + \dfrac{7}{3\times 5} + \dfrac{7}{5\times 7} + \cdots + \dfrac{7}{99\times 101}\\
\quad = \dfrac72\times \left( \dfrac{2}{1\times 3} + \dfrac{2}{3\times 5} + \dfrac{2}{5\times 7} + \cdots + \dfrac{2}{99\times 101}\right)\\
\quad = \dfrac72\times \left( \dfrac{3-1}{1\times 3} + \dfrac{5-3}{3\times 5} + \dfrac{7-5}{5\times 7} + \cdots + \dfrac{101-99}{99\times 101}\right)\\
\quad = \dfrac72\times \left(\dfrac{3}{1\times 3} - \dfrac{1}{1\times 3} + \dfrac{5}{3\times 5} - \dfrac{3}{3\times 5} + \cdots + \dfrac{101}{99\times 101} - \dfrac{99}{99\times 101}\right)\\
\quad = \dfrac72\times \left(1 - \dfrac13  + \dfrac13 - \dfrac15 + \cdots + \dfrac{1}{99} - \dfrac{1}{101} \right)\\
\quad = \dfrac72\times \left(1 - \dfrac{1}{101}\right)\\
\quad = \dfrac72\times \left(\dfrac{101}{101} - \dfrac{1}{101}\right)\\
\quad = \dfrac72\times \dfrac{100}{101}\\
\quad = \dfrac{350}{101}\\
\text{Bài 5:}\\
E = \dfrac{2}{3\times 7} + \dfrac{2}{7\times 11} + \dfrac{2}{11\times 15} + \cdots + \dfrac{2}{99\times 103}\\
\quad = \dfrac12\times \left(\dfrac{4}{3\times 7} + \dfrac{4}{7\times 11} + \dfrac{4}{11\times 15} + \cdots + \dfrac{103-99}{99\times 103}\right)\\
\quad = \dfrac12\times \left(\dfrac{7-3}{3\times 7} + \dfrac{11-7}{7\times 11} + \dfrac{15-11}{11\times 15} + \cdots + \dfrac{4}{99\times 103} \right)\\
\quad = \dfrac12\times \left(\dfrac{7}{3\times 7} -\dfrac{3}{3\times 7}+ \dfrac{11}{7\times 11} -\dfrac{7}{7\times 11}+ \cdots + \dfrac{103}{99\times 103}-\dfrac{99}{99\times 103}\right)\\
\quad = \dfrac12\times \left(\dfrac13 - \dfrac17 + \dfrac17 - \dfrac{1}{11} + \cdots + \dfrac{1}{99} - \dfrac{1}{103}\right)\\
\quad = \dfrac12\times \left(\dfrac13 - \dfrac{1}{103}\right)\\
\quad = \dfrac12\times \left(\dfrac{103}{309} - \dfrac{3}{309}\right)\\
\quad = \dfrac12\times \dfrac{100}{309}\\
\quad = \dfrac{50}{309}\\
\text{Bài 6:}\\
F = \dfrac{5}{3\times 6} + \dfrac{5}{6\times 9} + \dfrac{5}{9\times 12} + \cdots + \dfrac{5}{96 \times 99}\\
\quad = \dfrac53\times \left(\dfrac{3}{3\times 6} + \dfrac{3}{6\times 9} + \dfrac{3}{9\times 12} + \cdots + \dfrac{3}{96 \times 99} \right)\\
\quad = \dfrac53\times \left(\dfrac{6-3}{3\times 6} + \dfrac{9-6}{6\times 9} + \dfrac{12-9}{9\times 12} + \cdots + \dfrac{99-96}{96 \times 99} \right)\\
\quad = \dfrac53\times \left(\dfrac{6}{3\times 6} - \dfrac{3}{3\times 6} + \dfrac{9}{6\times 9} - \dfrac{6}{6\times 9} +\cdots + \dfrac{99}{96\times 99} - \dfrac{96}{96\times 99} \right)\\
\quad = \dfrac53\times \left(\dfrac13 - \dfrac16 + \dfrac16 - \dfrac19 + \cdots + \dfrac{1}{96} - \dfrac{1}{99} \right)\\
\quad = \dfrac53\times \left(\dfrac13 - \dfrac{1}{99}\right)\\
\quad = \dfrac53\times \left(\dfrac{33}{99} - \dfrac{1}{99}\right)\\
\quad = \dfrac53\times \dfrac{32}{99}\\
\quad = \dfrac{160}{297}
\end{array}\)

Chúc bạn học tốt và xin hay nhất ạ ##