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\(1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}...1\frac{1}{120}\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}...\frac{121}{120}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{11.11}{10.12}\)
\(=\frac{2.3.4...11}{1.2.3...10}.\frac{2.3.4...11}{3.4.5...12}\)
\(=11.\frac{2}{12}=11.\frac{1}{6}=\frac{11}{6}\)


\(1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}.....1\frac{1}{99}\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}......\frac{100}{99}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.....\frac{10.10}{9.11}\)
\(=\frac{\left(2.3.4...10\right)\left(2.3.4....10\right)}{\left(1.2.3...9\right)\left(3.4.5....11\right)}\)
\(=\frac{10.2}{11}=\frac{20}{11}\)

Ta nhận thấy mẫu số của các phân số có qui luật 1x3; 2x4; 3x5; 4x6...... => mẫu số của phân số thứ 98 là 98x100
\(\Rightarrow A=\frac{4}{3}x\frac{9}{8}x\frac{16}{15}x\frac{25}{24}x\frac{36}{35}x...x\frac{9801}{9800}\)
\(A=\frac{2x2x3x3x4x4x5x5x6x6x...x99x99}{1x2x3x3x4x4x5x5x...x96x96x97x97x98x98x99x100}=\frac{2x99}{100}=\frac{99}{50}=1\frac{49}{50}\)
10/11
\(\dfrac{1}{3}+\dfrac{1}{8}+\dfrac{1}{15}+\dfrac{1}{24}+...+\dfrac{1}{120}\)
\(=\left(\dfrac{1}{3}+\dfrac{1}{15}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{8}+\dfrac{1}{24}+...+\dfrac{1}{120}\right)\)
\(=\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{9.11}\right)+\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{10.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{11}\right)+\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{10}-\dfrac{1}{12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{11}\right)+\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{12}\right)\)
\(=\dfrac{175}{264}\)