Bài 7: Phép cộng phân số

\(S=\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{30}\)

\(S>\dfrac{1}{30}+\dfrac{1}{30}+...+\dfrac{1}{30}=\dfrac{10}{30}=\dfrac{1}{3}\)

\(\dfrac{4}{16}=\dfrac{100}{x^2}-41\)

\(\Leftrightarrow\dfrac{4}{16}=\dfrac{100}{x^2}-\dfrac{41x^2}{x^2}\)

\(\Leftrightarrow\dfrac{4}{16}=\dfrac{100-41x^2}{x^2}\)

\(\Rightarrow4x^2=16\left(100-41x^2\right)\)

\(\Leftrightarrow4x^2=1600-656x^2\)

\(\Leftrightarrow1600-656x^2-4x^2=0\)

\(\Leftrightarrow-660x^2+1600=0\)

\(\Leftrightarrow-660x^2=-1600\)

\(\Leftrightarrow x^2=\dfrac{80}{33}\)

\(\Leftrightarrow x=\pm\dfrac{4\sqrt{165}}{33}\)

$\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}=\frac{2}{3.5}+\frac{2}{5.7}+$\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}$

=$\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{11}-\frac{1}{13}$

=$\frac{1}{3}-\frac{1}{13}=\frac{10}{39}$

Vậy....

\(S=\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}+\dfrac{2}{143}\)

\(S=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)

\(S=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)

\(S=\dfrac{1}{3}-\dfrac{1}{13}\)

\(S=\dfrac{13}{39}-\dfrac{3}{39}\)

\(S=\dfrac{10}{39}\)

Vậy \(S=\dfrac{10}{39}\)

\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)

\(=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)

\(=\dfrac{1}{2}-\dfrac{1}{8}\)

\(=\dfrac{4}{8}-\dfrac{1}{8}\)

\(=\dfrac{3}{8}\)

\(=\dfrac{2}{20}+\dfrac{2}{30}+...+\dfrac{2}{240}\)

\(=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(=2\cdot\dfrac{3}{16}=\dfrac{3}{8}\)

\(=\dfrac{1}{8}\left(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}\right)\)

\(=\dfrac{1}{8}\cdot2\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)

\(=\dfrac{1}{4}\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{6}-\dfrac{1}{7}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{6}{7}=\dfrac{3}{14}\)

Gọi các phân số cần tìm có tử là x

Theo đề, ta có: \(\dfrac{-1}{2}< \dfrac{x}{12}< \dfrac{-1}{12}\)

\(\Leftrightarrow-6< x< -1\)

=>\(x\in\left\{-5;-4;-3;-2\right\}\)

Tổng các phân số thỏa mãn yêu cầu đề bài là:

\(\dfrac{-5}{12}+\dfrac{-4}{12}+\dfrac{-3}{12}+\dfrac{-2}{12}=-\dfrac{14}{12}=\dfrac{-7}{6}\)