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bạn lấy mẩu so sánh với lần lược là 100, 101,...,200
mà nó bé hơn 99 nên xong
\(a)\dfrac{5}{77}-\dfrac{-4}{7}=\dfrac{5}{77}+\dfrac{44}{77}=\dfrac{49}{77}=\dfrac{7}{11}\\b)\dfrac{4}{33}-\dfrac{6}{-11}=\dfrac{4}{33}+\dfrac{18}{33}=\dfrac{22}{33}=\dfrac{2}{3} \\c)\dfrac{25}{7}-\dfrac{61}{21}=\dfrac{75}{21}-\dfrac{61}{21}=\dfrac{14}{21}=\dfrac{2}{3} \\d)\dfrac{5}{10}-\dfrac{1}{2}=\dfrac{1}{2}-\dfrac{1}{2}=0\)
\(e)\dfrac{2}{3}-\dfrac{-5}{6}=\dfrac{4}{6}+\dfrac{5}{6}=\dfrac{9}{6}=\dfrac{3}{2} \\f)\dfrac{5}{6}-\dfrac{-1}{3}=\dfrac{5}{6}+\dfrac{2}{6}=\dfrac{7}{6}\)
\(3n-2\inƯ\left(15\right)\) \(=\left\{1;-1;3;-3;5;-5;15;-15\right\}.\)
\(\Leftrightarrow n\in\left\{1;\dfrac{1}{3};\dfrac{5}{3};\dfrac{-1}{3};\dfrac{7}{3};-1;\dfrac{17}{3};\dfrac{-13}{3}\right\}.\)
Mà \(n\ne\dfrac{2}{3};n\in Z.\)
\(\Rightarrow n\in\left\{1;-1\right\}.\)
a: \(=\left(-\dfrac{3}{5}+\dfrac{-4}{5}+\dfrac{7}{5}\right)+\dfrac{1}{3}=\dfrac{1}{3}\)
b: \(=\dfrac{-3}{17}+\dfrac{2}{3}+\dfrac{3}{17}=\dfrac{2}{3}\)
e: \(=\dfrac{-5}{21}-\dfrac{16}{21}+1=0\)
g: \(=\dfrac{-4}{11}\cdot\dfrac{-11}{4}\cdot\dfrac{1}{3}=\dfrac{1}{3}\)
h: \(=\dfrac{7}{36}+\dfrac{8}{9}-\dfrac{2}{3}=\dfrac{7}{36}+\dfrac{32}{36}-\dfrac{24}{36}=\dfrac{15}{36}=\dfrac{5}{12}\)
i: \(=\dfrac{4}{7}-\dfrac{5}{8}-\dfrac{3}{28}=\dfrac{32}{56}-\dfrac{35}{56}-\dfrac{6}{56}=\dfrac{-9}{56}\)