- A = \(2+\dfrac{17}{20}-1-\dfrac{11}{15}+\left(6+\dfrac{9}{20}\right):3\)
= \(2+\dfrac{17}{20}-1-\dfrac{11}{15}+2+\dfrac{3}{20}\)
= \(3+\left(\dfrac{17}{20}+\dfrac{3}{20}\right)-\dfrac{11}{15}=4-\dfrac{11}{15}=\dfrac{49}{15}\)
- B = \(\dfrac{19+\dfrac{19}{13}+\dfrac{19}{19}+\dfrac{19}{101}}{7+\dfrac{7}{13}+\dfrac{7}{19}+\dfrac{7}{101}}\)
= \(\dfrac{19\left(1+\dfrac{1}{13}+\dfrac{1}{19}+\dfrac{1}{101}\right)}{7\left(1+\dfrac{1}{13}+\dfrac{1}{19}+\dfrac{1}{101}\right)}=\dfrac{19}{7}\)
- C = \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{10.11}\)
= \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{10}-\dfrac{1}{11}\)
= \(\dfrac{1}{2}-\dfrac{1}{11}=\dfrac{9}{22}\)
