a, \(\dfrac{14}{13}-\dfrac{1}{13}-\dfrac{19}{20}=1-\dfrac{19}{20}=\dfrac{1}{20}\)

b, \(-\dfrac{24}{17}+\dfrac{7}{17}+\dfrac{1}{16}=\dfrac{-17}{17}+\dfrac{1}{16}=-1+\dfrac{1}{16}=-\dfrac{15}{16}\)

\(\dfrac{14}{13}+\left(\dfrac{-1}{13}-\dfrac{19}{20}\right)=\left(\dfrac{14}{13}+\dfrac{-1}{13}\right)-\dfrac{19}{20}=\\ \dfrac{13}{13}-\dfrac{19}{20}=1-\dfrac{19}{20}=\dfrac{20}{20}-\dfrac{19}{20}=\dfrac{1}{20}\)

Câu b em làm tương tự nhé

\(a.\dfrac{14}{13}+\left(\dfrac{-1}{13}+\dfrac{19}{20}\right)=\left(\dfrac{14}{13}+\dfrac{-1}{13}\right)-\dfrac{19}{20}=1+\dfrac{19}{20}=\dfrac{20}{20}-\dfrac{19}{20}=\dfrac{1}{20}\\ b.\dfrac{-24}{17}+\left(\dfrac{-7}{17}-\dfrac{1}{16}\right)=\left(\dfrac{-24}{17}+\dfrac{7}{17}\right)+\dfrac{1}{16}=-1+\dfrac{1}{16}=\dfrac{-16}{16}+\dfrac{1}{16}=\dfrac{-15}{16}\)

\(\dfrac{-19}{23}\cdot\dfrac{13}{14}+\dfrac{13}{14}\cdot\dfrac{-15}{23}-\dfrac{13}{14}\cdot\dfrac{1}{23}\\ =\dfrac{13}{14}\cdot\left(\dfrac{-19}{23}+\dfrac{-15}{23}-\dfrac{1}{23}\right)\\ =\dfrac{13}{14}\cdot\dfrac{-35}{23}=\dfrac{-65}{46}\)

Đổi \(\dfrac{-33}{2013}=\dfrac{-1}{61}\)

\(\dfrac{-20}{-19}>\dfrac{13}{14}>\dfrac{-3}{61}>\dfrac{-1}{61}\)

\(\Rightarrow\dfrac{-33}{2013}\)

\(=\left(\dfrac{19}{13}+\dfrac{7}{13}\right)+\left(\dfrac{14}{6}+\dfrac{4}{6}\right)+\left(\dfrac{1}{9}+\dfrac{17}{9}\right)\)

\(=2+3+2\)

\(=7\)

=(19/13+7/13)+(14/6+4/6)+(1/9+17/9)

=2+3+2

=7

\(-\dfrac{3}{2013}\)

\(\dfrac{-3}{2013}\)

Ta có:\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{16}>4\cdot\dfrac{1}{16}=\dfrac{1}{4}\)

\(\dfrac{1}{17}+\dfrac{1}{18}+\dfrac{1}{19}+\dfrac{1}{20}>4\cdot\dfrac{1}{20}=\dfrac{1}{5}\)

=>\(\dfrac{1}{13}+\dfrac{1}{14}+...+\dfrac{1}{20}>\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{9}{20}\)

=>A>\(\dfrac{1}{12}+\dfrac{9}{20}\)

\(\dfrac{1}{12}>\dfrac{1}{20}\)

=>\(A>\dfrac{1}{20}+\dfrac{9}{20}=\dfrac{1}{2}\)

Vậy...

\(P=\dfrac{1+19+\dfrac{19}{13}+\dfrac{19}{101}}{7+\dfrac{7}{13}+\dfrac{7}{19}+\dfrac{7}{101}}\)

\(=\dfrac{19\left(1+\dfrac{1}{3}+\dfrac{1}{19}+\dfrac{1}{101}\right)}{7\left(1+\dfrac{1}{13}+\dfrac{1}{19}+\dfrac{1}{101}\right)}=\dfrac{19}{7}\)

\(B=\left|157\dfrac{13}{27}-273\dfrac{7}{19}\right|-96\dfrac{14}{27}+15\dfrac{12}{19}\)

\(=273\dfrac{7}{19}-153\dfrac{13}{27}-96\dfrac{14}{27}+15\dfrac{12}{19}\)

\(=\left(273+15+\dfrac{7}{19}+\dfrac{12}{19}\right)-\left(153+96+\dfrac{13}{27}+\dfrac{14}{27}\right)\)

\(=289-250=39\)

Ta có :

\(\dfrac{1}{11}>\dfrac{1}{20}\\ \dfrac{1}{12}>\dfrac{1}{20}\\ ..........\\ \dfrac{1}{20}=\dfrac{1}{20}\)

\(\Rightarrow\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\\ \Rightarrow S>\dfrac{10}{20}\\ \Rightarrow S>\dfrac{1}{2}\)