\(\dfrac{14}{27}\)...\(\dfrac{15}{26}\)
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\(B=\left|157\dfrac{13}{27}-273\dfrac{7}{19}\right|-96\dfrac{14}{27}+15\dfrac{12}{19}\)
\(=273\dfrac{7}{19}-153\dfrac{13}{27}-96\dfrac{14}{27}+15\dfrac{12}{19}\)
\(=\left(273+15+\dfrac{7}{19}+\dfrac{12}{19}\right)-\left(153+96+\dfrac{13}{27}+\dfrac{14}{27}\right)\)
\(=289-250=39\)
13)\(\dfrac{45}{8}\left(\dfrac{4}{15}-\dfrac{7}{8}\right)+\dfrac{45}{8}\left(\dfrac{11}{15}+\dfrac{9}{8}\right)\)
=\(\dfrac{45}{8}\left(\dfrac{4}{15}-\dfrac{7}{8}+\dfrac{11}{15}+\dfrac{9}{8}\right)\)
=\(\dfrac{45}{8}\left[\left(\dfrac{4}{15}+\dfrac{11}{15}\right)-\left(\dfrac{7}{8}-\dfrac{9}{8}\right)\right]\)
=\(\dfrac{45}{8}.\dfrac{5}{4}\)=\(\dfrac{225}{32}\)
14)\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right):\dfrac{7}{15}\)
=\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right).\dfrac{15}{7}\)
=\(\dfrac{15}{7}\left[\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right)\right]\)
=\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}-\dfrac{2}{5}+\dfrac{27}{4}\right)\)
=\(\dfrac{15}{7}.\dfrac{35}{8}\)=\(\dfrac{75}{8}\)
\(\dfrac{3}{4}=\dfrac{12}{16};\dfrac{5}{6}>\dfrac{7}{9};\dfrac{4}{15}< \dfrac{4}{13};\dfrac{5}{4}-\dfrac{5}{6}>\dfrac{3}{4}-\dfrac{2}{3}\)
\(\dfrac{15}{100}=0,15\)
\(115\%=\dfrac{115}{100}=1,15\)
Mà \(0,15< 1,15\)
Vậy \(\dfrac{15}{100}< 115\%\)
____________
\(\dfrac{5}{2}=2,5\)
Vậy \(\dfrac{5}{2}=2,5\)
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