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Ta có 75,95 = \(\frac{1519}{20}\)
3,5 = \(\frac{7}{2}\)
=> \(\frac{1519}{20}\): \(\frac{7}{2}\)= \(\frac{217}{10}\)= 21,7
\(a,\dfrac{2}{7}+\dfrac{1}{4}=\dfrac{8}{28}+\dfrac{7}{28}=\dfrac{15}{28}\\ b,\dfrac{3}{5}+\dfrac{3}{8}=\dfrac{3.8}{5.8}+\dfrac{3.5}{5.8}=\dfrac{6}{5}\\ c,=\dfrac{4.3}{9.3}+\dfrac{10}{27}=\dfrac{12+10}{27}=\dfrac{22}{27}\\ d,=\dfrac{2.3}{3.3}+\dfrac{7}{9}=\dfrac{6+7}{9}=\dfrac{13}{9}\\ e,=\dfrac{5.5}{12.5}+\dfrac{7.4}{15.4}=\dfrac{25+28}{60}=\dfrac{53}{60}\)
\(\dfrac{2}{7}+\dfrac{1}{4}=\dfrac{8}{28}+\dfrac{7}{28}=\dfrac{15}{28};\dfrac{3}{5}+\dfrac{3}{8}=\dfrac{24}{40}+\dfrac{15}{40};\dfrac{4}{9}+\dfrac{10}{27}=\dfrac{12}{27}+\dfrac{10}{27}=\dfrac{22}{27};\dfrac{2}{3}+\dfrac{7}{9}=\dfrac{18}{27}+\dfrac{21}{27}=\dfrac{39}{27};\dfrac{5}{12}+\dfrac{7}{15}=\dfrac{75}{180}+\dfrac{84}{180}=\dfrac{159}{180}\)
\(\frac{1}{3}\times\frac{3}{5}+\frac{1}{3}\times\frac{2}{5}\)
\(=\frac{1}{3}\times\left(\frac{3}{5}+\frac{2}{5}\right)\)
\(=\frac{1}{3}\)
\(21978:54=407\)
21978 : 54 = 407