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Bài giải
a, \(\frac{7}{12}+\frac{5}{6}+\frac{1}{4}-\frac{3}{7}-\frac{5}{12}\)
\(=\left(\frac{7}{12}-\frac{5}{12}+\frac{5}{6}+\frac{1}{4}\right)-\frac{3}{7}=\left(\frac{7}{12}-\frac{5}{12}+\frac{10}{12}+\frac{3}{12}\right)-\frac{3}{7}=\frac{5}{4}-\frac{3}{7}=\frac{23}{28}\)
b, \(\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=\frac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}=\frac{3^{29}\left(11-3\right)}{3^{28}\cdot4}=\frac{3\cdot8}{4}=6\)
a) \(\left(\frac{-1}{6}+\frac{5}{-12}\right)+\frac{7}{12}=\left(\frac{-2}{12}+\frac{-5}{12}\right)+\frac{7}{12}=\left(\frac{-7}{12}\right)+\frac{7}{12}=0\)
b)\(\frac{7}{36}-\frac{8}{-9}+\frac{-2}{3}=\frac{7}{36}+\frac{32}{36}-\frac{24}{36}=\frac{15}{36}=\frac{5}{12}\)
c) \(\frac{3}{5}-\frac{2}{5}.\frac{10}{12}=\frac{3}{5}-\frac{2}{5}.\frac{5}{6}=\frac{3}{5}-\frac{1}{3}=\frac{9}{15}-\frac{5}{15}=\frac{4}{15}\)
d) \(\frac{2}{\left(-3\right)^2}+\frac{5}{-13}-\frac{-3}{4}=\frac{2}{9}-\frac{5}{13}+\frac{3}{4}=\frac{8}{36}-\frac{15}{36}+\frac{27}{36}=\frac{5}{9}\)
\(a.\frac{-3}{5}< \frac{1}{-2}< \frac{-5}{-12}< \frac{2}{3}< \frac{3}{2}\)
\(b.\frac{6}{-5}< \frac{7}{-6}< \frac{9}{-10}< \frac{-2}{-5}< \frac{3}{4}\)
\(c.\frac{4}{-9}< \frac{-7}{21}< \frac{4}{-15}< \frac{8}{12}< \frac{24}{15}\)
Hok tốt :D
a) \(\frac{32}{7}\)
b)\(\frac{1}{15}\)