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\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
a) \(\left(\frac{3}{7}+\frac{1}{2}\right)^2=\left(\frac{6}{14}+\frac{7}{14}\right)^2=\left(\frac{13}{14}\right)^2=\frac{169}{196}\)
b) \(\left(\frac{3}{4}-\frac{5}{6}\right)^2=\left(\frac{9}{12}-\frac{10}{12}\right)^2=\left(-\frac{1}{2}\right)^2=\frac{1}{4}\)
d) \(\left(-\frac{10}{3}\right)^5.\left(-\frac{6}{5}\right)^4=-\frac{2560}{3}\)
e) \(\left(1+\frac{2}{3}-\frac{1}{4}\right).\left(\frac{4}{5}-\frac{3}{4}\right)^2=\frac{17}{12}.\left(\frac{1}{20}\right)^2=\frac{17}{12}.\frac{1}{400}=\frac{17}{4800}\)
f) \(2:\left(\frac{1}{2}-\frac{2}{3}\right)^3=2:\left(-\frac{1}{6}\right)^3=2:-\frac{1}{216}=-432\)
3,
a) (−23+37):45+(−13+47):45
= \(-\frac{5}{21}:\frac{4}{5}+\frac{5}{21}:\frac{4}{5}\)
= \(\left(-\frac{5}{21}+\frac{5}{21}\right):\frac{4}{5}\)
= \(0:\frac{4}{5}=0\)
2,
a) \(\frac{-3}{4}\).\(\frac{12}{-5}\).(\(\frac{-25}{6}\))
= \(\frac{-3.4.3.\left(-5\right).5}{4.\left(-5\right).3.3}\)
= \(-5\)
b) (−2).\(\frac{-38}{21}\).\(\frac{-7}{4}\).(\(\frac{-3}{8}\))
= \(\frac{-2.\left(-38\right)\left(-7\right)\left(-3\right)}{\left(-7\right)\left(-3\right)\left(-2\right)\left(-2\right).8}\)
= \(\frac{19}{8}\)
c) (\(\frac{11}{12}:\frac{33}{16}\)).\(\frac{3}{5}\)
= \(\left(\frac{11}{12}.\frac{16}{33}\right).\frac{3}{5}\)
= \(\frac{4}{9}.\frac{3}{5}\)
= \(\frac{4}{15}\)
d) \(\frac{7}{23}\left[\left(\frac{-8}{6}\right)-\frac{45}{18}\right]\)
= \(\frac{7}{23}.\left(\frac{-41}{10}\right)\)
= \(\frac{-287}{203}\)
3. Tính:
a) (\(\frac{-2}{3}+\frac{3}{7}\)):\(\frac{4}{5}\)+(\(\frac{-1}{3}+\frac{4}{7}\)):\(\frac{4}{5}\)
= (\(\frac{-2}{3}+\frac{3}{7}\)\(+\)\(\frac{-1}{3}+\frac{4}{7}\)) : \(\frac{4}{5}\)
= 0 : \(\frac{4}{5}\)
= 0
b) \(\frac{5}{9}\):(\(\frac{1}{11}-\frac{5}{22}\))+\(\frac{5}{9}\):(\(\frac{1}{15}-\frac{2}{3}\))
= \(\frac{5}{9}\): \(\frac{-3}{22}\)+ \(\frac{5}{9}\): \(\frac{-3}{5}\)
= \(\frac{5}{9}\): \(\frac{-81}{110}\)
= \(\frac{-550}{729}\)
Ta có: \(A=\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
\(=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)
\(=\left(6-5-3\right)+\left(-\dfrac{2}{3}-\dfrac{5}{3}+\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)
\(=-2-\dfrac{1}{2}=-\dfrac{5}{2}\)
ngu