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a) \(4\frac{5}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)=\frac{41}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)\)
\(=\frac{41}{9}\cdot\left(-\frac{7}{5}\right)+\frac{49}{9}\cdot\left(-\frac{7}{5}\right)=\left(\frac{41}{9}+\frac{49}{9}\right)\cdot\left(-\frac{7}{5}\right)=10\cdot\left(-\frac{7}{5}\right)=-14\)
b) \(\left(\frac{-3}{5}+\frac{4}{9}\right):\frac{7}{11}+\left(\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(\frac{-3}{5}+\frac{4}{9}+\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(\frac{-3}{5}+\frac{-2}{5}+\frac{4}{9}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(-1+1\right):\frac{7}{11}=0\cdot\frac{11}{7}=0\)
c) \(\left(\frac{3}{4}\right)^4\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\right)^2\cdot\left(\frac{3}{4}\right)^2\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{8}{9}\right)^2\)
\(=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)
d) \(\left(-\frac{3}{5}\right)^6\cdot\left(-\frac{5}{3}\right)^5=\left(-\frac{3}{5}\right)^5\cdot\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)^5=\left[\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)\right]^5\cdot\left(-\frac{3}{5}\right)\)
\(=1^5\cdot\left(-\frac{3}{5}\right)=1\cdot\left(-\frac{3}{5}\right)=-\frac{3}{5}\)
e) \(\frac{8^{14}}{4^4\cdot64^5}=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^4\cdot\left(2^6\right)^5}=\frac{2^{42}}{2^8\cdot2^{30}}=\frac{2^{42}}{2^{38}}=2^4=16\)
f) \(\frac{9^{10}\cdot27^7}{81^7\cdot3^{15}}=\frac{\left(3^2\right)^{10}\cdot\left(3^3\right)^7}{\left(3^4\right)^7\cdot3^{15}}=\frac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}=\frac{3^{41}}{3^{43}}=3^{-2}=\frac{1}{3^2}=\frac{1}{9}\)
a,\(\frac{-2}{5}+\frac{7}{21}=\frac{-2}{5}+\frac{1}{3}=\frac{-6}{15}+\frac{5}{15}=\frac{-1}{15}\)
b,\(\left(\frac{1}{3}\right)^5.3^5-2020^0=\left(\frac{1}{3}.3\right)^5-1=1^5-1=1-1=0\)
c,\(\left(-\frac{1}{4}\right).6\frac{2}{11}+3\frac{9}{11}.\left(-\frac{1}{4}\right)\)
\(=\left(-\frac{1}{4}\right).\left(6\frac{2}{11}+3\frac{9}{11}\right)=\left(-\frac{1}{4}\right).\left[\left(6+3\right)+\left(\frac{2}{11}+\frac{9}{11}\right)\right]\)
\(=\left(-\frac{1}{4}\right).\left[9+1\right]=\frac{-1}{4}.10=\frac{\left(-1\right).10}{4}=\frac{\left(-1\right).5}{2}=\frac{-5}{2}\)
\(\frac{\left(-6\right).5.\left(-15\right).\left(11\right)}{11.7.34.\left(-30\right)}\)
\(\frac{4950}{-78540}\)
\(\frac{-495}{7854}\)