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\(a,3^{16}:3=3^{16-1}=3^{15}\)
\(b,3^6.3^4.3^2.3=3^{6+4+2+1}=3^{13}\)
\(c,\left(-\frac{1}{4}\right).\left(6\frac{2}{11}\right)+\left(3\frac{9}{11}\right).\left(-\frac{1}{4}\right)=\left(-\frac{1}{4}\right).\frac{68}{11}+\frac{42}{11}.\left(-\frac{1}{4}\right)\)
\(=\left(-\frac{1}{4}\right)\left(\frac{68}{11}+\frac{42}{11}\right)\)
\(=\left(-\frac{1}{4}\right).10\)
\(=-\frac{10}{4}=-\frac{5}{2}\)
\(d,\left(-\frac{1}{2}\right)^3+\frac{1}{2}:5=\left(-\frac{1}{2}\right)\left(\left(\frac{1}{2}\right)^2-\frac{1}{5}\right)\)
\(=-\frac{1}{2}.\left(\frac{1}{4}-\frac{1}{5}\right)\)
\(=-\frac{1}{2}.\frac{1}{20}\)
\(=-\frac{1}{40}\)
\(g,1\frac{1}{25}+\frac{2}{21}-\frac{1}{25}+\frac{19}{21}=\frac{26}{25}+\frac{2}{21}-\frac{1}{25}+\frac{19}{21}\)
\(=\left(\frac{26}{25}-\frac{1}{25}\right)+\left(\frac{2}{21}+\frac{19}{21}\right)\)
\(=1+1\)
\(=2\)
a) \(8^{15}.4^{13}=\left(2^3\right)^{15}.\left(2^2\right)^{13}=2^{45}.2^{26}=2^{71}\)
b) \(\left(\frac{1}{2}\right)^{18}.\left(\frac{1}{4}\right)^{28}=\left(\frac{1}{2}\right)^{18}.\left(\frac{1}{2^2}\right)^{28}=\frac{1}{2^{18}}.\frac{1}{2^{56}}=\frac{1}{2^{74}}=\left(\frac{1}{2}\right)^{74}\)
c) \(9^{12}.27^{10}=\left(3^2\right)^{12}.\left(3^3\right)^{10}=3^{24}.3^{30}=3^{54}\)
a) = (23)15. (22)13 = 245.226 = 271
b) = \(\left(\frac{1}{2}\right)^{18}.\left(\left(\frac{1}{2}\right)^2\right)^{28}=\left(\frac{1}{2}\right)^{18}.\left(\frac{1}{2}\right)^{56}=\left(\frac{1}{2}\right)^{18+56}=\left(\frac{1}{2}\right)^{74}\)
c) = (32)12.(33)10 = 324.330 = 324+30 = 354
\(\frac{8^{11}.3^{17}}{27^{10}.9^{15}}=\frac{8^{11}.3^{17}}{3^{30}.3^{30}}=\frac{8^{11}}{3^{13}.3^{30}}=\frac{8^{11}}{3^{43}}\)
\(\frac{\left(5^4-5^3\right)^3}{125^4}=\frac{[\left(5-1\right).5^3]^3}{5^{12}}=\frac{\left(4.5^3\right)^3}{5^{12}}=\frac{64.5^9}{5^{12}}=\frac{64}{5^3}=\left(\frac{4}{5}\right)^3\)
\(\frac{4^{20}-2^{20}+6^{20}}{6^{20}-3^{20}+9^{20}}=\frac{2^{40}-2^{20}+6^{20}}{6^{20}-3^{20}+3^{40}}=\frac{2^{20}.\left(2^{20}-1+3^{30}\right)}{3^{20}.\left(2^{20}-2+3^{20}\right)}=\frac{2^{20}}{3^{20}}=\left(\frac{2}{3}\right)^{20}\)