a) \(\frac{205.5^{10}}{100^5}=\frac{41.5^{11}}{5^2.4}=\frac{41.5^9}{4}\)

b) \(\frac{\left(0,9\right)^5}{\left(0,3\right)^5}=\left(0,9:0,3\right)^5=3^5=243\)

c) \(\frac{6^2+3.6+3^2}{-13}=\frac{2^2.3^2+3.6+3^2}{-13}=\frac{3\left(2^2.3+6+3\right)}{-13}=\frac{3.21}{-13}=\frac{63}{-13}\)

d) \(\frac{4^6.9^5.6^9.120}{8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^5.\left(2.3\right)^9.2^3.3.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}=\frac{2^{24}.3^{20}.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{24}.3^{20}.5}{2^{11}.3^{11}.\left(2.3-1\right)}=\frac{2^{13}.3^9.5}{5}=2^{13}.3^9\)

a. \(\frac{20^5.5^{10}}{100^5}\)= \(\frac{20^5.5^{10}}{20^5.5^5}\)= \(5^5\)=\(3125\)

b. \(\frac{0,9^5}{0,3^6}\)= \(\frac{0,9^5}{0,3^5.0,3}\)= \(\left(\frac{0,9}{0,3}\right).\frac{1}{0,3}\)= \(243.\frac{1}{0,3}\)= \(810\)

c.\(\frac{6^3+3.6^2+3^3}{-13}=\frac{\left(3.2\right)^3+3.\left(3.2\right)^{^2}+3^3}{-13}=\frac{3^3.2^3+3.3^2.2^2+3^3}{-13}\)\(=\frac{3^3\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}=3^3.\left(-1\right)=-27\)

a. \(\frac{20^5.5^{10}}{100^5}\)

\(=\frac{20^5.\left(5^2\right)^5}{100^5}\)

\(=\frac{20^5.25^5}{100^5}\)

\(=\frac{500^5}{100^5}\)

\(=\left(\frac{500}{100}\right)^5\)

\(=5^5=3125\)

b. \(\frac{\left(0,9\right)^5}{\left(0,3\right)^6}\)

\(=\frac{\left(0,9\right)^5}{\left(0,3\right)^5.0,3}\)

\(=\left(\frac{0,9}{0,3}\right)^5.\frac{1}{0,3}\)

\(=3^5.\frac{1}{0,3}\)

\(=810\)

c. \(\frac{6^3+3.6^2+3^3}{-13}\)

\(=\frac{\left(3.2\right)^3+3.\left(3.2\right)^2+3^3}{-13}\)

\(=\frac{3^3\left(2^3+2^2+1\right)}{-13}\)

\(=\frac{3^3.13}{-13}\)

\(=\left(-3\right)^3\)

\(=-27\)

a) \(\left(\dfrac{3}{4}\right)^{-2}\cdot3^2\cdot12^0=16\)

b) \(\left(\dfrac{1}{12}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-2}=27\)

c) \(\left(2^{-2}\cdot5^2\right)^{-2}:\left(5\cdot5^{-5}\right)=16\)

a)\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{3^3\cdot2^3+3^3\cdot2^2+3^3}{-13}=\frac{3^3\left(2^3+2^2+1\right)}{-13}=-3^3=-27\)

b) \(\frac{2^3+3\cdot2^6-4^3}{2^3+3^2}=\frac{8+3\cdot64-64}{8+9}=\frac{8+192-64}{17}=\frac{136}{17}=8\)

c) \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{11}\cdot3^{10}\left(2+2\cdot5\right)}{2^{11}\cdot3^{10}\cdot\left(2\cdot3^2-3\right)}=\frac{12}{18-3}=\frac{12}{15}\)

d) \(\frac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}=\frac{5^5\cdot20^3-5^4\cdot20^3+20^3\cdot20^2\cdot5^2}{5^6\cdot4^5}=\frac{20^3\left(5^5-5^4+5^4\cdot4^2\right)}{20^5\cdot5}\)\(=\frac{5^4\left(5-1+16\right)}{20^2\cdot5}=\frac{5^4\cdot20}{20^2\cdot5}=\frac{5^3}{20}=\frac{5^3}{5\cdot4}=\frac{25}{4}\)

                                Bài giải

a)\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{3^3\cdot2^3+3^3\cdot2^2+3^3}{-13}=\frac{3^3\left(2^3+2^2+1\right)}{-13}=-3^3=-27\)

b) \(\frac{2^3+3\cdot2^6-4^3}{2^3+3^2}=\frac{8+3\cdot64-64}{8+9}=\frac{8+192-64}{17}=\frac{136}{17}=8\)

c) \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{11}\cdot3^{10}\left(2+2\cdot5\right)}{2^{11}\cdot3^{10}\cdot\left(2\cdot3^2-3\right)}=\frac{12}{18-3}=\frac{12}{15}\)

d) \(\frac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}=\frac{5^5\cdot20^3-5^4\cdot20^3+20^3\cdot20^2\cdot5^2}{5^6\cdot4^5}=\frac{20^3\left(5^5-5^4+5^4\cdot4^2\right)}{20^5\cdot5}\)\(=\frac{5^4\left(5-1+16\right)}{20^2\cdot5}=\frac{5^4\cdot20}{20^2\cdot5}=\frac{5^3}{20}=\frac{5^3}{5\cdot4}=\frac{25}{4}\)