\(\dfrac{27^{10}+9^5}{9^{13}+27^2}\)

\(=\dfrac{\left(3^3\right)^{10}+\left(3^2\right)^5}{\left(3^2\right)^{13}+\left(3^3\right)^2}\)

\(=\dfrac{3^{30}+3^{10}}{3^{26}+3^6}\)

\(=\dfrac{3^{10}\cdot\left(3^{20}+1\right)}{3^6\cdot\left(3^{20}+1\right)}\)

\(=\dfrac{3^{10}}{3^6}\)

\(=3^{10-6}\)

\(=3^4\)

1

\(a,\dfrac{5^{16}\cdot27^7}{125^5\cdot9^{11}}=\dfrac{5^{16}\cdot\left(3^3\right)^7}{\left(5^3\right)^5\cdot\left(3^2\right)^{11}}\)

\(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)

\(b,\left(-0,2\right)^2\cdot5-\dfrac{2^{13}\cdot27^3}{4^6\cdot9^5}\)

\(=0,04\cdot5-\dfrac{2^{13}\cdot\left(3^3\right)^3}{\left(2^2\right)^6\cdot\left(3^2\right)^5}\)

\(=0,2-\dfrac{2^{13}\cdot3^9}{2^{12}\cdot3^{10}}\)

\(=0,2-\dfrac{2}{3}\)

\(=-\dfrac{7}{15}\)

\(c,\dfrac{5^6+2^2\cdot25^3+2^3\cdot125^2}{26\cdot5^6}\)

\(=\dfrac{5^6+2^2\cdot\left(5^2\right)^3+2^3\cdot\left(5^3\right)^2}{5^6\cdot26}\)

\(=\dfrac{5^6+4\cdot5^6+8\cdot5^6}{5^6\cdot26}\)

\(=\dfrac{5^6\left(1+4+8\right)}{5^6\cdot26}\)

\(=\dfrac{13}{26}\)

\(=\dfrac{1}{2}\)

#\(Toru\)

\(a,\dfrac{5^{16}.27^7}{125^5.9^{11}}=\dfrac{\left(5^2\right)^8.9^7.3^7}{25^5.5^5.9^{11}}\\ =\dfrac{25^8.9^7.\left(3^2\right)^3.3}{25^5.\left(5^2\right)^2.5.9^{11}}=\dfrac{25^8.9^7.9^3.3}{25^5.25^2.5.9^{11}}\\ =\dfrac{25^8.9^{10}.3}{25^7.5.9^{11}}=\dfrac{25^7.9^{10}.25.3}{25^7.9^{10}.5.9}\\ =\dfrac{25.3}{5.9}=\dfrac{5.5.3}{5.3.3}=\dfrac{5}{3}\)

\(a,=\dfrac{3^6\cdot5^4\cdot9^4-5^{13}\cdot3^{13}\cdot5^{-9}}{3^{12}\cdot5^6+9^6\cdot5^6}=\dfrac{3^{14}\cdot5^4-5^4\cdot3^{13}}{3^{12}\cdot5^6+3^{12}\cdot5^6}\\ =\dfrac{3^{13}\cdot5^4\cdot2}{2\cdot3^{12}\cdot5^6}=\dfrac{3}{5^2}=\dfrac{3}{25}\)

\(b,=\dfrac{\left(\dfrac{2}{5}\cdot5\right)^7+\left(\dfrac{9}{4}\cdot\dfrac{16}{3}\right)^3}{2^7\cdot5^2+2^9}=\dfrac{2^7+12^3}{2^7\left(5^2+2^2\right)}=\dfrac{2^7+4^3\cdot3^3}{2^7\cdot29}=\dfrac{2^6\left(2+3^3\right)}{2^7\cdot29}=\dfrac{1}{2}\)

a: \(-\dfrac{4}{7}-\dfrac{5}{13}\cdot\dfrac{-39}{25}+\dfrac{-1}{42}:\dfrac{-5}{6}\)

\(=\dfrac{-4}{7}+\dfrac{3}{5}+\dfrac{1}{35}\)

\(=\dfrac{-20}{35}+\dfrac{21}{35}+\dfrac{1}{35}\)

\(=\dfrac{2}{35}\)

giúp mk ik, năn nỉ đó.

\(=\dfrac{2^{19}\cdot3^9-3\cdot3^8\cdot2^{18}\cdot5}{2^{19}\cdot3^9+2^{20}\cdot3^{10}}=\dfrac{-3^{10}\cdot2^{18}}{2^{19}\cdot3^9\cdot7}=-\dfrac{3}{14}\)

a: \(A=\dfrac{3^3\cdot2^3+3^3\cdot2^2+3^3\cdot1}{-13}=\dfrac{27\left(2^3+2^2+1\right)}{-13}=-27\)

b: \(B=\dfrac{2\cdot2^{12}\cdot3^6+2^{11}\cdot3^9}{2^3\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot3^8}\)

\(=\dfrac{2^{13}\cdot3^6+2^{11}\cdot3^9}{2^{10}\cdot3^7+2^{10}\cdot5\cdot3^8}\)

\(=\dfrac{2^{11}\cdot3^6\left(2^2+3^3\right)}{2^{10}\cdot3^7\left(1+5\cdot3\right)}=\dfrac{2}{3}\cdot\dfrac{4+27}{1+15}=\dfrac{2}{3}\cdot\dfrac{31}{16}=\dfrac{31}{24}\)

c: \(C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{35}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{29}\cdot3^{18}\left(5\cdot2^6-7\right)}=\dfrac{10-9}{5\cdot64-7}=\dfrac{1}{313}\)

\(a.\dfrac{-4}{7}-\dfrac{5}{13}\times\dfrac{-39}{25}+\dfrac{-1}{42}:\dfrac{-5}{6}\) 

\(=\dfrac{-4}{7}+\dfrac{3}{5}+\dfrac{1}{35}\) \(=\dfrac{1}{35}+\dfrac{1}{35}=\dfrac{2}{35}\) 

\(b.\dfrac{2}{9}\times\left[\dfrac{4}{5}:\left(\dfrac{1}{5}-\dfrac{2}{15}\right)+1\dfrac{2}{3}\right]-\dfrac{-5}{27}\) 

\(=\dfrac{2}{9}\times\left[\dfrac{4}{5}:\dfrac{1}{15}+\dfrac{5}{3}\right]-\dfrac{-5}{27}\) 

\(=\dfrac{2}{9}\times\left(12+\dfrac{5}{3}\right)-\dfrac{-5}{27}\) 

\(=\dfrac{2}{9}\times\dfrac{41}{3}-\dfrac{-5}{27}=\dfrac{82}{27}-\dfrac{-5}{27}=\dfrac{29}{9}\)

\(a,=\dfrac{1}{6}-\dfrac{5}{6}=-\dfrac{4}{6}=-\dfrac{2}{3}\\ b,=\dfrac{3^2}{3^{12}\cdot3^{18}}=\dfrac{1}{3^{28}}\)