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

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

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

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

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

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

\(=3^4\)

1

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

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

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

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

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

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

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

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

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

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

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

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

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

#\(Toru\)

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

\(\dfrac{4}{27}+\dfrac{5}{9}+\dfrac{13}{9}-2\\ =\dfrac{4}{27}+\dfrac{18}{9}-2\\ =\dfrac{4}{27}+\dfrac{18\times3}{9\times3}-\dfrac{2\times27}{27}\\ =\dfrac{4}{27}+\dfrac{54}{27}-\dfrac{54}{27}\\ =\dfrac{4}{27}\)

a: \(A=21\cdot100-11\cdot100+90\cdot100+100\cdot125\cdot16\)

\(=100\left(21-11+90\right)+100\cdot2000\)

\(=100\left(10+90+2000\right)=2100\cdot100=210000\)

b: \(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{28}\cdot3^{18}\left(5\cdot3-7\cdot2\right)}=2\)

\(\dfrac{2}{3}+\dfrac{1}{5}.\dfrac{10}{7}=\dfrac{2}{3}+\dfrac{10}{35}=\dfrac{70}{105}+\dfrac{30}{105}=\dfrac{100}{105}=\dfrac{50}{21}\)

a) Ta có: \(\dfrac{2}{3}+\dfrac{1}{5}\cdot\dfrac{10}{7}\)

\(=\dfrac{2}{3}+\dfrac{2}{7}\)

\(=\dfrac{14}{21}+\dfrac{6}{21}\)

\(=\dfrac{20}{21}\)

\(\dfrac{7}{12}-\dfrac{27}{7}.\dfrac{1}{8}=\dfrac{7}{12}-\dfrac{27}{56}=\dfrac{392}{672}-\dfrac{324}{672}=\dfrac{68}{672}=\dfrac{17}{168}\)

a, \(=\dfrac{2}{9}-\dfrac{10}{10}=\dfrac{2}{9}-1=-\dfrac{7}{9}\)

b, \(=-\dfrac{12}{6}+\dfrac{2}{5}=-2+\dfrac{2}{5}=-\dfrac{8}{5}\)

c, \(=\dfrac{27}{13}-1=\dfrac{14}{13}\)

d, \(=\dfrac{12}{11}+\dfrac{7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+1=\dfrac{23}{11}\)

a) \(\dfrac{2}{9}+\dfrac{-3}{10}+\dfrac{-7}{10}\)

\(=\dfrac{2}{9}+\dfrac{-10}{10}=\dfrac{2}{9}-1=-\dfrac{7}{9}\)

b) \(\dfrac{-11}{6}+\dfrac{2}{5}+\dfrac{-1}{6}\)

\(=-2+\dfrac{2}{5}=-\dfrac{8}{5}\)

a,\(=\dfrac{2}{9}+\left(\dfrac{-3}{10}+\dfrac{-7}{10}\right)=\dfrac{2}{9}+\left(-1\right)=\dfrac{-7}{9}\)

b,\(=\left(\dfrac{-11}{6}+\dfrac{-1}{6}\right)+\dfrac{2}{5}=-2+\dfrac{2}{5}=\dfrac{-8}{5}\)

c,\(=\dfrac{27}{13}-\left(\dfrac{106}{111}+\dfrac{-5}{111}\right)=\dfrac{27}{13}-1=\dfrac{14}{13}\)

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

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

a) \(\dfrac{3^6\cdot3^8\cdot5^4-5^{13}\cdot3^{13}\cdot5^{-9}}{3^{12}\cdot5^6+3^{12}\cdot5^6}=\dfrac{3^{14}\cdot5^4-3^{13}\cdot5^4}{3^{12}\cdot5^6\cdot2}=\dfrac{3^{12}\cdot5^4\left(3^2-3\right)}{3^{12}\cdot5^6\cdot2}=\dfrac{3^2-3}{5^2\cdot2}=\dfrac{6}{50}=\dfrac{3}{25}\)