H = 3^2.4^4.2^32/11.2^13.(2^2)^11-(2^4)^9

   = 3^2.(2^2)^4.2^32/11.2^13.2^22-2^36

   = 3^2.2^8.2^32/11.2^35-2^36

   = 3^2.2^40/2^35.(11-2)

   = 9.2^40/2^35.9 = 2^5 = 32

k mk nha

\(I=\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\frac{5.2^{30}.3^{27}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)

\(=\frac{5.2^{30}.3^{27}-3^{30}.2^{29}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)

\(=\frac{2^{29}.3^{27}.\left(5.2-3^3\right)}{2^{28}.3^{18}.\left(5.3-2.7\right)}\)

\(=\frac{2^{29}.3^{27}.-17}{2^{18}.3^{18}}\)

\(=\frac{2^9.3^9.-17}{1}\)

Ta có \(H=\frac{\left(3.4.2^{16}\right)}{11.2^{13}.4^{11}-16^9}\)

\(=\frac{3.4.2^{16}}{11.2^{13}.2^{22}-2^{36}}\)

\(=\frac{3.2^{18}}{11.2^{35}-2^{36}}\)

\(=\frac{3.2^{18}}{2^{35}.\left(11-2\right)}\)

\(=\frac{3.2^{18}}{2^{35}.3^2}\)

\(=\frac{1}{2^{17}.3}\)

giúp em câu c) với ạ

c) \(\frac{\left(3\cdot4\cdot2^{16}\right)}{11\cdot2^{13}\cdot4^{11}-16^9}=\frac{\left(3\cdot2^2\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot2^{22}-2^{36}}\)

\(=\frac{9\cdot2^4\cdot2^{32}}{11\cdot2^{35}-2^{26}}\)

\(=\frac{9\cdot2^4\cdot2^{32}2^{ }}{\left(11-2\right)\cdot2^{35}}\)

\(=\frac{9\cdot2^4\cdot2^{32}}{9\cdot2^{35}}\)

\(=\frac{9\cdot1\cdot2^{32}}{9\cdot2^{31}}=\frac{2^{32}}{2^{31}}=2\)

\(c,1.2.3...9-1.2.3...8-1.2.3...7.8^2\)

\(=1.2.3...8\left(9-1-8\right)\)

\(=1.2.3...8.0\)

\(=0\)

\(d,\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)

\(=\frac{3^2.4^2.2^{32}}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)

\(=\frac{3^2.2^4.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)

\(=\frac{3^2.2^{36}}{11.2^{35}-2^{36}}\)

\(=\frac{3^2.2^{36}}{2^{35}\left(11-2\right)}\)

\(=\frac{3^2.2^{36}}{2^{35}.9}\)

\(=\frac{3^2.2^{36}}{2^{35}.3^2}\)

\(=2\)

\(\frac{\left(3\cdot2^{18}\right)^2}{11\cdot2^{13}\cdot4^{11}-16^9}\)

\(=\frac{3^2+2^{18\cdot2}}{11\cdot2^{13}\cdot4^{11}-16^9}\)

\(=\frac{3^2\cdot2^{36}}{11\cdot2^{13}\cdot\left(2^2\right)^{11}-\left(2^4\right)^9}\)

 \(=\frac{3^2\cdot2^{36}}{11\cdot2^{13}\cdot2^{22}-2^{36}}\)

\(=\frac{3^2\cdot2^{36}}{11\cdot2^{35}-2^{36}}\)

\(=\frac{3^2\cdot2^{36}}{11\cdot2^{35}-2^{35}\cdot2}\)

\(=\frac{9\cdot2^{35}\cdot2}{2^{35}\left(11-2\right)}\)

\(=\frac{9\cdot2^{35}\cdot2}{2^{35}\cdot9}\)

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

\(A=\frac{3^7\cdot17-3^9}{2^3\cdot3^5}=\frac{3^7\left(17-3^2\right)}{2^3\cdot3^5}=\frac{3^7\cdot2^3}{2^3\cdot3^5}=9\)

\(B=\frac{3^2\cdot4^2\cdot2^{32}}{11\cdot2^{13}\cdot4^{11}-16^9}=\frac{3^2\cdot2^{36}}{2^{35}\cdot11-2^{36}}=\frac{3^2\cdot2^{36}}{2^{35}\left(11-2\right)}=\frac{3^2\cdot2^{36}}{2^{35}\cdot3^2}=2\)

\(\frac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}=\frac{3^{29}\left(11-3\right)}{2^2\cdot3^{28}}=\frac{3^{29}\cdot8}{2^2\cdot3^{28}}=6\)