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g) \(\frac{16^{12}.8}{32^5.64^4}=\frac{\left(2^4\right)^{12}.2^3}{\left(2^5\right)^5.\left(2^6\right)^4}=\frac{2^{48}.2^3}{2^{25}.2^{24}}=\frac{2^{51}}{2^{49}}=2^2=4\)
k) \(\frac{\left(0,8\right)^4}{\left(0,4\right)^5}=\frac{\left(0,4.2\right)^4}{\left(0,4\right)^5}=\frac{\left(0,4\right)^4.2^4}{\left(0,4\right)^5}=\frac{2^4}{0,4}=40\)
f) \(\frac{25^2.20^4}{5^{10}.4^5}=\frac{\left(5^2\right)^5.\left(4.5\right)^4}{5^{10}.4^5}=\frac{5^{10}.5^4.4^4}{5^{10}.4^5}=\frac{5^{14}.4^4}{5^{10}.4^5}=\frac{5^4}{4}\)
i) \(\frac{9^{15}.81^4}{27^8.3^{20}}=\frac{\left(3^2\right)^{15}.\left(3^4\right)^4}{\left(3^3\right)^8.3^{20}}=\frac{3^{30}.3^{16}}{3^{24}.3^{20}}=\frac{3^{46}}{3^{44}}=3^2=9\)
f) Ta có: \(\frac{25^2.20^4}{5^{10}.4^5}\)= \(\frac{\left(5^2\right)^2.\left(4.5\right)^4}{5^{10}.4^5}\)= \(\frac{5^4.4^4.5^4}{5^{10}.4^5}\)= \(\frac{5^8.4^4}{5^{10}.4^5}\)= \(\frac{1}{5^2.4}\)=\(\frac{1}{100}\).
i) Ta có: \(\frac{9^{15}.81^4}{27^8.3^{20}}\)= \(\frac{\left(3^2\right)^{15}.\left(3^4\right)^4}{\left(3^3\right)^8.3^{20}}\)= \(\frac{3^{30}.3^8}{3^{24}.3^{20}}\)= \(\frac{3^{38}}{3^{44}}\)=\(\frac{1}{3^6}\)= \(\frac{1}{729}\)
b) \(\frac{36^7}{2^{15}.27^5}=\frac{\left(2^2.3^2\right)^7}{2^{15}.\left(3^3\right)^5}=\frac{2^{14}.3^{14}}{2^{15}.3^{15}}=\frac{1.1}{2.3}=\frac{1}{6}\)
h) \(\frac{2^{18}.9^4}{6^6.8^4}=\frac{2^{18}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^4}=\frac{2^{18}.3^8}{2^6.3^6.2^{12}}=\frac{2^{18}.3^8}{2^{18}.3^6}=\frac{1.3^2}{1.1}=9\)
o) \(\frac{3^3+3.6^2+6^3}{13}=\frac{3^3+6^2\left(3+6\right)}{13}=\frac{3^3+6^2.3^2}{13}\)
\(=\frac{3^2\left(3+6^2\right)}{13}=\frac{9.3.13}{13}=\frac{9.3.1}{1}=27\)
C = \(\frac{2}{3}\sqrt{144}-\left(-\frac{3}{4}\right)\div\sqrt{\frac{225}{144}}\)
C = \(\frac{2}{3}.12+\frac{3}{4}\div\frac{5}{4}\)
C = \(8+\frac{3}{5}\)
C = \(8\frac{3}{5}\)
D = \(\frac{4^6.25^5-2^{12}.25^4}{2^{12}.5^8-10^8.64}\)
D = \(\frac{\left(2^2\right)^6.\left(5^2\right)^5-2^{12}.\left(5^2\right)^4}{2^{12}.5^8-\left(2.5\right)^8.2^6}\)
D = \(\frac{2^{12}.5^{10}-2^{12}.5^8}{2^{12}.5^8-2^8.5^8.2^6}\)
D = \(\frac{2^{12}.5^8.\left(25-1\right)}{2^{12}.5^8.\left(1-2^2\right)}\)
D = \(\frac{24}{-3}\)
D = \(-8\)
\(C=\frac{2}{3}\sqrt{144}-\left(\frac{-3}{4}\right):\sqrt{\frac{225}{144}}\)
\(=\frac{2}{3}\cdot12+\frac{3}{4}:\frac{5}{4}\)
\(=8+\frac{3}{4}\cdot\frac{4}{5}\)
\(=8+\frac{3}{5}\)
\(=\frac{40}{5}+\frac{3}{4}=\frac{43}{5}\)
\(D=\frac{4^6\cdot25^5-2^{12}\cdot25^4}{2^{12}\cdot5^8-10^8\cdot64}=\frac{\left(2^2\right)^6\cdot\left(5^2\right)^5-2^{12}\cdot\left(5^2\right)^4}{2^{12}\cdot5^8-\left(2\cdot5\right)^8\cdot2^6}\)
\(=\frac{2^{12}\cdot5^{10}-2^{12}\cdot5^8}{2^{12}\cdot5^8-2^{14}\cdot5^8}=\frac{5^8\left(2^{12}\cdot5^2-2^{12}\right)}{5^8\left(2^{12}-2^{14}\right)}\)
\(=\frac{2^{12}\cdot5^2-2^{12}}{2^{12}-2^{14}}=\frac{2^{12}\left(5^2-1\right)}{2^{12}\left(1-2^2\right)}=\frac{24}{-3}=-8\)
\(a)=\frac{7}{25}+\frac{4}{13}-\frac{5}{2}+\frac{18}{25}-\frac{17}{13}\)
\(=1-1-\frac{5}{2}\)
\(=-\frac{5}{2}\)
\(A=\frac{4^6.9^5+69.120}{8^4.3^{12}+6^{11}}=\frac{2^{12}.3^{10}+2^3.3^2.115}{2^{12}.3^{12}+\left(2.3\right)^{11}}=\frac{3^2.2^3\left(115+2^9.3^8\right)}{6^{11}\left(6+1\right)}=\frac{115+2^9.3^8}{6^8.3.7}\)
\(B=\frac{10^4+5.10^3+5^4}{25}=\frac{\left(10^2\right)^2+2.5^2.10^2+\left(5^2\right)^2}{25}=\frac{\left(10^2+5^2\right)^2}{25}=\frac{125^2}{25}=\frac{25.625}{25}=625\)
\(C=\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{4^{10}.2^{10}+4^{10}}{4^4.4^7+4^4.2^4}=\frac{4^{10}\left(2^{10}+1\right)}{4^4.2^4\left(2^{10}+1\right)}=\frac{4^6}{2^4}=256\)
f) \(\frac{25^2.20^4}{5^{10}.4^5}=\frac{\left(5^2\right)^2.\left(4.5\right)^4}{5^{10}.4^5}=\frac{5^4.5^4.4^4}{5^{10}.4^5}=\frac{5^8.4^4}{5^{10}.4^5}=\frac{1}{5^2.4}=\frac{1}{100}\)
g) \(\frac{16^{12}.8}{32^5.64^4}=\frac{\left(2^4\right)^{12}.2^3}{\left(2^5\right)^5.\left(2^6\right)^4}=\frac{2^{48}.2^3}{2^{25}.2^{24}}=\frac{2^{51}}{2^{49}}=2^2=4\)
h) \(\frac{2^{18}.9^4}{6^6.8^4}=\frac{2^{18}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^4}=\frac{2^{18}.3^8}{2^6.3^6.2^{12}}=\frac{2^{18}.3^8}{2^{18}.3^6}=3^2=9\)