a) 3-2 . 9n = 3n
b) \(\left(\frac{9}{25}\right)^n=\left(\frac{3}{5}\right)^{-4}\)
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\(\frac{\left(\frac{2}{3}\right)^3.\left(\frac{-3}{4}\right)^2.\left(-1\right)^{2019}}{\left(\frac{2}{5}\right)^2.\left(\frac{-5}{12}\right)^3}:\sqrt{\frac{9}{25}}\)\(=\frac{\frac{2^3}{3^3}.\frac{-3^2}{4^2}.\left(-1\right)}{\frac{2^2}{5^2}.\frac{-5^3}{12^3}}:\frac{3}{5}\)
\(=\frac{\frac{2^3}{5^3}.\frac{-3^2}{2^4}.\left(-1\right)}{\frac{2^2}{5^2}.\frac{-5^3}{2^6.3^3}}:\frac{3}{5}=\frac{\frac{-1}{3.2}}{\frac{-5}{2^4.3^3}}:\frac{3}{5}\)\(=\frac{-1}{3.2}.\frac{-2^4.3^3}{5}.\frac{5}{3}\)
\(=\frac{2^3.3^2}{5}.\frac{5}{3}=24\)
Ta có: \(\frac{0,8:\left(\frac{4}{5}\cdot1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right)\cdot2\frac{2}{17}}+\frac{\left(1,2\cdot0,5\right)}{\frac{4}{5}}\)
\(=\frac{\frac{4}{5}:\left(\frac{4}{5}\cdot\frac{5}{4}\right)}{\frac{16}{25}-\frac{1}{25}}+\frac{\left(\frac{27}{25}-\frac{2}{25}\right)\cdot\frac{7}{4}}{\left(\frac{59}{9}-\frac{13}{4}\right)\cdot\frac{36}{17}}+\frac{6}{5}\cdot\frac{1}{2}\cdot\frac{5}{4}\)
\(=\frac{\frac{4}{5}}{\frac{3}{5}}+\frac{\frac{7}{4}}{\frac{119}{36}\cdot\frac{36}{17}}+\frac{3}{4}\)
\(=\frac{4}{5}\cdot\frac{5}{3}+\frac{7}{4}\cdot\frac{1}{7}+\frac{3}{4}=\frac{4}{3}+\frac{1}{4}+\frac{3}{4}=\frac{7}{3}\)
\(a,A=\left[\frac{4}{11}.\left(\frac{1}{25}\right)^0+\frac{7}{22}.2\right]^{2010}-\left(\frac{1}{2^2}:\frac{8^2}{4^4}\right)^{2009}\)
\(A=\left(\frac{4}{11}.1+\frac{7}{11}\right)^{2010}-\left(\frac{1}{2^2}.2^2\right)^{2009}\)
\(A=1-1=0\)
\(b,B=\frac{0,8:\left(\frac{4}{5}.1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right).2\frac{2}{17}}+\left(1,2.0,5\right):\frac{4}{5}\)
\(B=\frac{0,8:1}{\frac{3}{5}}+\frac{\left(1\right):\frac{4}{7}}{\left(\frac{59}{9}-\frac{13}{4}\right).36}\)
\(B=0,8.\frac{5}{3}+\frac{\frac{7}{4}}{\frac{119}{36}.36}\)
\(B=\frac{4}{3}+\frac{7}{4}.\frac{1}{119}\)
\(B=\frac{4}{3}+\frac{1}{68}=\frac{275}{204}\)