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\(\frac{4^5.9^5-2.6^{10}}{2^{10}.3^9+6^9.20}=\frac{\left(4.9\right)^5-2.2^{10.}.3^{10}}{2^{10}.3^9+2^9.3^9.2^2.5}\)
\(\frac{2^{10}.3^{10}^{ }-2^{11}.3^{10}}{2^{10}.3^9+2^{11}.3^9.5}=\frac{2^{10}.3^{10}\left(1-2\right)}{2^{10}.3^9\left(1+2.5\right)}\)
\(\frac{2^{10}.3^{10}.\left(-1\right)}{2^{10}.3^9.\left(-9\right)}=\frac{-3}{-9}=\frac{1}{3}\)
\(A=-\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}\)
\(=\frac{-6.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}\)
\(=-\frac{6}{9}=-\frac{2}{3}\)
Có: \(5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9=5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}=5\cdot2^{30}\cdot3^{18}-3^{20}\cdot2^{29}\)
\(=3^{18}\cdot2^{29}\cdot\left(5\cdot2-3^2\right)=3^{18}\cdot2^{29}\)
Lại có: \(5\cdot2^{10}\cdot6^{19}-7\cdot2^{29}\cdot27^6=5\cdot2^{10}\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}=5\cdot2^{29}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}\)
\(=2^{19}\cdot3^{18}\cdot\left(5\cdot3-7\right)=2^{19}\cdot3^{18}\cdot2^3=2^{22}\cdot3^{18}\)
Vậy \(\frac{3^{18}\cdot2^{29}}{2^{22}\cdot3^{18}}=2^7=128\)
1.\(\left(-\frac{6}{5}+\frac{6}{16}-\frac{6}{23}\right):\left(\frac{9}{5}-\frac{9}{16}+\frac{9}{23}\right)\)
\(=6\left(-\frac{1}{5}+\frac{1}{16}-\frac{1}{23}\right):\left(-9\right)\left(\frac{-1}{5}+\frac{1}{16}-\frac{1}{23}\right)\)
\(=6:\left(-9\right)=-\frac{2}{3}\)
2. \(\frac{\frac{3}{7}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{0.5-\frac{1}{3}+\frac{1}{4}}{-\frac{3}{2}+1-\frac{3}{4}}\)
\(=\frac{3\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{-3\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}-\frac{1}{3}\)
\(=\frac{9}{13}-\frac{5}{15}=\frac{4}{15}\)
ai làm nhanh , trình bày rõ ràng 10 tk
\(\frac{\frac{6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{6\left(\frac{1}{5}+\frac{1}{19}-\frac{1}{23}\right)}{9\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{6}{9}=\frac{2}{3}\)