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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 \(\frac{3^{18}\cdot2^{29}}{2^{22}\cdot3^{18}}=2^7=128\)
\(\frac{4^5.9^5-2.6^{10}}{2^{10}.3^9+6^9.20}=\frac{\left(4.9\right)^5-2.2^{10.}.3^{10}}{2^{10}.3^9+2^9.3^9.2^2.5}\)
\(\frac{2^{10}.3^{10}^{ }-2^{11}.3^{10}}{2^{10}.3^9+2^{11}.3^9.5}=\frac{2^{10}.3^{10}\left(1-2\right)}{2^{10}.3^9\left(1+2.5\right)}\)
\(\frac{2^{10}.3^{10}.\left(-1\right)}{2^{10}.3^9.\left(-9\right)}=\frac{-3}{-9}=\frac{1}{3}\)
\(\sqrt{10^2-6^2}-\sqrt{13^2-12^2}+\sqrt{13^2}-\sqrt{12^2}\)
\(=\sqrt{100-36}-\sqrt{169-144}+\sqrt{13^2}-\sqrt{12^2}\)
\(=\sqrt{64}-\sqrt{25}+\sqrt{13^2}-\sqrt{12^2}\)
\(=\sqrt{8^2}-\sqrt{5^2}+\sqrt{13^2}-\sqrt{12^2}\)
\(=8-5+13-12=4\)
\(a,7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)\)
\(=\frac{68}{9}-\frac{11}{4}-\frac{32}{9}\)
\(=4-\frac{11}{4}\)
\(=\frac{5}{4}\)
\(b,\left[\frac{3}{5}.\left(\frac{-7}{10}\right)+\frac{2}{10}.\frac{3}{5}\right]:\left(-3\right)\)
\(=\left(\frac{-21}{50}+\frac{3}{25}\right):\left(-3\right)\)
\(=\frac{-3}{10}:\left(-3\right)\)
\(=\frac{1}{10}\)
\(B=\frac{10}{7.12}+\frac{10}{12.17}+\frac{10}{17.22}+...+\frac{10}{2017.2022}\)
\(=2\left(\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{17.22}+...+\frac{5}{2017.2022}\right)\)
\(=2\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{2017}-\frac{1}{2022}\right)\)
\(=2\left(\frac{1}{7}-\frac{1}{2022}\right)=\frac{2015}{7077}\)
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