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a) \(\frac{2^{11}.9^2}{3^5.16^2}=\frac{2^{11}.3^4}{3^5.2^8}=\frac{2^3}{3}=\frac{8}{3}\)
\(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^9.2^6.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\frac{2^9.2^6.3^8}{2^9.2^6.3^6}=\frac{2^9.2^6.3^6.3^2}{2^9.2^6.3^6}=3^2=9\)
a, \(\frac{2^{15}.\left(-9\right)^4}{-6^3.8^3}=\frac{2^{15}.\left(-3.3\right)^4}{-\left(2.3\right)^3.\left(2^3\right)^3}=\frac{2^{15}.3^4.3^4}{-2^3.3^3.2^9}=\frac{2^{15}.3^8}{-2^{12}.3^3}=\frac{2^3.3^5}{-1}=-8.243=-1944\)
b, \(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(1+2^{10}\right)}=\frac{2^{20}}{2^{12}}=\frac{1}{2^8}=\frac{1}{256}\)
\(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^8}{2^{15}.3^6}=\frac{3^8}{3^6}=3^2=9\)
a.=[ \(\frac{1}{5}\)x 5 ]5 b. \(\frac{102^3}{40^3}\)=[ \(\frac{102}{40}\)]3 c.=\(\frac{75^{10}.5^{20}}{75^{10}.75^5}\)=\(\frac{5^{20}}{75^5}\)
=15 =[ \(\frac{51}{20}\)]3
=1
1. sai dấu nhé
2.a, \(\frac{45^{10}.5^{20}}{75^{15}}=\frac{\left(3^2.5\right)^{10}.5^{20}}{\left(5^2.3\right)^{15}}=\frac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5=243\)
b, \(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(\frac{4}{5}\right)^5}{\left(\frac{2}{5}\right)^6}=\frac{\left(\frac{2}{5}\cdot2\right)^5}{\left(\frac{2}{5}\right)^6}=\frac{\left(\frac{2}{5}\right)^5\cdot2^5}{\left(\frac{2}{5}\right)^5\cdot\frac{2}{5}}=2^5\div\frac{2}{5}=32\cdot\frac{5}{2}=80\)
c, \(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^2}{2^{15}}=3^2=9\)
:V Làm sai hết rồi sai ngay từ bước đầu tiên.
\(\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-....-\frac{1}{9.10}\)
\(=\frac{1}{3.4}-\left(\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{9.10}\right)\)
\(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(=\frac{1}{12}-\frac{3}{20}\)
\(=\frac{-11}{12}\)
\(\frac{1}{3.4}-\frac{1}{4.5}-...-\frac{1}{9.10}\)
= \(-\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
= \(-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
= \(-\left(\frac{1}{3}-\frac{1}{10}\right)\)
= \(-\frac{7}{30}\)