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1 tháng 8 2018

a. \(\frac{20^5.5^{10}}{100^5}\)

\(=\frac{20^5.\left(5^2\right)^5}{100^5}\)

\(=\frac{20^5.25^5}{100^5}\)

\(=\frac{500^5}{100^5}\)

\(=\left(\frac{500}{100}\right)^5\)

\(=5^5=3125\)

b. \(\frac{\left(0,9\right)^5}{\left(0,3\right)^6}\)

\(=\frac{\left(0,9\right)^5}{\left(0,3\right)^5.0,3}\)

\(=\left(\frac{0,9}{0,3}\right)^5.\frac{1}{0,3}\)

\(=3^5.\frac{1}{0,3}\)

\(=810\)

c. \(\frac{6^3+3.6^2+3^3}{-13}\)

\(=\frac{\left(3.2\right)^3+3.\left(3.2\right)^2+3^3}{-13}\)

\(=\frac{3^3\left(2^3+2^2+1\right)}{-13}\)

\(=\frac{3^3.13}{-13}\)

\(=\left(-3\right)^3\)

\(=-27\)

6 tháng 8 2016

a) \(C=\frac{\left(\frac{2}{3}\right)^3\times\left(-\frac{3}{4}\right)^2\times\left(-1\right)^5}{\left(\frac{2}{5}\right)^2\times\left(-\frac{5}{12}\right)^2}\)

\(C=\frac{\frac{2^3}{3^3}.\frac{\left(-3\right)^2}{4^2}.\left(-1\right)^5}{\frac{2^2}{5^2}.\frac{\left(-5\right)^2}{12^2}}\)

\(C=\frac{\frac{-\left(2^3.3^2\right)}{3^3.2^4}}{\frac{2^2.5^2}{5^2.2^4.3^2}}\)

\(C=\frac{\frac{-1}{3.2}}{\frac{1}{2^2.3^2}}\)

\(C=\frac{\frac{-1}{6}}{\frac{1}{36}}\)

\(C=-6\)

b) \(D=\frac{6^6+6^3\times3^3+3^6}{-73}\)

\(D=\frac{2^6.3^6+2^3.3^3.3^3+3^6}{-73}\)

\(D=\frac{2^6.3^6+2^3.3^6+3^6}{-73}\)

\(D=\frac{3^6\left(2^6+2^3+1\right)}{-73}\)

\(D=\frac{3^6.73}{\left(-1\right).73}\)

\(D=-3^6=-729\)

17 tháng 9 2020

a) \(4\frac{5}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)=\frac{41}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)\)

\(=\frac{41}{9}\cdot\left(-\frac{7}{5}\right)+\frac{49}{9}\cdot\left(-\frac{7}{5}\right)=\left(\frac{41}{9}+\frac{49}{9}\right)\cdot\left(-\frac{7}{5}\right)=10\cdot\left(-\frac{7}{5}\right)=-14\)

b) \(\left(\frac{-3}{5}+\frac{4}{9}\right):\frac{7}{11}+\left(\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)

\(=\left(\frac{-3}{5}+\frac{4}{9}+\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)

\(=\left(\frac{-3}{5}+\frac{-2}{5}+\frac{4}{9}+\frac{5}{9}\right):\frac{7}{11}\)

\(=\left(-1+1\right):\frac{7}{11}=0\cdot\frac{11}{7}=0\)

c) \(\left(\frac{3}{4}\right)^4\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\right)^2\cdot\left(\frac{3}{4}\right)^2\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{8}{9}\right)^2\)

\(=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)

d) \(\left(-\frac{3}{5}\right)^6\cdot\left(-\frac{5}{3}\right)^5=\left(-\frac{3}{5}\right)^5\cdot\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)^5=\left[\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)\right]^5\cdot\left(-\frac{3}{5}\right)\)

\(=1^5\cdot\left(-\frac{3}{5}\right)=1\cdot\left(-\frac{3}{5}\right)=-\frac{3}{5}\)

e) \(\frac{8^{14}}{4^4\cdot64^5}=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^4\cdot\left(2^6\right)^5}=\frac{2^{42}}{2^8\cdot2^{30}}=\frac{2^{42}}{2^{38}}=2^4=16\)

f) \(\frac{9^{10}\cdot27^7}{81^7\cdot3^{15}}=\frac{\left(3^2\right)^{10}\cdot\left(3^3\right)^7}{\left(3^4\right)^7\cdot3^{15}}=\frac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}=\frac{3^{41}}{3^{43}}=3^{-2}=\frac{1}{3^2}=\frac{1}{9}\)