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『Kuroba ム Tsuki Ryoo... · Accepted Answer

`#3107.\text{DN}`c)$\dfrac{27^3+9^3+9^4}{37}$$=\dfrac{9^3\cdot3^3+9^3+9^4}{37}\
=\dfrac{9^3\cdot\left(3^3+1\right)+9^4}{37}\
=\dfrac{9^3\cdot28+9^4}{37}\
=\dfrac{9^3\cdot\left(28+9\right)}{37}\
=\dfrac{9^3\cdot37}{37}\
=9^3=729$d)$\dfrac{10^9\cdot81^{10}}{8^4\cdot25^5\cdot9^{10}}$$=\dfrac{2^9\cdot5^9\cdot\left(3^2\right)^{10}}{\left(2^3\right)^4\cdot\left(5^2\right)^5\cdot\left(3^2\right)^{10}}$$=\dfrac{2^9\cdot5^9\cdot3^{20}}{2^{12}\cdot5^{10}\cdot3^{20}}$$=\dfrac{1}{2^3\cdot5}\
=\dfrac{1}{8\cdot5}\
=\dfrac{1}{40}$Câu trả lời: `#3107.\text{DN}`c)$\dfrac{27^3+9^3+9^4}{37}$$=\dfrac{9^3\cdot3^3+9^3+9^4}{37}\
=\dfrac{9^3\cdot\left(3^3+1\right)+9^4}{37}\
=\dfrac{9^3\cdot28+9^4}{37}\
=\dfrac{9^3\cdot\left(28+9\right)}{37}\
=\dfrac{9^3\cdot37}{37}\
=9^3=729$d)$\dfrac{10^9\cdot81^{10}}{8^4\cdot25^5\cdot9^{10}}$$=\dfrac{2^9\cdot5^9\cdot\left(3^2\right)^{10}}{\left(2^3\right)^4\cdot\left(5^2\right)^5\cdot\left(3^2\right)^{10}}$$=\dfrac{2^9\cdot5^9\cdot3^{20}}{2^{12}\cdot5^{10}\cdot3^{20}}$$=\dfrac{1}{2^3\cdot5}\
=\dfrac{1}{8\cdot5}\
=\dfrac{1}{40}$

『Kuroba ム Tsuki Ryoo... · Answer

e)$\dfrac{9^4\cdot4^5\cdot25^3}{8^3\cdot27^2\cdot5^7}$$=\dfrac{\left(3^2\right)^4\cdot\left(2^2\right)^5\cdot\left(5^2\right)^3}{\left(2^3\right)^3\cdot\left(3^3\right)^2\cdot5^7}\
=\dfrac{3^8\cdot2^{10}\cdot5^6}{2^9\cdot3^6\cdot5^7}\
=\dfrac{3^2\cdot2}{5}\
=\dfrac{18}{5}$f)$\dfrac{5}{4}\cdot\dfrac{8}{15}+\dfrac{-5}{16}\cdot\dfrac{8}{15}-1$$=\dfrac{8}{15}\cdot\left(\dfrac{5}{4}-\dfrac{5}{16}\right)-1\
=\dfrac{8}{15}\cdot\dfrac{15}{16}-1\
=\dfrac{8}{16}-1\
=\dfrac{1}{2}-1\
=-\dfrac{1}{2}$g)$4\dfrac{5}{9}\div\left(-\dfrac{5}{7}\right)+5\dfrac{4}{9}\div\left(-\dfrac{5}{7}\right)\
=4\dfrac{5}{9}\cdot\left(-\dfrac{7}{5}\right)+5\dfrac{4}{9}\cdot\left(-\dfrac{7}{5}\right)\
=\left(-\dfrac{7}{5}\right)\cdot\left(4\dfrac{5}{9}+5\dfrac{4}{9}\right)\
=\left(-\dfrac{7}{5}\right)\cdot\left[\left(4+5\right)+\left(\dfrac{5}{9}+\dfrac{4}{9}\right)\right]\
=\left(-\dfrac{7}{5}\right)\cdot\left(9+1\right)\
=\left(-\dfrac{7}{5}\right)\cdot10\
=-14$h)$\dfrac{2}{7}\cdot\dfrac{8}{19}+\dfrac{5}{7}\cdot\dfrac{8}{19}\
=\dfrac{8}{19}\cdot\left(\dfrac{2}{7}+\dfrac{5}{7}\right)\
=\dfrac{8}{19}\cdot\dfrac{7}{7}\
=\dfrac{8}{19}\cdot1\
=\dfrac{8}{19}$i)$\dfrac{9}{10}\cdot\dfrac{23}{11}-\dfrac{1}{11}\cdot\dfrac{9}{10}+\dfrac{9}{10}\
=\dfrac{9}{10}\cdot\left(\dfrac{23}{11}-\dfrac{1}{11}+1\right)\
=\dfrac{9}{10}\cdot\left(\dfrac{23}{11}-\dfrac{1}{11}+\dfrac{11}{11}\right)\
=\dfrac{9}{10}\cdot\left(\dfrac{23-1+11}{11}\right)\
=\dfrac{9}{10}\cdot\dfrac{33}{11}\
=\dfrac{9}{10}\cdot3\
=\dfrac{27}{10}$Câu trả lời: e)$\dfrac{9^4\cdot4^5\cdot25^3}{8^3\cdot27^2\cdot5^7}$$=\dfrac{\left(3^2\right)^4\cdot\left(2^2\right)^5\cdot\left(5^2\right)^3}{\left(2^3\right)^3\cdot\left(3^3\right)^2\cdot5^7}\
=\dfrac{3^8\cdot2^{10}\cdot5^6}{2^9\cdot3^6\cdot5^7}\
=\dfrac{3^2\cdot2}{5}\
=\dfrac{18}{5}$f)$\dfrac{5}{4}\cdot\dfrac{8}{15}+\dfrac{-5}{16}\cdot\dfrac{8}{15}-1$$=\dfrac{8}{15}\cdot\left(\dfrac{5}{4}-\dfrac{5}{16}\right)-1\
=\dfrac{8}{15}\cdot\dfrac{15}{16}-1\
=\dfrac{8}{16}-1\
=\dfrac{1}{2}-1\
=-\dfrac{1}{2}$g)$4\dfrac{5}{9}\div\left(-\dfrac{5}{7}\right)+5\dfrac{4}{9}\div\left(-\dfrac{5}{7}\right)\
=4\dfrac{5}{9}\cdot\left(-\dfrac{7}{5}\right)+5\dfrac{4}{9}\cdot\left(-\dfrac{7}{5}\right)\
=\left(-\dfrac{7}{5}\right)\cdot\left(4\dfrac{5}{9}+5\dfrac{4}{9}\right)\
=\left(-\dfrac{7}{5}\right)\cdot\left[\left(4+5\right)+\left(\dfrac{5}{9}+\dfrac{4}{9}\right)\right]\
=\left(-\dfrac{7}{5}\right)\cdot\left(9+1\right)\
=\left(-\dfrac{7}{5}\right)\cdot10\
=-14$h)$\dfrac{2}{7}\cdot\dfrac{8}{19}+\dfrac{5}{7}\cdot\dfrac{8}{19}\
=\dfrac{8}{19}\cdot\left(\dfrac{2}{7}+\dfrac{5}{7}\right)\
=\dfrac{8}{19}\cdot\dfrac{7}{7}\
=\dfrac{8}{19}\cdot1\
=\dfrac{8}{19}$i)$\dfrac{9}{10}\cdot\dfrac{23}{11}-\dfrac{1}{11}\cdot\dfrac{9}{10}+\dfrac{9}{10}\
=\dfrac{9}{10}\cdot\left(\dfrac{23}{11}-\dfrac{1}{11}+1\right)\
=\dfrac{9}{10}\cdot\left(\dfrac{23}{11}-\dfrac{1}{11}+\dfrac{11}{11}\right)\
=\dfrac{9}{10}\cdot\left(\dfrac{23-1+11}{11}\right)\
=\dfrac{9}{10}\cdot\dfrac{33}{11}\
=\dfrac{9}{10}\cdot3\
=\dfrac{27}{10}$

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