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\(\frac{\left(-2\right)^3.3^3.5^3.7.8}{3.2^4.5^3.14}=\frac{2^5.3^3.5^3.7.\left(-2\right)}{3.2^5.5^3.7}=\frac{3^2.\left(-2\right)}{1}=-18\)
Ta có; \(\frac{\left(2^3.3^4\right)}{\left(2^2.3^3.5\right)}\Leftrightarrow\frac{\left(2^2.2^1.3^3.3^1\right)}{2^2.3^3.5}\Leftrightarrow\frac{\left(2.3\right)}{5}=\frac{6}{5}\)
Đs: \(\frac{6}{5}\)
\(a,\)\(\frac{2^5\times3^{12}\times7^8}{2^7\times3^{10}\times7^9}=\frac{3^2\times\left(2^5\times3^{10}\times7^8\right)}{2^2\times7\times\left(2^5\times3^{10}\times7^8\right)}\)\(=\frac{3^2}{2^2\times7}=\frac{9}{28}\)
\(b,\)Tương tự
\(\dfrac{9^{14}\cdot25^5\cdot8^7}{18^2\cdot625^3\cdot24^3}\)
\(=\dfrac{3^{42}\cdot5^{10}\cdot2^{21}}{2^2\cdot3^4\cdot5^{12}\cdot2^9\cdot3^3}=\dfrac{3^{42}\cdot5^{10}\cdot2^{21}}{2^{11}\cdot3^7\cdot5^{12}}\)
\(=\dfrac{3^{35}}{5^2}\cdot2^{10}\)
\(\dfrac{-2^3\cdot3^3\cdot5^3\cdot7\cdot8}{3\cdot5^3\cdot2^4\cdot42}\)
\(=\dfrac{-2^6\cdot3^3\cdot5^3\cdot7}{3\cdot5^3\cdot2^4\cdot2\cdot3\cdot7}\)
\(=\dfrac{-2^6\cdot3^3\cdot5^3}{2^5\cdot3^2\cdot5^3}=-2\cdot3=-6\)