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\(=\frac{3^7\cdot\left(2^4\right)^3}{\left(2^2\cdot3\right)^5\cdot\left(3^3\right)^2}\)
\(=\frac{3^7\cdot2^{12}}{2^{10}\cdot3^5\cdot3^6}\)
\(=\frac{3^7\cdot2^{12}}{2^{10}\cdot3^{11}}\)
\(=\frac{2^2}{3^4}\)
\(=\frac{4}{81}\)
\(\dfrac{3}{5}.\dfrac{6}{7}+\dfrac{3}{7}:\dfrac{5}{3}-\dfrac{2}{7}:1\dfrac{2}{3}.\)
\(=\dfrac{3}{5}.\dfrac{6}{7}+\dfrac{3}{7}.\dfrac{3}{5}-\dfrac{2}{7}.\dfrac{3}{5}.\)
\(=\dfrac{3}{5}\left(\dfrac{6}{7}+\dfrac{3}{7}-\dfrac{2}{7}\right).\)
\(=\dfrac{3}{5}.1=\dfrac{3}{5}.\)
Vậy.....
\(\dfrac{3^7.16^3}{12^5.27^2}=\dfrac{3^7.\left(2^4\right)^3}{\left(2^2.3\right)^5.\left(3^3\right)^2}.\)
\(=\dfrac{3^7.2^{12}}{\left(2^2\right)^5.3^5.3^6}=\dfrac{3^7.2^{12}}{2^{10}.3^{11}}.\)
\(=\dfrac{2^2}{3^4}=\dfrac{4}{81}.\)
Vậy.....
\(\frac{2^{12}\cdot3^5-\left(2^2\right)^6.3^5.3}{2^{12}.\left(3^2\right)^3+\left(2^3\right)^4.3^5}\)
=\(\frac{2^{12}\cdot3^5-2^{12}.3^5.3}{2^{12}.3^5+2^{12}.3^5}\)
=3
Bài 1 : Bài giải
\(\frac{28^{15}\cdot3^{17}}{84^{16}}=\frac{\left(2^2\cdot7\right)^{15}\cdot3^{17}}{\left(2^2\cdot3\cdot7\right)^{16}}=\frac{2^{30}\cdot7^{15}\cdot3^{17}}{2^{32}\cdot3^{16}\cdot7^{16}}=\frac{3}{2^2\cdot7}=\frac{3}{4\cdot7}=\frac{3}{28}\)
Bài 2 : Bài giải
\(\frac{3^6\cdot21^{12}}{175^9\cdot7^3}=\frac{3^6\cdot\left(3\cdot7\right)^{12}}{\left(5^2\cdot7\right)^9\cdot7^3}=\frac{3^6\cdot3^{12}\cdot7^{12}}{5^{18}\cdot7^9\cdot7^3}=\frac{3^{18}\cdot7^{12}}{5^{18}\cdot7^{12}}=\frac{3^{18}}{5^{18}}\)
\(\frac{3^{10}\cdot6^7\cdot4}{10^9\cdot5^8}=\frac{3^{10}\cdot\left(2\cdot3\right)^7\cdot2^2}{\left(2\cdot5\right)^9\cdot5^8}=\frac{3^{10}\cdot2^7\cdot3^7\cdot2^2}{2^9\cdot5^9\cdot5^8}=\frac{3^{17}\cdot2^9}{2^9\cdot5^{17}}=\frac{3^{17}}{5^{17}}\)
Ta có : \(3^{17}\cdot5^{18}=3^{17}\cdot5^{17}\cdot5=\left(3\cdot5\right)^{17}\cdot5=15^{17}\cdot5\)
\(3^{18}\cdot5^{17}=3\cdot3^{17}\cdot5^{17}=3\cdot\left(3\cdot5\right)^{17}=3\cdot15^{17}\)
\(\text{ Vì }5\cdot15^{17}>3\cdot15^{17}\text{ }\Rightarrow\text{ }3^{17}\cdot5^{18}>3^{18}\cdot5^{17}\text{ }\Rightarrow\text{ }\frac{3^{18}}{5^{18}}< \frac{3^{17}}{5^{17}}\)
\(\begin{array}{l}a)\frac{{{3^{12}} + {3^{15}}}}{{1 + {3^3}}}\\ = \frac{{{3^{12}} + {3^{12}}{{.3}^3}}}{{1 + {3^3}}}\\ = \frac{{{3^{12}}.(1 + {3^3})}}{{1 + {3^3}}}\\ = {3^{12}}\\b)2:{\left( {\frac{1}{2} - \frac{2}{3}} \right)^2} + 0,{125^3}{.8^3} - {( - 12)^4}:{6^4}\\ = 2:{\left( {\frac{3}{6} - \frac{4}{6}} \right)^2} + {(0,125.8)^3} - {12^4}:{6^4}\\ = 2:{\left( {\frac{{ - 1}}{6}} \right)^2} + {1^3} - {(\frac{{12}}{6})^4}\\ = 2:\frac{1}{{36}} + 1 - {2^4}\\ = 2.36 + 1 - 16\\ = 72 + 1 - 16=57\end{array}\)
\(\frac{3^7.16^3}{12^5.27^2}=\frac{3^7.\left(2^4\right)^3}{\left(2^2.3\right)^5.\left(3^3\right)^2}=\frac{3^7.2^{12}}{2^{10}.3^5.3^6}=\frac{3^7.2^{12}}{2^{10}.3^{11}}=\frac{2^2}{3^4}=\frac{4}{81}\)
\(\frac{3^7.16^3}{12^5.27^2}=\frac{3^7.\left(2^4\right)^3}{\left(2^2.3\right)^5.\left(3^3\right)^2}=\frac{3^7.2^{12}}{2^{10}.3^{11}}=\frac{2^2}{3^4}\)