(6*45^4–15^13*5^–9)/(27^4*25^3+45^6)
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\(3^6\cdot45^4-15^{13}\cdot5^{-9}\)
\(=3^6\cdot\left(3^2\cdot5\right)^4-\left(3\cdot5\right)^{13}\cdot5^{-9}\)
\(=3^6\cdot3^8\cdot5^4-3^{13}\cdot5^{13}\cdot5^{-9}=3^{14}\cdot5^4-3^{13}\cdot5^4=3^{13}\cdot5^4\left(3-1\right)=3^{13}\cdot5^4\cdot2\)
\(27^4\cdot25^3+45^6\)
\(=\left(3^3\right)^4\cdot\left(5^2\right)^3+\left(3^2\cdot5\right)^6\)
\(=3^{12}\cdot5^6+3^{12}\cdot5^6=2\cdot3^{12}\cdot5^6\)
Ta có: \(\frac{3^6\cdot45^4-15^{13}\cdot5^{-9}}{27^4\cdot25^3+45^6}\)
\(=\frac{3^{13}\cdot5^4\cdot2}{2\cdot3^{12}\cdot5^6}\)
\(=\frac{3}{5^2}=\frac{3}{25}\)
a: \(-\dfrac{4}{7}-\dfrac{5}{13}\cdot\dfrac{-39}{25}+\dfrac{-1}{42}:\dfrac{-5}{6}\)
\(=\dfrac{-4}{7}+\dfrac{3}{5}+\dfrac{1}{35}\)
\(=\dfrac{-20}{35}+\dfrac{21}{35}+\dfrac{1}{35}\)
\(=\dfrac{2}{35}\)
\(A=\dfrac{3^6\cdot3^8\cdot5^4-5^{13-9}\cdot3^{13}}{3^{12}\cdot5^6+3^{12}\cdot5^6}\)
\(=\dfrac{3^{14}\cdot5^4-5^4\cdot3^{13}}{3^{12}\cdot5^6\cdot2}\)
\(=\dfrac{3^{13}\cdot5^4\cdot2}{3^{12}\cdot5^6\cdot2}=\dfrac{3}{25}\)
\(B=\left(\dfrac{2}{5}\cdot5\right)^7+\left(\dfrac{9}{4}:\dfrac{3}{16}\right)^3=1+12^3=1729\)