Ta có: \(\frac{2^{12}\cdot3^5-4^6\cdot81}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}\)
\(=\frac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}\)
\(=\frac{2^{12}\cdot3^4\cdot\left(3-1\right)}{2^{12}\cdot3^5\left(3+1\right)}=\frac13\cdot\frac24=\frac13\cdot\frac12=\frac16\)
Ta có: \(\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
\(=\frac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)
\(=\frac{5^{10}\cdot7^3\cdot\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\frac{-6}{9}=\frac{-30}{9}=\frac{-10}{3}\)
Ta có: \(A=\frac{2^{12}\cdot3^5-4^6\cdot81}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
\(=\frac16+\frac{10}{3}=\frac16+\frac{20}{6}=\frac{21}{6}=\frac72\)
