\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3\right)^3.7^3+5^9.2^3.7^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(=\frac{2^{12}.3^5.\left(3-1\right)}{2^{12}.3^6.\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}\)
\(=\frac{2^{12}.3^5.2}{2^{12}.3^6.4}-\frac{5^{10}.7^3.\left(-6\right)}{5^9.7^3.9}\)
\(=\frac{1}{3.2}-\frac{5.\left(-2\right)}{3}\)
\(=\frac{1}{6}-\frac{-10}{3}\)
\(=\frac{1}{6}+\frac{10}{3}\)
\(=\frac{1}{6}+\frac{20}{6}=\frac{21}{6}=\frac{7}{2}\)
\(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}\) = \(\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}\) = \(\frac{2^{12}.\left(3^5-3^4\right)}{2^{12}.\left(3^6+3^5\right)}\) = \(\frac{3^5-3^4}{3^5+3^6}\) = \(\frac{243-81}{243+729}\) = \(\frac{1}{6}\)
\(\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\) = \(\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.7^9}\) = \(\frac{5^{10}.\left(7^3.7^4\right)}{5^9.\left(7^3+7^9\right)}\) = \(\frac{5.\left(343+2401\right)}{343+40353607}\) = \(\frac{13720}{40353950}\) = \(\frac{4}{11765}\)
A = \(\frac{1}{6}\) - \(\frac{4}{11765}\) = \(\frac{11741}{70590}\)
Bn tự rút gọn nha
