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\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6}-\frac{5^{10}.7^4-25^5.49^2}{\left(125.7\right)3+5^9.\left(14\right)^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{125^3.7^3+5^9.\left(2.7\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.7^3.2^3}\)
\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3\left(1-7\right)}{5^9.7^3\left(1+8\right)}\)
\(=\frac{2}{3.4}-\frac{5.\left(-6\right)}{9}=\frac{2}{12}-\frac{-30}{9}\)
\(=\frac{1}{6}+\frac{10}{3}=\frac{1}{6}+\frac{20}{6}=\frac{21}{6}=\frac{7}{2}\)
Bạn ơi cho mk hỏi chỗ đoạn kia bạn lấy 1-7 ở đâu và 1 + 8 ở đâu
\(\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\\ =\dfrac{2^{12}\cdot3^5-\left(2^2\right)^6\cdot\left(3^2\right)^2}{2^{12}\cdot3^6}-\dfrac{5^{10}\cdot7^3-\left(5^2\right)^5\cdot\left(7^2\right)^2}{\left(5^3\cdot7\right)^3+5^9\cdot\left(2\cdot7\right)^3}\\ =\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot2^3\cdot7^3}\\ =\dfrac{2^{12}\cdot3^4\left(3-1\right)}{2^{12}\cdot3^6}-\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}\\ =\dfrac{2}{9}-\dfrac{-6}{1+8}=\dfrac{2}{9}+\dfrac{6}{9}=\dfrac{8}{9}\)
\(\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\\ =\dfrac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\\ =\dfrac{2^{12}.3^5.\left(1-3\right)}{2^{12}.3^6}-\dfrac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}\\ =\dfrac{2^{12}.3^5.\left(-2\right)}{2^{12}.3^6}-\dfrac{5^{10}.7^3.\left(-6\right)}{5^9.7^3.9}\\ =\dfrac{-2}{3}-\dfrac{5.\left(-6\right)}{9}\\ =\dfrac{-2}{3}+\dfrac{30}{9}\\ =\dfrac{8}{3}\)
