Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(\Rightarrow A=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3.7\right)^3+5^9.\left(2.7\right)^3}\)
\(\Rightarrow A=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.7^3.2^3}\)
\(\Rightarrow A=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3\left(1-4\right)}{5^9.7^8\left(1+2^3\right)}\)
\(\Rightarrow A=\frac{2}{3.4}-\frac{5.\left(-3\right)}{9}\)
\(\Rightarrow A=\frac{1}{3}-\frac{-15}{9}\)
\(\Rightarrow A=\frac{1}{3}+\frac{5}{3}\)
\(\Rightarrow A=\frac{6}{3}=2\)
Vậy \(A=2\)
\(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3.7\right)^3+5^9.\left(2.7\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{2^{12}.3^4.2}{2^{12}.3^5.4}-\frac{5^{10}.7^3.\left(-6\right)}{5^9.7^3.9}\)
\(=\frac{1}{6}-\frac{-10}{3}=\frac{1}{6}+\frac{10}{3}=\frac{1}{6}+\frac{20}{6}=\frac{21}{6}=\frac{7}{2}\)
\(A=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(=\dfrac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(=\dfrac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5.\left(3+1\right)}-\dfrac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}\)
\(=\dfrac{2^{12}.3^4.2}{2^{12}.3^5.4}-\dfrac{5^{10}.7^3.\left(-6\right)}{5^9.7^3.9}\)
\(=\dfrac{1}{6}-\dfrac{-10}{3}\)
\(=\dfrac{7}{2}\)
\(A=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)
\(=\dfrac{3^4\left(3-1\right)}{3^5\left(3+1\right)}-\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\cdot9}\)
\(=\dfrac{1}{3\cdot2}-\dfrac{1}{5}\cdot\dfrac{-6}{9}=\dfrac{1}{6}+\dfrac{6}{45}=\dfrac{45+36}{270}=\dfrac{81}{270}=\dfrac{3}{10}\)
\(\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(=\dfrac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.7^3.2^3}\)
\(=\dfrac{2^{12}(3^5-3^4)}{2^{12}(3^6+3^5)}-\dfrac{5^{10}(7^3-7^4)}{5^9.7^3(1+2^3)}\)
\(=\dfrac{2^{12}.162}{2^{12}.972}-\dfrac{5^{10}(-2058)}{5^9.7^3.9}\)
\(=\dfrac{2^{12}.162}{2^{12}.972}-\dfrac{5^{10}(-2058)}{5^9.7^3.9}\)
\(=\dfrac{162}{972}-\dfrac{5(-2058)}{7^3.9}\)
\(=\dfrac{2.3^4}{2^2.3^5}-\dfrac{5.2.7^3.\left(-3\right)}{7^3.3^2}\)
\(=\dfrac{1}{2.3}-\left(\dfrac{-\left(5.2\right)}{3}\right)\)
\(=\dfrac{1}{6}-\left(\dfrac{-10}{3}\right)\)
\(=\dfrac{7}{2}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^4.2^3.7^3}\)
\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3\left(1-7\right)}{5^4.7^3\left(5^5+2^3\right)}\)
\(=\frac{1}{6}+\frac{93750}{3133}\)
\(B=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}=\dfrac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}=\dfrac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}=\dfrac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}=\dfrac{2}{3.4}=\dfrac{1}{3}\)
sai r
