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\(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)+8^4.3^5}\)
\(\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^2.3+2^4.3^4.3^5}\)
\(\frac{2^{12}.3^5-2^{12}.3^4}{2^2.3+2^4.3^{20}}\)
\(\frac{2^{12}.3^4\left(3-1\right)}{2^2.3\left(1+2^2.3^{19}\right)}\)
\(\frac{2^{10}.3^2.4}{1+2^2.3^{19}}\)
bn tự giải tiếp nha mk phải có việc rồi nếu ko lm đc thì kb với mk tối mk giải cho nha
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6}-\frac{5^{10}.7^4-25^5.49^2}{\left(125.7\right)3+5^9.\left(14\right)^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{125^3.7^3+5^9.\left(2.7\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.7^3.2^3}\)
\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3\left(1-7\right)}{5^9.7^3\left(1+8\right)}\)
\(=\frac{2}{3.4}-\frac{5.\left(-6\right)}{9}=\frac{2}{12}-\frac{-30}{9}\)
\(=\frac{1}{6}+\frac{10}{3}=\frac{1}{6}+\frac{20}{6}=\frac{21}{6}=\frac{7}{2}\)
Bạn ơi cho mk hỏi chỗ đoạn kia bạn lấy 1-7 ở đâu và 1 + 8 ở đâu
\(\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^4.7^4}{5^9.7^3+5^92^3.7^3}\)\(=\frac{2^{12}\left(3^5-3^4\right)}{2^{12}\left(3^6+3^5\right)}-\frac{7^3\left(5^{10}-5^4.7^4\right)}{7^3\left(5^9+5^9.2^3\right)}\)
\(=\frac{3^5-3^4}{3^6+3^5}-\frac{5^{10}-5^4.7^4}{5^9+5^9.2^3}\)\(=\frac{3^4\left(3-1\right)}{3^4\left(3^2+3\right)}-\frac{5^4\left(5^6-7^4\right)}{5^4\left(5^5+5^5.2^3\right)}\)
\(=\frac{1}{6}-\frac{5^6-7^4}{5^5+5^5.2^3}\)cậu cứ phân tích từ từ
không vt lại đề
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^6.7^3+5^9.2^3.7^3}\)
\(=\frac{2^{12}\cdot3^4.\left(3-1\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3\left(1-7\right)}{5^6.7^3\left(1+2^3\right)}\)
\(=\frac{1}{6}-\frac{-10}{3}=\frac{1}{6}+\frac{20}{6}=\frac{21}{6}\)
a) \(\dfrac{2727-101}{3.303+404}=\dfrac{2626}{909+404}=\dfrac{2626}{1313}=2\)
b) \(\dfrac{8.9-4.15}{12.7-180}=\dfrac{72-60}{84-180}=\dfrac{12}{-96}=\dfrac{-1}{8}\)
c) \(\dfrac{-19}{3^2.7.11}=\dfrac{-19}{9.7.11}=\dfrac{-19}{63.11}=\dfrac{-19}{693}\)
d) \(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\dfrac{2^{12}.3^{10}+120.6^9}{2^{12}.3^{12}-6^{11}}=\dfrac{2^2.6^{10}+20.6.6^9}{6^{12}-6^{11}}=\dfrac{4.6^{10}+20.6^{10}}{6^{11}\left(6-1\right)}=\dfrac{\left(4+20\right).6^{10}}{5.6^{11}}=\dfrac{24}{30}=\dfrac{4}{5}\)
mk ko viết lại đề
\(A=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}+\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{12}.3^{12}}\)
\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}+\frac{2^{12}.3^{10}\left(1+5\right)}{2.\left(2^{12}.3^{12}\right)}\)
\(=\frac{2}{3.4}+\frac{2^{12}.3^{10}.6}{2.2^{12}.3^{12}}=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}\)
Vậy A= \(\frac{1}{2}\)