NC
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
30 tháng 3 2018
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= \(\frac{1}{2}\)
DP
27 tháng 6 2017
\(M=\frac{9^4.27^5.3^6.3^4}{3^8.81^4.234.8^2}=\frac{\left(3^2\right)^4.\left(3^3\right)^5.3^6.3^4}{3^8.\left(3^4\right)^4.3^5.\left(2^3\right)^2}\)
\(M=\frac{3^8.3^{15}.3^6.3^4}{3^{18}.3^{16}.3^5.2^6}=\frac{81}{64}\)
\(N=\frac{4^6.9^5.6^9.120}{8^4.3^{12}-6^{11}}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(N=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}=\frac{2.6}{3.5}\)
\(N=\frac{4}{5}\)
TQ
2
17 tháng 6 2016
Câu 1 xem kỉ đề
\(B,\frac{49^6.5-7^{11}}{\left(-7\right)^{10}.5-2.49^5}=\frac{7^{12}.5-7^{11}}{7^{10}.5-2.7^{10}}=\frac{7^{11}.\left(7.5-1\right)}{7^{10}.\left(5-2\right)}=\frac{7.34}{3}=\frac{238}{3}\)
17 tháng 6 2016
a) A=212.35-\(\frac{2^{12}.3^6}{2^{12}}\)+93+84.35
=212.35-36+36+212.35
=213.35
b)B=496.5-5.\(\frac{7^{11}}{\left(-7\right)^{10}}-2.49^5\)
=496.5-7.5-2.495
=712.5-7.5-2.710
21 tháng 10 2018
ban nao co chuyen shin ko cho minh muon minh giai cho 10 bai nhu the i love pac pac
LN
28 tháng 1 2019
\(A=\frac{2^{12}.3^4-4^5.9^2}{\left(2^2.3\right)^6+8^4.3^5}\)
\(A=\frac{2^{12}.3^4-2^{10}.3^4}{2^{12}.3^6+2^{12}.3^5}\)
\(A=\frac{2^{10}.3^4\left(2^2-1\right)}{2^{10}.3^4\left(2^2.3^2+2^2.3\right)}\)
\(A=\frac{2^2-1}{2^2.3^2+2^2.3}\)
\(A=\frac{4-1}{36+12}\)
\(A=\frac{3}{48}=\frac{1}{16}\)
MT
6 tháng 6 2015
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6}-\frac{5^{10}.7^4-25^5.49^2}{\left(125.7\right)3+5^9.\left(14\right)^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{125^3.7^3+5^9.\left(2.7\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.7^3.2^3}\)
\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3\left(1-7\right)}{5^9.7^3\left(1+8\right)}\)
\(=\frac{2}{3.4}-\frac{5.\left(-6\right)}{9}=\frac{2}{12}-\frac{-30}{9}\)
\(=\frac{1}{6}+\frac{10}{3}=\frac{1}{6}+\frac{20}{6}=\frac{21}{6}=\frac{7}{2}\)
27 tháng 1 2016
Bạn ơi cho mk hỏi chỗ đoạn kia bạn lấy 1-7 ở đâu và 1 + 8 ở đâu
9 tháng 7 2015
\(\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^4.7^4}{5^9.7^3+5^92^3.7^3}\)\(=\frac{2^{12}\left(3^5-3^4\right)}{2^{12}\left(3^6+3^5\right)}-\frac{7^3\left(5^{10}-5^4.7^4\right)}{7^3\left(5^9+5^9.2^3\right)}\)
\(=\frac{3^5-3^4}{3^6+3^5}-\frac{5^{10}-5^4.7^4}{5^9+5^9.2^3}\)\(=\frac{3^4\left(3-1\right)}{3^4\left(3^2+3\right)}-\frac{5^4\left(5^6-7^4\right)}{5^4\left(5^5+5^5.2^3\right)}\)
\(=\frac{1}{6}-\frac{5^6-7^4}{5^5+5^5.2^3}\)cậu cứ phân tích từ từ
\(\frac{2^{12}.3^5-4^6.3^6}{2^{12}.9^3+8^4.3^5}\)
\(=\frac{2^{12}.3^5-4^6.3.3^5}{2^{12}.3.3^5+4^6.3^5}\)
\(=\frac{1-3}{3+1}\)
\(=\frac{-1}{2}\)
P=212.35-46.36/212.93+84.35
P=212.35-(22)6.36/212.(33)3+(23)4.35
P=212.35-212.36/212.39+212.35
P=212.(35-36)/212.(39+35)
P=35-36/39+35