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3 tháng 3 2023
\(\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\\ =\dfrac{2^{12}\cdot3^5-\left(2^2\right)^6\cdot\left(3^2\right)^2}{2^{12}\cdot3^6}-\dfrac{5^{10}\cdot7^3-\left(5^2\right)^5\cdot\left(7^2\right)^2}{\left(5^3\cdot7\right)^3+5^9\cdot\left(2\cdot7\right)^3}\\ =\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot2^3\cdot7^3}\\ =\dfrac{2^{12}\cdot3^4\left(3-1\right)}{2^{12}\cdot3^6}-\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}\\ =\dfrac{2}{9}-\dfrac{-6}{1+8}=\dfrac{2}{9}+\dfrac{6}{9}=\dfrac{8}{9}\)
VT
3 tháng 3 2023
\(\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\\ =\dfrac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\\ =\dfrac{2^{12}.3^5.\left(1-3\right)}{2^{12}.3^6}-\dfrac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}\\ =\dfrac{2^{12}.3^5.\left(-2\right)}{2^{12}.3^6}-\dfrac{5^{10}.7^3.\left(-6\right)}{5^9.7^3.9}\\ =\dfrac{-2}{3}-\dfrac{5.\left(-6\right)}{9}\\ =\dfrac{-2}{3}+\dfrac{30}{9}\\ =\dfrac{8}{3}\)
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= \(\frac{1}{2}\)
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
S
12 tháng 1 2018
không vt lại đề
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^6.7^3+5^9.2^3.7^3}\)
\(=\frac{2^{12}\cdot3^4.\left(3-1\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3\left(1-7\right)}{5^6.7^3\left(1+2^3\right)}\)
\(=\frac{1}{6}-\frac{-10}{3}=\frac{1}{6}+\frac{20}{6}=\frac{21}{6}\)
24 tháng 8 2015
a=2^12.3^5-2^12.3^4/2^12.3^6+2^12.3^5 - 5^10.7^3-5^10.7^4/5^9.7^3+5^9.7^3.2^3
a=2^12.3^4.(3-1)/2^12.3^5.(3+1)-5^10.7^3.(1-7)/5^9.7^3.(1+8)
a=2/12-30/9
a=1/6-10/3=-19/6
a=
Ta có : A = \(\frac{\left(2^2.3\right)^6+8^4.3^5}{2^{12}.3^5-4^6.9^2}\)
= \(\frac{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}\)
= \(\frac{2^{12}.3^6+2^{12}.3^5}{2^{12}.3^5-2^{12}.3^4}\)
= \(\frac{2^{12}.\left(3^6+3^5\right)}{2^{12}.\left(3^5-3^4\right)}\)
= \(\frac{3^6+3^5}{3^5-3^4}\)
= \(6\)
\(A=\frac{\left(2^2.3\right)^6+8^4.3^5}{2^{12}.3^5-4^6.9^2}\)
\(A=\frac{2^{12}.3^6+\left(2^3\right)^4.3^5}{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}\)
\(A=\frac{2^{12}.3^6+2^{12}.3^5}{2^{12}.3^5-2^{12}.3^4}\)
\(A=\frac{2^{12}.\left(3^6+3^5\right)}{2^{12}.\left(3^5-3^4\right)}\)
\(\Rightarrow A=\frac{3^5.\left(3+1\right)}{3^4.\left(3-1\right)}=\frac{3^5.4}{3^4.2}=\frac{3.3^4.2.2}{3^4.2}=\frac{3.2}{1}=6\)