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VT
0
VT
0
14 tháng 11 2018
\(B=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(=\frac{2^{12}.3^5.\left(1-3\right)}{2^{12}.3^6.\left(3+1\right)}-\frac{5^9.7^3.\left(1-7\right)}{5^9.7^3\left(1+2^3.1\right)}\)
\(=\frac{-1}{2}-\frac{-2}{3}=\frac{-7}{6}\)
10 tháng 2 2022
\(A=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot8}\)
\(=\dfrac{3^4\cdot\left(3-1\right)}{3^5\left(3+1\right)}-\dfrac{5^{10}\cdot7^3\cdot\left(-6\right)}{5^9\cdot7^3\cdot9}=\dfrac{2}{3\cdot4}-\dfrac{5\cdot\left(-2\right)}{3}\)
\(=\dfrac{1}{6}+\dfrac{10}{3}=\dfrac{1}{6}+\dfrac{20}{6}=\dfrac{21}{6}=\dfrac{7}{2}\)
28 tháng 6 2018
\(B=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{16}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{16}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3.7\right)^3+5^9.\left(2.7\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{16}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3\left(5^6-7\right)}{5^9.7^3.\left(1+2^3\right)}\)
\(=\frac{2}{3.4}-\frac{5.\left(15625-7\right)}{1+8}\)
\(=\frac{1}{6}-\frac{\text{78090}}{9}\)
đề ghi vậy , ko ra kq nhé
TG
5 tháng 4 2020
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3\right)^3.7^3+5^9.2^3.7^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(=\frac{2^{12}.3^5.\left(3-1\right)}{2^{12}.3^6.\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}\)
\(=\frac{2^{12}.3^5.2}{2^{12}.3^6.4}-\frac{5^{10}.7^3.\left(-6\right)}{5^9.7^3.9}\)
\(=\frac{1}{3.2}-\frac{5.\left(-2\right)}{3}\)
\(=\frac{1}{6}-\frac{-10}{3}\)
\(=\frac{1}{6}+\frac{10}{3}\)
\(=\frac{1}{6}+\frac{20}{6}=\frac{21}{6}=\frac{7}{2}\)
5 tháng 4 2020
\(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}\) = \(\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}\) = \(\frac{2^{12}.\left(3^5-3^4\right)}{2^{12}.\left(3^6+3^5\right)}\) = \(\frac{3^5-3^4}{3^5+3^6}\) = \(\frac{243-81}{243+729}\) = \(\frac{1}{6}\)
\(\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\) = \(\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.7^9}\) = \(\frac{5^{10}.\left(7^3.7^4\right)}{5^9.\left(7^3+7^9\right)}\) = \(\frac{5.\left(343+2401\right)}{343+40353607}\) = \(\frac{13720}{40353950}\) = \(\frac{4}{11765}\)
A = \(\frac{1}{6}\) - \(\frac{4}{11765}\) = \(\frac{11741}{70590}\)
Bn tự rút gọn nha
DH
2 tháng 4 2020
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3\left(1-7\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{2^{12}.3^4.2}{2^{12}.3^5.4}-\frac{5^{10}.7^3\left(-6\right)}{5^9.7^3.9}\)
\(=\frac{1}{6}-\frac{\left(-10\right)}{3}\)
\(=\frac{7}{2}\)
LT
0
NP
1
1 tháng 6 2022
\(=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)
\(=\dfrac{5^{10}\cdot7^4\cdot\left(1-7\right)}{5^9\cdot7^3\cdot\left(1+2^3\right)}=\dfrac{5\cdot7\cdot\left(-6\right)}{9}=\dfrac{-2\cdot5\cdot7}{3}=\dfrac{-70}{3}\)
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