Cho E=2^15.0,5^5+3.2^10/2^11.2^3-2^15:2^3 Cho F=2.6^7+6^8/2^5.9^4-2.3^7+3^8 Cho G=4^5.9^4-2.6^9/2^10.3^8+6^8.20
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a: \(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=\dfrac{-1}{3}\)
b: \(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)
\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)
\(=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)
\(=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}\)
\(=\frac{-2}{6}=\frac{-1}{3}\)
\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1-5\right)}=\dfrac{-2}{-4}=\dfrac{1}{2}\)
\(A=\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+3^8\cdot2^{10}\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\)
\(=-\dfrac{2}{6}=-\dfrac{1}{3}\)
\(E=\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)
\(=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
\(A=\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
\(=\dfrac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)
\(=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}\)
\(=-\dfrac{2}{6}=-\dfrac{1}{3}\)
Ta có
\(E=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\cdot\frac{5^4.20^4}{25^5.4^5}\)
\(=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\cdot\frac{2^8.5^8}{5^{10}.2^{10}}\)
\(=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}\cdot\frac{1}{5^2.2^2}\)
\(=\frac{\left(-2\right)}{6}\cdot\frac{1}{100}=-\frac{1}{3}\cdot\frac{1}{100}=-\frac{1}{300}\)
Vậy : \(E=-\frac{1}{300}\)
Bài làm
\(E=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}.\frac{5^4.20^4}{25^5.4^5}\)
\(\Rightarrow E=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}.\frac{5^4.4^4.5^4}{5^{10}.4^5}\)
\(\Rightarrow E=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}.\frac{5^8.4^4}{5^{10}.4^5}\)
\(\Rightarrow E=\frac{2^{10}\left(3^8-3^9\right)}{2^{10}\left(3^8+3^8.5\right)}.\frac{1}{5^2.4}\)
\(\Rightarrow E=\frac{3^8-3^9}{3^8\left(1+5\right)}.\frac{1}{100}\)
\(\Rightarrow E=\frac{3^8\left(1-3\right)}{3^8\left(1+5\right)}.\frac{1}{100}\)
\(\Rightarrow E=-\frac{2}{6}.\frac{1}{100}\)
\(\Rightarrow E=-\frac{1}{300}\)
Đề bài yêu cầu gì?