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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
\(E=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}.\frac{5^4.20^4}{25^5.4^5}\)
\(\Rightarrow E=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}.\frac{5^4.4^4.5^4}{5^{10}.4^5}\)
\(\Rightarrow E=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}.\frac{5^8.4^4}{5^{10}.4^5}\)
\(\Rightarrow E=\frac{2^{10}\left(3^8-3^9\right)}{2^{10}\left(3^8+3^8.5\right)}.\frac{1}{5^2.4}\)
\(\Rightarrow E=\frac{3^8-3^9}{3^8\left(1+5\right)}.\frac{1}{100}\)
\(\Rightarrow E=\frac{3^8\left(1-3\right)}{3^8\left(1+5\right)}.\frac{1}{100}\)
\(\Rightarrow E=-\frac{2}{6}.\frac{1}{100}\)
\(\Rightarrow E=-\frac{1}{300}\)
a) Ta có: \(\left(-2\right)^3+\frac{1}{2}:\frac{1}{8}-\sqrt{25}+\left|-64\right|\)
\(=-8+\frac{1}{2}\cdot8-5+64\)
\(=-8+4-5+64=55\)
b) Ta có: \(\left(\frac{-3}{4}+\frac{2}{7}\right):\frac{2}{3}+\left(\frac{-1}{4}+\frac{5}{7}\right):\frac{2}{3}\)
\(=\left(\frac{-3}{4}+\frac{2}{7}\right)\cdot\frac{3}{2}+\left(\frac{-1}{4}+\frac{5}{7}\right)\cdot\frac{3}{2}\)
\(=\left(\frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{5}{7}\right)\cdot\frac{3}{2}\)
\(=0\cdot\frac{3}{2}=0\)
c) Ta có: \(\frac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\frac{2^{10}\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot20}=\frac{2\left(2^9\cdot9^4-6^9\right)}{6^8\left(2^2+20\right)}=\frac{-1}{3}\)
a) ( -2 )3 + \(\frac{1}{2}:\frac{1}{8}\) - √25 + \(|-64|\)
= \(\frac{-8}{1}\) + \(\frac{1}{2}.\frac{8}{1}\) - \(\frac{5}{1}\) + \(\frac{64}{1}\)
= \(\frac{-16}{2}+\frac{1}{2}.\frac{8}{1}-\frac{10}{2}+\frac{128}{2}\)
= \(\frac{-16}{2}+\frac{8}{2}-\frac{10}{2}+\frac{128}{2}\)
= \(\frac{-16+8-10+128}{2}\) = \(\frac{110}{2}\) = 55
a) \(\frac{5^4.20^4}{25^5.4^5}=\frac{5^4.\left(5.2^2\right)^4}{\left(5^2\right)^5.\left(2^2\right)^5}=\frac{5^4.5^4.2^8}{5^{10}.2^{10}}=\frac{5^8.2^8}{5^{10}.2^{10}}=\frac{1}{5^2.2^2}=\frac{1}{25.4}=\frac{1}{100}.\)
Chúc bạn học tốt!
Ta có: \(\frac{\frac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}}{\sqrt{\frac{25}{9}}}=\frac{\frac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+3^8\cdot2^8\cdot20}}{\frac{5}{3}}=\frac{2^{10}\left(3^8-3^9\right)}{2^83^8\left(2^2+20\right)}\cdot\frac{3}{5}=\frac{2^83^8\left(1-3\right)\cdot2^2}{2^83^8\cdot24}\cdot\frac{3}{5}\)
\(=\frac{-2\cdot2^2}{24}\cdot\frac{3}{5}=\frac{-8}{24}\cdot\frac{3}{5}=\frac{-24}{120}=\frac{-1}{5}\)
\(\frac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}:\sqrt{\frac{25}{9}}\\ =\frac{2^{10}\cdot3^8-2\cdot\left(2\cdot3\right)^9}{2^{10}\cdot3^8+\left(2\cdot3\right)^8\cdot\left(2^2\cdot5\right)}:\frac{5}{3}\\ =\frac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\cdot\frac{3}{5}\\ =\frac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\cdot\frac{3}{5}\\ =\frac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}\cdot\frac{3}{5}\\ =\frac{1-3}{1+5}\cdot\frac{3}{5}\\ =\frac{-2}{6}\cdot\frac{3}{5}=\frac{-1}{5}\)
1: \(=5^{20}\cdot\left(\dfrac{1}{5}\right)^{20}+\left(\dfrac{-3}{4}\cdot\dfrac{-4}{3}\right)^8-1\)
=1+1-1=1
2: \(=\dfrac{15-8}{6}\cdot\dfrac{6}{7}+\left(-\dfrac{3}{2}\right)^2\)
=1+9/4
=13/4
3: \(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{3^8\cdot2^{10}+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{3^8\cdot2^{10}\cdot6}=\dfrac{-2}{6}=\dfrac{-1}{3}\)
a) \(A=\frac{2^4.25^4}{10^5.5^5}=\frac{2^4.\left(5^2\right)^4}{\left(2.5\right)^5.5^5}=\frac{2^4.5^8}{2^5.5^{10}}=\frac{1}{2.5^2}=\frac{1}{50}\)
b) \(B=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
\(B=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}=\frac{-2}{6}=\frac{-1}{3}\)
\(A=\frac{2^4.25^4}{10^5.5^5}\)
\(A=\frac{2^4.\left(5^2\right)^4}{\left(2.5\right)^5.5^5}\)
\(A=\frac{2^4.5^8}{2^5.5^5.5^5}\)
\(A=\frac{5^8}{2.5^{10}}\)
\(A=\frac{1}{2.5^2}=\frac{1}{2.25}=\frac{1}{50}\)
\(B=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
\(B=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)
\(B=\frac{2^{10}.3^8-2^{10}.2.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(B=\frac{2^{10}.\left(3^8-2.3^9\right)}{2^{10}.\left(3^8+3^8.5\right)}\)
\(\Rightarrow B=\frac{3^8.2.3^9}{3^8+3^8.5}\) ( \(2^{10}\ne0\))
\(B=\frac{3^8.\left(1-2.3\right)}{3^8.\left(1+5\right)}\)
\(B=\frac{1-6}{1+5}\left(3^8\ne0\right)\)
\(B=\frac{-5}{6}\)
Tham khảo nhé~
\(A=\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=6\)
\(B=\frac{\left(3\cdot4\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot4^{11}-16^9}=2\)
\(C=\frac{4^5\cdot9^{4-2\cdot6^9}}{2^{10}\cdot3^8+6^8\cdot20}=0\)
A=\(\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=6\)