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\(\left(-\frac{10}{3}\right)^5.\left(-\frac{6}{5}\right)^4=\frac{-\left(2.5\right)^5}{3^5}.\frac{\left(3.2\right)^4}{5^4}=\frac{-2^5.5^5.3^4.2^4}{3^5.5^4}=\frac{-2^9.5}{3}\)
Ta có: \(\left(-\frac{10}{3}\right)^5.\left(-\frac{6}{5}\right)^4=\frac{\left(-10\right)^5}{3^5}.\frac{\left(-6\right)^4}{5^4}=\frac{\left(-2.5\right)^5}{3^5}.\frac{\left(-2.3\right)^4}{5^4}=\frac{\left(-2\right)^5.5^5}{3^5}.\frac{\left(-2\right)^5.3^4}{5^4}=\frac{\left(-2\right)^5.5^5.\left(-2\right)^4.3^4}{3^5.5^4}\)\(=\frac{\left[\left(-2\right)^5.\left(-2\right)^4\right].5^5.3^4}{3^5.5^4}=\frac{\left(-2\right)^9.5}{3}=\frac{-512.5}{3}=-\frac{2560}{3}\)
Chuk bn hk tốt! 
Số số hạng là (2012-4):4+1=503(số)
=>\(A=503\cdot\dfrac{\left(2012+4\right)}{2}\cdot\dfrac{1}{1000}\cdot\dfrac{6+9+10}{12}\)
\(=503\cdot1008\cdot\dfrac{1}{1000}\cdot\dfrac{25}{12}=\dfrac{21}{10}\cdot503=\dfrac{10563}{10}\)
Đặt 4+6+8+10+...+2012 là A
Ta có: số số hạng A là:(2012-4)/2+1=1005
tổng A là:(2012+4).1005/2=1013040
=1013040.\(\frac{1}{1000}\) .(\(\frac{1}{2}+\frac{3}{4}+\frac{5}{6}\))
=1013,04.(\(\frac{6}{12}+\frac{9}{12}+\frac{10}{12}\))
=1013,04.\(\frac{25}{12}\)
=2110,5
\(\frac{5^4.20^4}{25^5.4^5}=\frac{5^4.5^4.4^4}{5^{10}.4^5}=\frac{5^8.4^4}{5^{10}.4^5}=\frac{1}{5^2.4}=\frac{1}{25.4}=\frac{1}{100}\)
\(\frac{5^4.20^4}{25^5.4^5}=\frac{5^4.4^4.5^4}{\left(5^2\right)^5.4^4.4}=\frac{5^4.4^4.5^4}{5^{10}.4^4.4}=\frac{5^4.4^4.5^4}{5^4.5^4.5^2.4^4.4}=\frac{1}{5^2.4}=\frac{1}{100}\)