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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!
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2 , 5 - x = 1 , 3
x = 2 , 5 - 1 , 3
x = 1 , 2
Chúc bạn chăm ngoan học giỏi !
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\(1\cdot2+2\cdot3+3\cdot4+...+n\left(n+1\right)\\ =\dfrac{1}{3}\left[1\cdot2\cdot3+2\cdot3\cdot3+...+3n\left(n+1\right)\right]\\ =\dfrac{1}{3}\left[1\cdot2\left(3-0\right)+2\cdot3\left(4-1\right)+...+n\left(n+1\right)\left(n+2-n+1\right)\right]\\ =\dfrac{1}{3}\left[1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4-...-\left(n-1\right)n\left(n+1\right)+n\left(n+1\right)\left(n+2\right)\right]\\ =\dfrac{n\left(n+1\right)\left(n+2\right)}{3}\)
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\(\frac{5^4.20^4}{25^5.4^5}=\frac{5^4.5^4.4^4}{\left(5^2\right)^5.4^5}=\frac{5^8.4^4}{5^{10}.4^5}=\frac{1}{25.4}=\frac{1}{100}\)
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\(\frac{5^4.20^4}{25^5.4^5}=\frac{5^4.5^4.4^4}{5^{10}.4^5}=\frac{5^8.4^4}{5^{10}.4^5}=\frac{1}{5^2.4}=\frac{1}{25.4}=\frac{1}{100}\)
\(\frac{5^4.20^4}{25^5.4^5}=\frac{5^4.4^4.5^4}{\left(5^2\right)^5.4^4.4}=\frac{5^4.4^4.5^4}{5^{10}.4^4.4}=\frac{5^4.4^4.5^4}{5^4.5^4.5^2.4^4.4}=\frac{1}{5^2.4}=\frac{1}{100}\)