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Lời giải:
\(\frac{20^4.15^5}{12^5.25^4}=\frac{(2^2.5)^4.(3.5)^5}{(2^2.3)^5.(5^2)^4}=\frac{2^8.5^4.3^5.5^5}{2^{10}.3^5.5^8}\)
\(=\frac{2^8.5^9.3^5}{2^{10}.3^5.5^8}=\frac{2^8.5^8.5.3^5}{2^8.2^2.3^5.5^8}=\frac{5}{2^2}=\frac{5}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\dfrac{3}{4}.\dfrac{-8}{5}+\dfrac{1}{4}=-\dfrac{6}{5}+\dfrac{1}{4}=-\dfrac{19}{20}\)
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\(\left(\dfrac{1}{2020}-\dfrac{2018}{2019}\right):\dfrac{2}{9}-\left(\dfrac{-2019}{2020}+\dfrac{1}{2019}\right):\dfrac{2}{3}=\)
\(\left(\dfrac{1}{2020}-\dfrac{2018}{2019}\right)\cdot\dfrac{3}{2}-\left(\dfrac{-2019}{2020}+\dfrac{1}{2019}\right)\cdot\dfrac{3}{2}=\)
\(\left[\left(\dfrac{1}{2020}-\dfrac{2018}{2019}\right)-\left(\dfrac{-2019}{2020}+\dfrac{1}{2019}\right)\right]\cdot\dfrac{3}{2}=\)
\(\left[\dfrac{1}{2020}-\dfrac{2018}{2019}-\dfrac{2019}{2020}-\dfrac{1}{2019}\right]\cdot\dfrac{3}{2}=\)
\(0\)
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\(\left(\dfrac{4}{13}-\dfrac{13}{5}\right):\dfrac{4}{5}+\left(\dfrac{9}{13}+\dfrac{3}{5}\right):\dfrac{4}{5}\)
\(=\left(\dfrac{4}{13}-\dfrac{13}{5}\right)\cdot\dfrac{5}{4}+\left(\dfrac{9}{13}+\dfrac{3}{5}\right)\cdot\dfrac{5}{4}\)
\(=\left[\dfrac{4}{13}-\dfrac{13}{5}+\dfrac{9}{13}+\dfrac{3}{5}\right]\cdot\dfrac{5}{4}\)
\(=\dfrac{-5}{4}\)
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\(\left(\frac{-5}{3}+\frac{3}{5}\right):\left|\frac{-1}{2}-0,6\right|\)
= \(\left(\frac{-25}{15}+\frac{9}{15}\right):\left|\frac{-1}{2}-\frac{6}{10}\right|.\)
= \(\frac{-16}{15}:\left|\frac{-5}{10}-\frac{6}{10}\right|.\)
= \(\frac{-16}{15}:\left|\frac{-11}{10}\right|.\)
= \(\frac{16}{15}:\frac{11}{10}.\)
= \(\frac{16}{15}\cdot\frac{10}{11}.\)
= \(\frac{32}{33}.\)