Ý bạn đề bài thế này đúng ko: \(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}\right)^2\cdot\left(-1\right)^{2019}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\).
Nếu vậy mình làm như thế này nhé!
Ta có: B = \(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}\right)^2\cdot\left(-1\right)^{2019}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)
= \(\frac{\frac{8}{27}\cdot\frac{9}{16}\cdot\left(-1\right)}{\frac{4}{25}\cdot\left(-\frac{125}{1728}\right)}\)
= \(\frac{\frac{-2^3\cdot3^2}{3^3\cdot2^4}}{-\frac{5}{432}}\)
= \(\frac{\frac{1}{3\cdot2}}{\frac{5}{2^4\cdot3^3}}\)
= \(\frac{1}{3\cdot2}\div\left(\frac{5}{2^4\cdot3^3}\right)\)
= \(\left(\frac{1}{2\cdot3}\cdot\frac{2^4\cdot3^3}{5}\right)\)
= \(\left(\frac{2^3\cdot3^2}{5}\right)\)
= \(\frac{72}{5}.\)
Vậy B = \(\frac{72}{5}\).
