Question

nhanh giúp mình nhé mai mình đi học rùi

\(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{3}{4}\right)^2.\left(-1\right)}{\left(\dfrac{2}{3}\right)^2.\left(\dfrac{5}{12}\right)^3}\)

Answer 1

\(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{3}{4}\right)^2.\left(-1\right)}{\left(\dfrac{2}{3}\right)^2.\left(\dfrac{5}{12}\right)^3}\)

=\(\dfrac{\left(\dfrac{2}{3}\right)^2.\dfrac{2}{3}.\left(\dfrac{3}{4}\right)^2.\left(-1\right)}{\left(\dfrac{2}{3}\right)^2.\left(\dfrac{5}{12}\right)^3}\)

=\(\dfrac{\dfrac{2}{3}.\dfrac{9}{16}.\left(-1\right)}{\dfrac{125}{1728}}\)

=\(\dfrac{\dfrac{2}{3}.-\dfrac{9}{16}}{\dfrac{125}{1728}}\)

=\(-\dfrac{\dfrac{3}{8}}{\dfrac{125}{1728}}\)

=\(-\dfrac{648}{125}\)

Answer 2

mình sẽ tích nha