Đáp án:
e) $\dfrac{25}{81}$
d) $\dfrac{-72}{5}$
Giải thích các bước giải:
e) $(\dfrac{1}{3}-\dfrac{2}{9})^2+(\dfrac{1}{3})^3.\dfrac{8}{3}$
$=(\dfrac{1}{9})^2+\dfrac{1}{9}.\dfrac{8}{3}$
$=\dfrac{1}{9}.(\dfrac{1}{9}+\dfrac{8}{3})$
$=\dfrac{1}{9}.\dfrac{25}{9}$
$=\dfrac{25}{81}$
d) $\dfrac{(\dfrac{2}{3})^3.(\dfrac{-3}{4})^2.(-1)^{2020}}{(\dfrac{2}{5})^2.(\dfrac{-5}{12})^2}$
$=\dfrac{\dfrac{8}{27}.\dfrac{9}{16}.1}{\dfrac{4}{25}.\dfrac{-125}{1728}}$
$=\dfrac{\dfrac{1}{6}}{\dfrac{-5}{432}}$
$=\dfrac{1}{6}.\dfrac{-432}{5}$
$=\dfrac{-72}{5}$
