Question

Здраствуйте.
Задание: вычислите.

Даю 100 баллов​

Answer 1

$1)\ \ -0.027^{-1/3}+\bigg(\dfrac16\bigg)^{-1}-3^{-1}+(5.5)^0=\dfrac1{\sqrt[3]{-0.027}}+6-\dfrac13+1=\\\\=\dfrac1{-0.3}+\dfrac{18}3-\dfrac13+\dfrac33=\dfrac{-10+18-1+3}{3}=\dfrac{10}{3}=3\dfrac13$

$2)\ \ \Bigg{(\bigg(\dfrac34\bigg)^0}\Bigg)^{-0.5}-7.5-\Big(\sqrt[4]{4^3}\Big)^2-2\cdot(-2)^4={\bigg(\dfrac34\bigg)}^0-7.5-\Big(2^{6/4}\Big)^2-2\cdot16=\\\\=1-7.5-2^{6/2}-32=-38.5-2^3=-38.5-8=-46.5$

$3)\ \ (0.008)^{2/3}\cdot(0.64)^{0.5}:(0.04)^{-0.5}:(0.25)^{-1.5}=\\\\=\Big((0.2)^3\Big)^{2/3}\cdot\Big((0.8)^2\Big)^{1/2}:\dfrac1{\Big((0.2)^2\Big)^{1/2}}:\dfrac1{\Big((0.5)^2\Big)^{3/2}}=(0.2)^2\cdot0.8\cdot0.2\cdot(0.5)^3=\\\\=(0.2)^2\cdot0.16\cdot(0.5)^3=0.04\cdot0.16\cdot0.125=0.0008$

$4)\ \ 0.125^{-1/3}-\bigg(-\dfrac16\bigg)^{-2}+256^{0.75}+(1.2)^0=\\\\=\dfrac1{\Big((0.5)^3\Big)^{1/3}}-\dfrac1{\bigg(-\dfrac16\bigg)^2}+\Big(4^4\Big)^{3/4}+1=\dfrac1{0.5}-\dfrac{\quad1\quad}{\dfrac1{36}}+4^3+1=$

$=2-36+64+1=31$