Степень с рациональным показателем! Помогите! 10 класс

Question 1

Question 2

Решите задачу:

$A1.\;a)\;28\cdot32^{\frac25}=28\cdot(32^{\frac15})^2=28\cdot(\sqrt[5]{32})^2=28\cdot2^2=28\cdot4=112\\b)\;\frac{3^\frac12\cdot2^\frac12}{\sqrt[4]{36}}=\frac{\sqrt3\cdot\sqrt2}{\sqrt6}=\frac{\sqrt{3\cdot2}}{\sqrt6}=\frac{\sqrt6}{\sqrt6}1\\c)\;\frac{\left(0,216^{\frac49}\right)^{\frac32}}{0,09^{\frac34}\cdot0,027^{\frac16}}=\frac{0,216^{\frac49\cdot\frac32}}{(0,3^2)^{\frac34}\cdot(0,3^3)^{\frac16}}=\frac{0,216^{\frac23}}{0,3^{\frac32}\cdot0,3^{\frac12}}=\frac{(\sqrt[3]{0,216})^2}{0,3^{\frac32+\frac12}}=$
$=\frac{0,6^2}{0,3^2}=\left(\frac{0,6}{0,3}\right)^2=2^2=4\\A2.\;a)\;y^{1,7}\cdot y^{2,8}\cdot y^{-1,5}=y^{1,7+2,8-1,5}=y^3\\b)\;(a^{-0,75})^{\frac43}=(a^{-\frac34})^\frac43=a^{\left(-\frac34\right)\cdot\frac43}=a^{-1}\\B1.\;625^{-\frac32}\cdot5^{-3}\cdot25+7\cdot(4^0)^4-25^{-3\frac12}+\left(\frac18\right)^{\frac13}=\\=(5^4)^{-\frac32}\cdot5^{-3}\cdot5^2+7\cdot1^4-25^{-\frac72}+(2^{-3})^{\frac13}=\\=5^{-6}\cdot5^{-3}\cdot5^2+7-(\sqrt{25})^{-7}+2^{-1}=5^{-7}+7-5^{-7}+\frac12=7\frac12$
$B2.\;\frac{a-1}{a^{\frac34}+a^\frac12}}\cdot\frac{a^\frac12+a^\frac14}{a^\frac12+1}\cdot a^\frac14=\frac{(a^\frac12-1)(a^\frac12+1)}{a^\frac34+a^\frac12}\cdot\frac{a^\frac14(a^\frac12+a^\frac14)}{a^\frac12+1}=\\=\frac{a^\frac12-1}{a^\frac34+a^\frac12}\cdot(a^\frac34+a^\frac12)=a^\frac12-1=\sqrt a-1\\a=25\\\sqrt{25}-1=5-1=4$