Помогите ришить 2 вариант

Question 1

Question 2

Решите задачу:

$1) \sqrt[3]{0,3}* \sqrt[3]{-0,09} = \sqrt[3]{0,3*(-0,09)} = \sqrt[3]{-0,027} =-0,3$
$2) \sqrt[4]{54*24}= \sqrt[4]{(2*3*3*3)*(2*2*2*3)}= \sqrt[4]{2^4*3^4}=2*3=6$
$3) \sqrt[3]{-20*25*128}=- \sqrt[3]{(2^2*5)*(5*5)*2^7} =- \sqrt[3]{2^9*5^3} =$ $-2^3*5=-40$
$4) \frac{ \sqrt[5]{160} *2^{-0,25}}{5^{ \frac{1}{5} } \sqrt[4]{8} } = \frac{\sqrt[5]{32*5}*2^{- \frac{1}{4} } }{ \sqrt[5]{5}* \sqrt[4]{2^3} } = \frac{\sqrt[5]{2^5}* \sqrt[5]{5} }{ \sqrt[5]{5}* \sqrt[4]{2^3} *2^{ \frac{1}{4} }} = \frac{2}{ \sqrt[4]{2^3} \sqrt[4]{2}}= \frac{2}{ \sqrt[4]{2^3*2} } = \frac{2}{ \sqrt[4]{2^4} } = \frac{2}{2}$ $=2$
$5) \sqrt[3]{27^2} *4 \frac{17}{27}*36^{- \frac{1}{2} } = \sqrt[3]{(3^3)^2} * \frac{71}{27} * \frac{1}{36^{ \frac{1}{2} }} = \sqrt[3]{3^6} * \frac{71}{27}* \frac{1}{ \sqrt{36} } = \frac{3^2*71*1}{27*6}=$ $\frac{71}{18}=3 \frac{17}{18}$
$6)8^{ \frac{1}{3} }:2^{-1}+3^{-2}*324^{ \frac{1}{2} }= \sqrt[3]{8} *2^1+ \frac{ \sqrt{324} }{3^2} = \sqrt[3]{2^3} *2+ \frac{ \sqrt{4*81}} {9} =$ $2*2+ \frac{2*9}{9} =4+2=6$
$7) \sqrt[5]{17- \sqrt{46} } *\sqrt[5]{17+ \sqrt{46} } =\sqrt[5]{(17- \sqrt{46} )(17+ \sqrt{46} )} =$ $\sqrt[5]{17^2-( \sqrt{46})^2 } = \sqrt[5]{289-46} = \sqrt[5]{243} = \sqrt[5]{3^5} =3$
$8) \frac{27^{-3}*256^4*81^2}{16^7*9^0*45^{-1}} = \frac{(2^8)^4*(3^4)^2*45}{27^3*16^7*1} = \frac{2^{32}*3^8}{(3^3)^3*(2^4)^7} = \frac{2^{32}*3^8}{3^9*2^{28}} = \frac{2^{32-28}}{3^{9-8}} = \frac{2^4}{3^1} = \frac{16}{3}$ $=5 \frac{1}{3}$