Решите что-нибудь пожалуйста

Question 1

Question 2

Решите задачу:

$\displaystyle 1)..\\a).16^{- 3/4}=\frac{1}{16^{3/4}}=\frac{1}{\sqrt[4]{16^{3}}}= \frac{1}{\sqrt[4]{2^{12}}}=\frac{1}{2^{3}}=\frac{1}{8}=0,125;\\\\b).(\frac{1}{125})^{-2/3}=125^{2/3}=(5^{3})^{2/3}=5^{6/3}=5^{2}=25;\\\\2)..\\a).\sqrt[6]{x^{3}}*\sqrt{x}=x^{3/6}*x^{1/2}=x^{1/2+1/2}=x;\\\\b).\frac{x^{0,5} }{( \sqrt[4]{x})^{2}}=x^{1/2}*x^{-2/4}=x^{1/2-1/2}=x^{0}=1;\\\\3)..\\ a).\frac{3^{2/3}*81^{3/4}}{3^{-1/3}}=3^{2/3}*(3^{4})^{3/4}*3^{1/3}=3^{2/3+3+1/3}=3^{4}=81;$

$\displaystyle b).(10^{-1/3}*0,01^{1/3})^{-1}=(10^{-1/3}*(10^{-2})^{1/3})^{-1}=(10^{-1})^{-1}=10; \\ \\ 4).. \\ a).(16x)^{3/4}*( \frac{1}{8}x^{3/8})^{-2/3}=8x^{3/4}*4x^{-6/24}=32 \sqrt{x}; \\ \\ b).(a+b^{1/4})(a-b^{1/4})+ \sqrt{b}=a^{2}- \sqrt{b}+ \sqrt{b}=a^{2}; \\ \\ c).\frac{ab^{1/3}-a^{1/3}b}{(ab)^{1/3}}=ab^{1/3}*(ab)^{-1/3}-a^{1/3}b*(ab)^{-1/3}= \\ \\ =a^{2/3}-b^{2/3};$