Question

Решите что-нибудь пожалуйста

Answer 1

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

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