Question

Решите пожалуйста), или 2 или 3

Answer 1

$2) left(frac{1-x^{frac{3}{2}}}{1-x^{frac{1}{2}}}+x^{frac{1}{2}}right)left(frac{1+x^{frac{3}{2}}}{1+x^{frac{1}{2}}}-x^{frac{1}{2}}right)= \\=left(frac{1-x^{frac{3}{2}}}{1-x^{frac{1}{2}}}right)left(frac{1+x^{frac{3}{2}}}{1+x^{frac{1}{2}}}right)-x^{frac{1}{2}}cdotleft(frac{1-x^{frac{3}{2}}}{1-x^{frac{1}{2}}}right)+x^{frac{1}{2}}cdotleft(frac{1+x^{frac{3}{2}}}{1+x^{frac{1}{2}}}right)-x^{frac{1}{2}}x^frac{1}{2}=$
$frac{left(1-x^{frac{3}{2}}right)left(1+{x^{frac{3}{2}}}right)}{left(1-x^{frac{1}{2}}right)left(1+x^{frac{1}{2}}right)}-frac{x^{frac{1}{2}}left(1-x^{frac{3}{2}}right)}{1-x^{frac{1}{2}}}+frac{x^{frac{1}{2}}left(1+x^{frac{3}{2}}right)}{1+x^{frac{1}{2}}}-x^{frac{1}{2}+frac{1}{2}}=$
$frac{1^2-left(x^{frac{3}{2}right)^2}}{1^2-left(x^{frac{1}{2}right)^2}}-frac{x^{frac{1}{2}}-x^{frac{1}{2}}x^{frac{3}{2}}}{1-x^{frac{1}{2}}}+frac{x^{frac{1}{2}}+x^{frac{1}{2}}x^{frac{3}{2}}}{1+x^{frac{1}{2}}}-x^1=\\=frac{1-x^{frac{3}{2}cdot2}}{1-x^{frac{1}{2}cdot2}}-frac{x^{frac{1}{2}}-x^{frac{1}{2}+frac{3}{2}}}{1-x^{frac{1}{2}}}+frac{x^{frac{1}{2}}+x^{frac{1}{2}+frac{3}{2}}}{1+x^{frac{1}{2}}}-x^1=$
$=frac{1-x^{3}}{1-x^1}-frac{x^{frac{1}{2}}-x^{frac{4}{2}}}{1-x^{frac{1}{2}}}+frac{x^{frac{1}{2}}+x^{frac{4}{2}}}{1+x^{frac{1}{2}}}-x=\\=frac{1-x^{3}}{1-x}-frac{x^{frac{1}{2}}-x^2}{1-x^{frac{1}{2}}}+frac{x^{frac{1}{2}}+x^2}{1+x^{frac{1}{2}}}-x=$
$=frac{1-x^{3}}{1-x}-frac{left(x^{frac{1}{2}}-x^2right)left(1+x^{frac{1}{2}}right)-left(x^{frac{1}{2}}+x^2right)left(1-x^{frac{1}{2}}right)}{left(1-x^{frac{1}{2}}right)left(1+x^{frac{1}{2}}right)}-x=\\=frac{1-x^{3}}{1-x}-frac{left(x^{frac{1}{2}}+x^{frac{1}{2}}x^{frac{1}{2}}-x^2-x^2x^{frac{1}{2}}right)-left(x^{frac{1}{2}}-x^{frac{1}{2}}x^{frac{1}{2}}+x^2-x^2x^{frac{1}{2}}right)}{left(1-x^{frac{1}{2}}right)left(1+x^{frac{1}{2}}right)}-x=$
$=frac{1-x^{3}}{1-x}-frac{x^{frac{1}{2}}+x^{frac{1}{2}}x^{frac{1}{2}}-x^2-x^2x^{frac{1}{2}}-x^{frac{1}{2}}+x^{frac{1}{2}}x^{frac{1}{2}}-x^2+x^2x^{frac{1}{2}}}{1^2-left(x^{frac{1}{2}}right)^2}-x=\\=frac{1-x^{3}}{1-x}-frac{2x^{frac{1}{2}+frac{1}{2}}-2x^2}{1-x}-x=frac{1-x^{3}}{1-x}-frac{2x-2x^2}{1-x}-x=\\=frac{1-x^{3}-2x+2x^2}{1-x}-x;$
$x=3,\\frac{1-3^{3}-2cdot3+2cdot3^2}{1-3}-3=frac{1-27-6+18}{-2}-3=frac{-14}{-2}-3=7-3=4.$


$3) frac{sqrt[3]{12sqrt[3]{24}+6sqrt[3]{375}}}{sqrt[3]{2sqrt[3]3}}=sqrt[3]{frac{{12sqrt[3]{24}+6sqrt[3]{375}}}{{2sqrt[3]3}}}=sqrt[3]{frac{{2left(6sqrt[3]{24}+3sqrt[3]{375}}right)}{{2sqrt[3]3}}}=\\=sqrt[3]{frac{{6sqrt[3]{24}+3sqrt[3]{375}}}{{sqrt[3]3}}}=sqrt[3]{frac{6sqrt[3]{24}}{sqrt[3]3}+frac{3sqrt[3]{375}}{sqrt[3]3}}=sqrt[3]{6sqrt[3]{frac{24}{3}}+3sqrt[3]{frac{375}{3}}}=$
$=sqrt[3]{6sqrt[3]{8}+3sqrt[3]{125}}=sqrt[3]{6sqrt[3]{2^3}+3sqrt[3]{5^3}}=sqrt[3]{6cdot2+3cdot5}=\\=sqrt[3]{12+15}=sqrt[3]{27}=sqrt[3]{3^3}=3.$