Решите пожалуйста), или 2 или 3

Question 1

Question 2

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

$3)\ \frac{\sqrt[3]{12\sqrt[3]{24}+6\sqrt[3]{375}}}{\sqrt[3]{2\sqrt[3]3}}=\sqrt[3]{\frac{{12\sqrt[3]{24}+6\sqrt[3]{375}}}{{2\sqrt[3]3}}}=\sqrt[3]{\frac{{2\left(6\sqrt[3]{24}+3\sqrt[3]{375}}\right)}{{2\sqrt[3]3}}}=\\\\=\sqrt[3]{\frac{{6\sqrt[3]{24}+3\sqrt[3]{375}}}{{\sqrt[3]3}}}=\sqrt[3]{\frac{6\sqrt[3]{24}}{\sqrt[3]3}+\frac{3\sqrt[3]{375}}{\sqrt[3]3}}=\sqrt[3]{6\sqrt[3]{\frac{24}{3}}+3\sqrt[3]{\frac{375}{3}}}=$
<img src="https://tex.z-dn.net/?f=%3D%5Csqrt%5B3%5D%7B6%5Csqrt%5B3%5D%7B8%7D%2B3%5Csqrt%5B3%5D%7B125%7D%7D%3D%