Решите 1 номер пожалуйста

$1)\ f(x)=\sqrt[4]{3-2x-x^2},$

ООФ: $3-2x-x^2\ge0,$

$-x^2-2x+3=0,\\D=(-2)^2-4\bullet(-1)\bullet3=4+12=16,\\x_{1,2}=\frac{-(-2)\pm\sqrt{16}}{-2}=\frac{2\pm4}{-2}=-1\mp2,\\x_1=-1-2=-3,\\x_2=-1+2=1,\\-x^2-2x+3=-(x+3)(x-1)\ge0,\\a=-1\ \textless \ 0,\to\ -3\le x\le1.\\\\OTBET:\ -3\le x\le 1.$

$2)\ f(x)=\sqrt{3-2x}-\sqrt[6]{9-x^2},$

ООФ: $\left \{ {{3-2x\ge0} \atop {9-x^2\ge0}} \right,$

$a)\ 3-2x\ge0,\\2x\le3,\\x\le\frac{3}{2},\\x\le1.5;\\\\b)\ 9-x^2\ge0,\\(3-x)(3+x)\ge0,\\a=-1\to\ -3\le x\le3.\\\\ \left \{ {{x\le1,5,} \atop -3\le x\le3} \right. \to \ -3\le x\le1,5.\\\\OTBET:\ -3\le x\le1,5.$