a) $\left\{ \begin{array}{l} (1-x)^{2}>5+3x+x^{2}\\ (x+2)^{2}<x^{3}+6x^{2}-7x-5\end{array} \right.$ ⇔{1−2x+x2−5−3x−x2>0x3+6x2−7x−5−x2−4x−4>0 ⇔{−4−5x>0x3+5x2−11x−9>0 ⇔$\left\{ \begin{array}{l} x<\frac{-4}{5}\\ \left[ {} \right.\begin{matrix} -6,483<x<-0,65\\2,13<x \end{matrix} \end{array} \right.$ ⇒ -6,483<x<−45
a)
\displaystyle{\left\{ \begin{array}{l} (1-x)^{2}>5+3x+x^{2}\\ (x+2)^{2}}\Leftrightarrow
{1−2x+x2−5−3x−x2>0x3+6x2−7x−5−x2−4x−4>0$$⇔\left\{
−4−5x>0x3+5x2−11x−9>0 \right.
\Leftrightarrow$$\left\{ \begin{array}{l} x<\frac{-4}{5}\\ \left[ {} \right.\begin{matrix} -6,483 $\Rightarrow$ -6,483