0\; ,\; x\ne -2;x\ne 1\\\\x^2 \geq 0\; ,\; \; x^2+2x+4\ \textgreater \ 0\; \; pri\; \; x\in (-\infty ,+\infty )\\\\+++(-2)---(-\sqrt3)+++(-1)---(1)+++(\sqrt3)---(\, 2\, )+++\\\\x\in (-\infty ,-2)\cup (-\sqrt3,-1)\cup (1,\sqrt3)\cup (2,+\infty )" alt=" \frac{(x^4-3x^2)(x^4+x^3-8x-8)}{(x+2)(1-x)} \ \textless \ 0\\\\ \frac{x^2(x^2-3)\cdot ((x^3(x+1)-8(x+1))}{-(x+2)(x-1)} \ \textless \ 0\\\\ \frac{x^2(x-\sqrt3)(x+\sqrt3)(x+1)(x^3-8)}{(x+2)(x-1)} \ \textgreater \ 0\\\\ \frac{x^2(x-\sqrt3)(x+\sqrt3)(x+1)(x-2)(x^2+2x+4)}{(x+2)(x-1)}>0\; ,\; x\ne -2;x\ne 1\\\\x^2 \geq 0\; ,\; \; x^2+2x+4\ \textgreater \ 0\; \; pri\; \; x\in (-\infty ,+\infty )\\\\+++(-2)---(-\sqrt3)+++(-1)---(1)+++(\sqrt3)---(\, 2\, )+++\\\\x\in (-\infty ,-2)\cup (-\sqrt3,-1)\cup (1,\sqrt3)\cup (2,+\infty )" align="absmiddle" class="latex-formula">