Question

Решите систему неравенств
1)$\left \{ {{x^2-x+6\ \textgreater \ 0} \atop {x^2\leq25 }} \right.$
2)$\frac{x^2(1-x)}{x^2-6x-9} \leq 0$

Answer 1

$1)\; \; \left \{ {{x^2-x+6>0} \atop {x^2\leq 25}} \right.\; \left \{ {{x\in (-\infty ,+\infty )} \atop {(x-5)(x+5)\leq 0}} \right. \; \left \{ {{x\in (-\infty ,+\infty )} \atop {x\in [-5.5\, ]}} \right. \; \; \Rightarrow \; \; x\in [-5,5\; ]\\\\2)\; \; \frac{x^2(1-x)}{x^2-6x-9}\leq 0\\\\x^2-6x-9=0\; ,\; D/4=9+9=18\; ,\; \; x_{1,2}=3\pm 3\sqrt2\\\\\frac{x^2(x-1)}{(x-3-3\sqrt2)(x+3-3\sqrt2)} \geq 0\\\\znaki:\; \; \; ---(3-3\sqrt2)+++[\, 0\; ]+++[\; 1\; ]---(3+3\sqrt2)+++\\\\x\in (3-3\sqrt2\, ;\, 1\; ]\cup (3+3\sqrt2\, ;\, +\infty \, )$