Найдите область определения функции :

Решите задачу:

$y= \frac{ \sqrt{x+12} }{ x^{2-1} }= \frac{ \sqrt{x+12} }{ x } \\\ \left \{ {{ x+12 \geq 0} \atop {x \neq 0}} \right. \\\ \left \{ {{ x \geq -12} \atop {x \neq 0}} \right. \\\ x\in(-12; 0)\cup(0; +\infty)$

$y= \frac{x- \sqrt{-x^{2}-7x+8 } }{1+ \sqrt{x+3} } \\\ \left \{ {{-x^{2}-7x+8 \geq 0} \atop {x+3 \geq 0}} \right. \\\ \left \{ {{x^2+7x-8 \leq 0} \atop {x \geq -3}} \right. \\\ \left \{ {{-8 \leq x \leq 1} \atop {x \geq -3}} \right. \\\ -3 \leq x \leq 1$