Question

X^2-16>0 решите неравенство с графиком

Answer 1

................................

Answer 2

$x^2-16 textgreater 0$
$x^2 textgreater 16$
$x_{1} textgreater 4$
$x_{2} textless -4$
\\\\\\\\\(-4)____(4)\\\\\\\\\
ответ: х∈(-∞;-4)∪(4;+∞)