Question

решить неравенство  х^2-16>0

Answer 1

$x^2-16>0\\\\(x-4)(x+4)>0\\\\x-4=0\; \; ,\; \; x_1=4\\\\x+4=0\; \; ,\; \; x_2=-4\\\\znaki:\; \; +++(-4)---(4)+++\\\\x\in (-\infty ;-4)\cup (4;+\infty )$