Question

1) (x ^ 2 - 1)/4 > 5 + x​
2) (x - 4)(x - 5) <= 2
помогите пожалуйста

Answer 1

$\displaystyle\bf\\1)\\\\\frac{x^{2}-1 }{4} > 5+x\\\\\\\frac{x^{2}-1 }{4} \cdot 4 > (5+x)\cdot 4\\\\\\x^{2} -1 > 20+4x\\\\x^{2}-4x-21 > 0\\\\(x+3)\cdot(x-7) > 0\\\\\\+ + + + + (-3) - - - - - (7) + + + + + \\\\\\Otvet \ : \ x\in(-\infty \ ; \ -3)\cup(7 \ ; \ +\infty)$

$\displaystyle\bf\\2)\\\\(x-4)(x-5)\leq 2\\\\x^{2} -5x-4x+20-2\leq 0\\\\x^{2}-9x+18\leq 0\\\\(x-3)\cdot(x-6)\leq 0\\\\\\+ + + + + \Big[3\Big] - - - - - \Big[6\Big]+ + + + + \\\\\\Otvet \ : \ x\in\Big[3 \ ; \ 6\Big]$