Решите уравнение X^3-3X^2-8X+24=0 и из фото тоже!)

$x^3-3x^2-8x+24=0\\ x^2(x-3)-8(x-3)=0\\ (x-3)(x^2-8)=0\\ x_1=3\\ x_{2,3}=\pm 2\sqrt{2}$

Пример 7.
$\displaystyle \frac{a^2-9b^2}{2a^2}\cdot \frac{a}{2a+6b}= \frac{(a-3b)(a+3b)}{2a^2}\cdot \frac{a}{2(a+3b)} = \frac{a-3b}{4a} \\ \\ \\ if\,\,\,a= \sqrt{2} ;\,\,\, b=\sqrt{50} \,\,\,\, then\,\,\,\,\, \frac{\sqrt{2} -3\sqrt{50} }{4\sqrt{2} } = \frac{\sqrt{2} (1-3\cdot 5)}{4\sqrt{2} } =-3.5$

Пример 8.
$7+4(-5x-7) \leq 4\\ \\ 7-20x-28 \leq 4\\ \\ -20x \leq 4+21\\ \\ -20x \leq 25\\ \\ x \geq -1.25$