Question

9z^2-16<(3z-1)(3z+1)-3z
3z-2/5-z-2/3 больше или равно 0

Answer 1

$9z^2 - 16 < (3z-1)(3z+1) - 3z\\9z^2 - 16 < 9z^2 - 3z - 1\\3z-15<0\\z-5<0\\z<5\\Answer: z \in (-\infty; 5)\\3z - \frac{2}{5} - z - \frac{2}{3} \geq 0\\ 2z - \frac{16}{15} \geq 0\\2z \geq \frac{16}{15}\\z \geq \frac{8}{15}\\Answer: z \in [\frac{8}{15}; +\infty)$