Question

Решите пожалуйста. СРОЧНО.

Answer 1

Ответ:

$\displaystyle \tt x\in \boxed{\displaystyle \tt (} \: \boxed{\displaystyle \tt -b} \: ; \: \boxed{\displaystyle \tt -2} \: \boxed{\displaystyle \tt ]}$

Объяснение:

$\displaystyle \tt (x+3)^2-x^2\leq 3x+3\\\displaystyle \tt x^2+6x+9-x^2\leq 3x+3\\\displaystyle \tt 6x+9\leq 3x+3\\\displaystyle \tt 3x\leq -6\\\displaystyle \tt x\leq -6\div3\\\displaystyle \tt \bold{x\leq -2}$