Question

Помогите решить -17/(х+3)^2-7>=0

Answer 1

$- \frac{17}{(x+3)^2-7} \ \geq \\ \\ \frac{17}{(x+3)^2-7} \leq 0 \\ \\ (x+3)^2-7\ \textless \ 0 \\ \\ (x+3- \sqrt{7} )(x+3+ \sqrt{7} )\ \textless \ 0 \\ \\ +++(-3- \sqrt{7} )----(-3+ \sqrt{7} )++++ \\ \\ otvet:x\in(-3- \sqrt{7};-3+ \sqrt{7})$