Question

Помогите решить неравенство!
$6 ^{x} \sqrt{15-x^{2}-2x } \geq 36 \sqrt{15-x^{2}-2x }$

Answer 1

$\displaystyle 6^{x} \sqrt{15-x^{2}-2x} \geq 36\sqrt{15-x^{2}-2x} \\ \\ odz: 15-x^{2}-2x \geq 0 \\ x^{2}+2x-15 \leq 0 \\ D=b^{2}-4ac=4+60=64 \\ \\ x_{1,2}= \frac{-bб \sqrt{D}}{2a} \\ \\ x_{1}=-5 \\ x_{2}=3 \\ \\ (x-3)(x+5) \leq 0 \\ \\ x \leq 3 \\ x \geq -5 \\ \\odz: [-5; 3]\\ \\ 6^{x} \geq 6^{2} \\ x \geq 2$

Ответ: [2; 3]