Question

Решите неравенство: log(внизу2)(x^2 -2x)< 3
СРОЧНО.

Answer 1

$\log_2 (x^2-3x)\ \textless \ 3 \\ \log_2 (x^2-3x)\ \textless \ \log_2 8\\ x^2 - 3x \ \textless \ 8 \\ x^2 - 3x - 8 \ \textless \ 0\\ D = 9 + 32 =( \sqrt{41})^2 \\ x_1 = \frac{3+\sqrt{41}}{2} \\ x_2 = \frac{3-\sqrt{41}}{2} \\ x \in ( \frac{3-\sqrt{41}}{2}; \frac{3+\sqrt{41}}{2}) \\ x^2 - 3x \ \textgreater \ 0 \\ x \in (-\infty;0) \cup (3; + \infty) \\ OTVET : x \in ( \frac{3-\sqrt{41}}{2}; 0) \cup(3;\frac{3+\sqrt{41}}{2})$