Question

Log2(x-5)+log2(x+2)> 3 , решить

Answer 1

ОДЗ
$\left \{ {{x-5\ \textgreater \ 0} \atop {x+2\ \textgreater \ 0}} \right. \\ \\ \left \{ {{x\ \textgreater \ 5} \atop {x\ \textgreater \ -2}} \right. \\ \\ x\ \textgreater \ 5$
$log_2(x-5)+log_2(x+2)\ \textgreater \ 3 \\ \\ log_2(x-5)(x+2)\ \textgreater \ 3 \\ \\ (x+2)(x-5)\ \textgreater \ 2^3 \\ \\ x^{2} -5x+2x-10\ \textgreater \ 8 \\ \\ x^{2} -3x-18\ \textgreater \ 0 \\ \\ D=3^2+4*18= 81 \\ \\ x_1= \frac{3+9}{2}=6 \\ \\ x_2=\frac{3-9}{2}=-3 \\ \\ x \in (-\infty; -3) U (6; + \infty)$

Объединим с ОДЗ
х ∈ (6; +∞)

ответ х ∈ (6; +∞)