Question

можете помочь, пожалуйста.?

Answer 1

$\left \{ {{x^2-x-6 \geq 0} \atop {x^2-4x-5\ \textless \ 0}} \right. \\x^2-x-6=0\\D=1+24=25\\x_1= \frac{6}{2} =3\\x_2=-2\\x^2-4x-5=0\\a-b+c=0\\x_1=-1\\x_2=5\\ \left \{ {{(x-3)(x+2) \geq 0} \atop {(x+1)(x-5)\ \textless \ 0}} \right. \left \{ {{x \leq -2,,,x \geq 3} \atop {-1\ \textless \ x\ \textless \ 5}} \right.$
x∈[3;5)