Question

найдите область определения функции ​

Answer 1

Ответ:

D(y)= [-4;-3) U (1;+∞)

Объяснение:

$\left \{ {{x+4\geq 0} \atop {x^{2} +2x-3>0}} \right. \\x^{2} +2x-3=0\\D=4+12=16\\x_{1}=\frac{-2+4}{2}=1\\x_{2}=\frac{-2-4}{2}=-3 \\\\(x-1)(x+3)=0\\\\\left \{ {{x\geq -4} \atop {(x-1)(x+3)>0}} \right.$

x≥-4

x∈(-∞;-3) U (1;+∞)

⇔ D(y)= [-4;-3) U (1;+∞)