Помогите пожалуйста...!.!

$\sqrt{8-7x- x^{2} } - \frac{3}{x+5}$

$\left \{ {{8-7x- x^{2} \geq 0} \atop {x+5 \neq 0}} \right.$

$\left \{ {{ x^{2}+7x-8 \leq 0} \atop {x+5 \neq 0}} \right.$

$\left \{ {{(x-1)(x+8) \leq 0} \atop {x \neq -5}} \right.$

+ - +
-------------------[-8]------------------[1]-----------------
///////////////////////
----------------------------(-5)-----------------------------

$x$ ∈ $[-8;-5)$ ∪ $(-5;1]$

P.S.

$x^{2}+7x-8=0$

$D=7^2-4*1*(-8)=49+32=81$

$x_1= \frac{-7+9}{2} =1$

$x_2= \frac{-7-9}{2} =-8$

$x^{2} +7x-8=(x-1)(x+8)$

Ответ: $[-8;-5)$ ∪ $(-5;1]$