Помогите с решением

$x^2+ \frac{4x^2}{|x|} -12=0 \\ \\ \frac{4x^2}{|x|}=12-x^2$
ОДЗ:
$\left \{ {{12-x^2 \geq 0 } \atop {x \neq 0}} \right. =\ \textgreater \ \left \{ {{x \leq 12} \atop {x \neq 0}} \right. ==\ \textgreater \ \left \{ {{x \leq \pm2 \sqrt{3} } \atop {x \neq 0}} \right. \\ \\ x\in [-2 \sqrt{3} ;0) \cup (0;2 \sqrt{3}]$

$4x^2=|x|(12-x^2) \\ \left \{ {{4x^2=x(12-x^2)} \atop {-4x^2=x(12-x^2)}} \right. \\ \\ 4x^2=x(12-x^2) \\ 4x^2=12x-x^3 \\ x^3+4x^2-12x=0 \\ x(x^2+4x-12)=0 \\ x_1 \neq 0 \\ \\ x^2+4x-12=0 \\ D=16+48=64=8^2 \\ x_1= \frac{-4-8}{2} \neq -6 \\ \\ x_2= \frac{-4+8}{2}=2 \\ x=2 \\ \\ -4x^2=x(12-x^2) \\ -4x^2=12x-x^3 \\ x^3-4x^2-12x=0 \\ x(x^2-4x-12)=0 \\ x \neq 0 \\ \\ x^2-4x-12=0 \\ D=16+48=8^2 \\ x_1= \frac{4+8}{2} \neq 6 \\ \\ x_2= \frac{4-8}{2}= -2 \\ x=-2$

x=2; x=-2

5)
$x^2-8 \sqrt{x^2}+15=0 \\ \sqrt{x^2} =|x| \\ x^2-8|x|+15=0 \\ 8|x|=x^2+15$
ОДЗ:
$x^2+15 \geq 0 \\ x^2 \geq -15 \\ x \in R$

$8|x|=x^2+15 \\ \\ \left \{ {{8x=x^2+15} \atop {8x=-x^2-15}} \right. \\ \\ 8x=x^2+15 \\ x^2-8x+15=0 \\ D=64-60=4=2^2 \\ x_1= \frac{8+2}{2}=5 \\ \\ x_2= \frac{8-2}{2} =3 \\ \\ 8x=-x^2-15 \\ x^2+8x+15=0 \\ D=64-60=4=2^2 \\ x_1= \frac{-8-2}{2}=-5 \\ \\ x_2= \frac{-8+2}{2} =-3$
ОТВЕТ: x1=5; x=-5; x3=3; x4=-3