[tex]\displaystyle 1)x^4-13x^2+36=0\\t^2-13t+36=0\\t=4,t=9\\x^2=4,x^2=9\\x_1=-3,x_2=-2,x_3=2,x_4=3\\\\2)x^4-x^2-12=0\\t^2-t-12=0\\t=-3,t=4\\x^2=-3,x^2=4\\x \notin R, x=-2,x=2\\x_1=-2,x_2=2\\\\3)(x^2-3)^2-4(x^2-3)-12=0\\t^2-4t-12=0\\t=-2,t=6\\x^2-3=-2,x^2-3=6\\x_1=-3,x_2=-1,x_3=1,x_4=3[/tex]

Объяснение:

[tex]1) x^4-13x^2+36=0\\(x^2)^2-13x^2+36=0\\x^2=t; t\geq 0\\t^2-13t+36=0\\D=(-13)^2-4*1*36=169-144=25\\t_1=\frac{13+5}{2}=\frac{18}{2}=9\\ t_2=\frac{13-5}{2}=\frac{8}{2} =4\\\\x^2=9\\x=3; x=-3\\x^2=4\\x=2; x=-2\\[/tex]

Ответ: 3;-3;2;-2

[tex]2) x^4-x^2-12=0\\(x^2)^2-x^2-12=0\\x^2=t; t\geq 0\\t^2-t-12=0\\D=(-1)^2-4*1*(-12)=1+48=49\\t_1=\frac{1+7}{2} =\frac{8}{2} =4\\t_2=\frac{1-7}{2} =\frac{-6}{2} =-3[/tex]

(t=-3 не подходит)

[tex]x^2=4\\x=2; x=-2[/tex]

Ответ: 2; -2

[tex]3) (x^2-3)^2-4(x^2-3)-12=0\\(x^2-3)=t\\t^2-4t-12=0\\D=(-4)^2-4*1*(-12)=16+48=64\\t_1=\frac{4+8}{2} =\frac{12}{2} =6\\t_2=\frac{4-8}{2} =\frac{-4}{2} =-2\\\\x^2-3=6\\x^2=9\\x=3; x=-3\\\\x2-3=-2\\x^2=1\\x=1; x=-1[/tex]

Ответ: 3; -3; 1; -1