Ответ:

[tex]\frac{-3 - \sqrt{17}}{2}; \ \frac{-3 + \sqrt{17}}{2}; \ 2 - \sqrt{6}; \ 2 + \sqrt{6}[/tex]

Объяснение:

[tex]x^2+\frac{4}{x^2}-(x-\frac{2}{x})-16=0 \\ \\ (x-\frac{2}{x})=t \ \Rightarrow \ t^2=x^2-4+\frac{4}{x^2} \ \Rightarrow \ x^2+\frac{4}{x^2}=t^2+4 \\ \\ t^2+4-t-16=0 \\ t^2-t-12=0 \\ t_1=-3 \\ t_2=4 \\ \\ 1) \ t=-3 \\ x-\frac{2}{x} =-3 \ |*x, \ x \neq 0 \\ x^2-2=-3x \\ x^2+3x-2=0 \\ D=9+8=17 \\ \\ x_{1,2}=\frac{-3 \pm \sqrt{17}}{2}[/tex]

[tex]2) \ t=4 \\ x-\frac{2}{x} =4 \ |*x, \ x \neq 0 \\ x^2-2=4x \\ x^2-4x-2=0 \\ D=16+8=24=6*4=(2\sqrt{6})^2 \\ \\ x_{3,4}=\frac{4 \pm 2\sqrt{6}}{2} =2 \pm \sqrt{6}[/tex]

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