Ответ:

х = -6; х = -1; х = 2; х = 3

Объяснение:

[tex]\displaystyle (\frac{x^2+6}{x} )^2+2*(\frac{x^2+6}{x} )=35[/tex]
Пусть [tex]\displaystyle \frac{x^2+6}{x} = t[/tex], тогда
[tex]\displaystyle t^2+2t-35=0;\\D = 2^2-4*1*(-35) = 4+140 = 144 = 12^2;\\t_{12} = \frac{-2 \pm 12}{2*1};\\ t_{1} = \frac{-2 + 12}{2*1}=\frac{10}{2} = 5 ;\\ t_{2} = \frac{-2 - 12}{2*1}=-\frac{14}{2} = -7 ;[/tex]
Вернёмся к замене
а) [tex]\displaystyle \left \{ {{\frac{x^2+6}{x} = 5} \atop {x\neq 0}} \right. < = > \left \{ {{x^2+6x=5x} \atop {x\neq 0}} \right. < = > \left \{ {{x^2-5x+6x=0} \atop {x\neq 0}} \right.[/tex]
Рассмотрим отдельно 1-ое уравнение системы
x²-5x+6 = 0;
D = (-5)²-4*1*6 = 25-24 = 1 = 1²
x₁₂ = (5±1)/(2*1);
x₁ = (5+1)/(2*1) = 6/2 = 3; x₂ = (5-1)/(2*1) = 4/2 = 2;
б) [tex]\displaystyle \left \{ {{\frac{x^2+6}{x} = -7} \atop {x\neq 0}} \right. < = > \left \{ {{x^2+6x=-7x} \atop {x\neq 0}} \right. < = > \left \{ {{x^2+7x+6x=0} \atop {x\neq 0}} \right.[/tex]
Рассмотрим отдельно 1-ое уравнение системы
x²+7x+6 = 0;
D = 7²-4*1*6 = 49-24 = 25 = 5²
x₁₂ = (-7±5)/(2*1);
x₁ = (-7+5)/(2*1) = -2/2 = -1; x₂ = (-7-5)/(2*1) = -12/2 = -6;