Verified answer

Ответ:

[tex]\frac{-1-\sqrt{145}}{6}\\\\\frac{-1+\sqrt{145}}{6}\\\\-1\\\\4[/tex]

Пошаговое объяснение:

[tex]\frac{x^2-4}{x}-\frac{x}{x^2-4}=\frac{8}{3}[/tex]

ОДЗ:

[tex]x \neq -2,\ x \neq 0,\ x \neq 2[/tex]

подставляем

[tex]\frac{x^2-4}{x}=t\\\\t-\frac{1}{t}=\frac{8}{3}\\\\t-\frac{1}{t}-\frac{8}{3}=0\\\\\frac{3(t^2-1)}{3t}-\frac{8t}{3t}=0\\\\\frac{3t^2-3t-8t}{3t}=0\\\\\frac{3t^2-3-8t}{3t}=0\\\\\frac{3t^2-8t-3}{3t}=0[/tex]

[tex]3t^2-8t-3=0\\\\D=(-8)^2-4\cdot3\cdot (-3)=64+36=100\\\\\sqrt{D}=\sqrt{100}=10\\\\t_1=\frac{8-10}{2\cdot 3}=\frac{-2}{6}=-\frac{1}{3}\\\\t_2=\frac{8+10}{2\cdot 3}=\frac{18}{6}=3[/tex]

[tex]\frac{x^2-4}{x}=-\frac{1}{3}\\\\3(x^2-4)=-x\\\\3x^2-12+x=0\\\\3x^2+x-12=0\\\\D=1^2-4\cdot3\cdot(-12)=1+144=145\\\\\sqrt{D}=\sqrt{145}\\\\x_1=\frac{-1-\sqrt{145}}{2\cdot 3}=\frac{-1-\sqrt{145}}{6}\\\\x_2=\frac{-1+\sqrt{145}}{2\cdot 3}=\frac{-1+\sqrt{145}}{6}[/tex]

[tex]\frac{x^2-4}{x}=3\\\\x^2-4=3x\\\\x^2-3x-4=0\\\\D=(-3)^2-4\cdot 1\cdot(-4)=9+16=25\\\\\sqrt{D}=\sqrt{25}=5\\\\x_3=\frac{3-5}{2\cdot 1}=\frac{-2}{2}=-1\\\\x_4=\frac{3+5}{2\cdot 1}=\frac{8}{2}=4[/tex]