[tex]\displaystyle\bf\\x^{2} -4x-\frac{12}{x^{2} -4x} -4=0\\\\\\x^{2} -4x=a \ \ \ ; \ \ \ x\neq 0 \ \ ; \ \ \ x\neq 4\\\\\\a-\frac{12}{a} -4=0 \ \ , \ \ a\neq 0\\\\\\a^{2} -4a-12=0\\\\D=(-4)^{2} -4\cdot(-12)=16+48=64=8^{2} \\\\\\a_{1}=\frac{4-8}{2} =\frac{-4}{2} =-2\\\\\\a_{2} =\frac{4+8}{2} =\frac{12}{2} =6\\\\1)\\\\x^{2} -4x=-2\\\\x^{2} -4x+2=0\\\\D=(-4)^{2} -4\cdot 2=16-8=8=(2\sqrt{2} )^{2} \\\\\\x_{1}=\frac{4-2\sqrt{2} }{2} =2-\sqrt{2}[/tex]

[tex]\displaystyle\bf\\x_{2}=\frac{4+2\sqrt{2} }{2} =2+\sqrt{2} \\\\\\2)\\\\x^{2} -4x=6\\\\x^{2} -4x-6=0\\\\D=(-4)^{2} -4\cdot(-6)=16+24=40=(2\sqrt{10})^{2} \\\\\\x_{3} =\frac{4-2\sqrt{10} }{2} =2-\sqrt{10} \\\\\\x_{4} =\frac{4+2\sqrt{10} }{2} =2-\sqrt{10} \\\\\\a_{1} +a_{2} +x_{1 } + x_{2} + x_{3}+x_{4} =-2+6+2-\sqrt{2} +2+\sqrt{2} +\\\\+2-\sqrt{10} +2+\sqrt{10} =12[/tex]

В поле ответа записать  :  12

Сумма корней уравнения с введённой переменной равна  4 .

Сумма корней исходного уравнения равна 4 .