[tex]\displaystyle\bf\\x^{2} -4x-\frac{12}{x^{2} -4x} -4=0\\\\x^{2} -4x=m \ \ \ ; \ \ m\neq 0\\\\m-\frac{12}{m} -4=0\\\\\frac{m^{2}-4m-12 }{m} =0\\\\m^{2} -4m-12=0\\\\Teorema \ Vieta:\\\\m_{1} +m_{2} =4\\\\m_{1} \cdot m_{2} =-12\\\\m_{1} =-2\\\\m_{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} \\\\x_{2} =\frac{4+2\sqrt{2} }{2} =2+\sqrt{2}[/tex]

[tex]\displaystyle\bf\\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}[/tex]

Сумма корней с введённой переменной равна :

[tex]\displaystyle\bf\\m_{1} +m_{2} =-2+6=\boxed4[/tex]

Сумма корней исходного уравнения равна :

[tex]\displaystyle\bf\\x_{1} +x_{2}+ x_{3} +x_{4} =2-\sqrt{2} +2+\sqrt{2} +2-\sqrt{10} +2+\sqrt{10} =8[/tex]

Сумма корней нового и исходного уравнений равна :

4 + 8 =  12

Ответ : 12