Ответ:

2; 6

Объяснение:

[tex]\frac{3}{x}+\frac{1}{x-4} =1\\\\\frac{3}{x}+\frac{1}{x-4} -1=0\\\\\frac{3(x-4)+1x-1(x(x-4))}{x(x-4)} =0\\\\\frac{3x-12+x-x^2+4x}{x(x-4)} =0\\\\\frac{-x^2+8x-12}{x(x-4)} =0\\\\x\neq 0\\\\x-4\neq 0= > x\neq 4\\\\\\-x^2+8x-12=0|:(-1)\\\\x^2-8x+12=0\\\\D=64-4*12=64-48=16\\\\x_1=\frac{8+4}{2} =\frac{12}{2}=6\\\\x_2=\frac{8-4}{2} =\frac{4}{2}=2[/tex]

Ответ:

2; 6

Объяснение: