ĐK: ` x\ne 2; x \ne 4`.
`2/(-x^2+6x-8)-(x-1)/(x-2)=(x+3)/(x-4)`
`<=> -2-(x-1)(x-4)=(x+3)(x-2)`
`<=> −x^2+5x−6=x^2+x−6`
`<=> 2x^2-4x=0`
`<=> 2x(x-2)=0`
`<=>` \(\left[{}\begin{matrix}x=0\\x=2\left(L\right)\end{matrix}\right.\)
Vậy `S={0}`.
ĐKXĐ: \(x\neq 2;x\neq 4\)
\(PT\Leftrightarrow\dfrac{-2}{\left(x-2\right)\left(x-4\right)}-\dfrac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}=\dfrac{\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}\)
\(\Rightarrow-2-\left(x-1\right)\left(x-4\right)=\left(x+3\right)\left(x-2\right)\)
\(\Leftrightarrow-2-\left(x^2-5x+4\right)=x^2+x-6\Leftrightarrow2x^2-4x=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=2\left(l\right)\end{matrix}\right.\)
Vậy x = 0
