\(ĐKXĐ: \begin{cases}x\neq2\\x\neq4\\\end{cases}\\\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\\\Leftrightarrow \dfrac{(x-3)(x-4)+(x-2)^2}{(x-2)(x-4)}= \dfrac{-(x-2)(x-4)}{(x-2)(x-4)}\\\Rightarrow (x-3)(x-4)+(x-2)^2=-(x-2)(x-4)\\\Leftrightarrow x^2-7x+12+x^2-4x+4=-(x^2-6x+8)\\\Leftrightarrow x^2-7x+12+x^2-4x+4+x^2-6x+8=0\\\Leftrightarrow 3x^2-17x+24=0\\\Leftrightarrow 3x^2-8x-9x+24=0\\\Leftrightarrow x(3x-8)-3(3x-8)=0\\\Leftrightarrow (3x-8)(x-3)=0\\\Leftrightarrow \left[ \begin{array}{l}3x-8=0\\x-3=0\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=\dfrac{8}{3}\\x=3\end{array} \right.(tmđk)\)
Vậy `S={8/3;3}`
