Sửa đề: \(\dfrac{2x}{x-1}=\dfrac{x^2-x+4}{\left(x-1\right)\left(x-4\right)}\)
ĐKXĐ: \(x\notin\left\{1;4\right\}\)
Ta có: \(\dfrac{2x}{x-1}=\dfrac{x^2-x+4}{\left(x-1\right)\left(x-4\right)}\)
\(\Leftrightarrow\dfrac{2x\left(x-4\right)}{\left(x-1\right)\left(x-4\right)}=\dfrac{x^2-x+4}{\left(x-1\right)\left(x-4\right)}\)
Suy ra: \(2x^2-8x-x^2+x-4=0\)
\(\Leftrightarrow x^2-7x-4=0\)
\(\Delta=\left(-7\right)^2-4\cdot1\cdot\left(-4\right)=49+16=65\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{7-\sqrt{65}}{2}\left(nhận\right)\\x_2=\dfrac{7+\sqrt{65}}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{7-\sqrt{65}}{2};\dfrac{7+\sqrt{65}}{2}\right\}\)