ĐKXĐ: \(x\ne\left\{1;3\right\}\)
\(\Leftrightarrow\frac{1}{x^2-4x+3}+x^2-4x+3+2=0\)
Đặt \(x^2-4x+3=t\ne0\)
\(\frac{1}{t}+t+2=0\Leftrightarrow t^2+2t+1=0\)
\(\Rightarrow t=-1\Leftrightarrow x^2-4x+3=-1\)
\(\Leftrightarrow\left(x-2\right)^2=0\Rightarrow x=2\)