(x-1)(x-3)+x^2-4x+5=0
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)+\left(x^2-4x+4\right)+1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)+\left(x-2\right)^2+1=0\)(*)
Đặt \(y=x-2\) khi đó (*) có dạng:
\(\left(y+1\right)\left(y-1\right)+y^2+1=0\)
\(\Leftrightarrow y^2-1+y^2+1=0\)
\(\Leftrightarrow2y^2=0\)
\(\Leftrightarrow y=0\)
Hay \(x+2=0\Rightarrow x=-2\)
Vậy x=-2