\(\Leftrightarrow\left(x^2+1\right)\left[x^2-2x\left(k-1\right)+\left(k-1\right)^2+k^2-4k+5\right]=2x\)
\(\Leftrightarrow\left(x^2+1\right)\left[\left(x-k+1\right)^2+\left(k-2\right)^2+1\right]=2x\)
Do \(VT>0\) \(\forall x\Rightarrow VP>0\Rightarrow x>0\)
Mặt khác \(\left\{{}\begin{matrix}x^2+1\ge2x\\\left(x-k+1\right)^2+\left(k-2\right)^2+1\ge1\end{matrix}\right.\)
\(\Rightarrow\left(x^2+1\right)\left[\left(x-k+1\right)^2+\left(k-2\right)^2+1\right]\ge2x\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}x^2+1=2x\\\left(x-k+1\right)^2+\left(k-2\right)^2+1=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\k=2\end{matrix}\right.\)
Vậy \(k=2\) thì pt có nghiệm \(x=1\)
