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$pt \Leftrightarrow (x^2+2)(x^2+k^2-4k+3)=2k+1$$\Leftrightarrow (x^2+2).x^2+(x^2+2)(k^2-4k+3)=2k+1$$\Leftrightarrow x^4+x^2(k^2-4k+5)+(2k^2-10k+5)=0$$\text{Do }k^2-4k+5>0\text{ nên pt có nghiệm}\Leftrightarrow \begin{cases}\Delta \ge0 \\ 2k^2-10k+5 \le0 \end{cases}\Leftrightarrow \begin{cases}k^4+18k^2+5 \ge 8k^3 (\text{đúng theo cosi)}\\ \dfrac{5-\sqrt{15}}{2} \le k \le \dfrac{5+\sqrt{15}}{2} \end{cases}$$Leftrightarrow k\in \left[ \frac{5-\sqrt{15}}{2};\frac{5+\sqrt{15}}{2}\right]$

$pt \Leftrightarrow (x^2+2)(x^2+k^2-4k+3)=2k+1$$\Leftrightarrow (x^2+2).x^2+(x^2+2)(k^2-4k+3)=2k+1$$\Leftrightarrow x^4+x^2(k^2-4k+5)+(2k^2-10k+5)=0$$\text{Do }k^2-4k+5>0\text{ nên pt có nghiệm}\Leftrightarrow \begin{cases}\Delta \ge0 \\ 2k^2-10k+5 \le0 \end{cases}\Leftrightarrow \begin{cases}k^4+18k^2+5 \ge 8k^3 (\text{đúng theo cosi)}\\ \dfrac{5-\sqrt{15}}{2} \le k \le \dfrac{5+\sqrt{15}}{2} \end{cases}$$\Leftrightarrow k\in \left[ \frac{5-\sqrt{15}}{2};\frac{5+\sqrt{15}}{2}\right]$