\(x^4-8x^3+22x^2-24x+7+2m=0\)
\(\Leftrightarrow\left(x^2-4x\right)^2+6\left(x^2-4x\right)+7+2m=0\)
\(đặt:x^2-4x=t\Rightarrow t^2+6t+7+2m=0\)
\(\Delta=8-8m>0\Leftrightarrow m< 1\Rightarrow\left[{}\begin{matrix}t1=\dfrac{-6+\sqrt{8-8m}}{2}=\sqrt{2-2m}-3\\t2=\dfrac{-6-\sqrt{8-8m}}{2}=-\sqrt{2-2m}-3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x^2-4x-\sqrt{2-2m}+3=0\left(1\right)\\x^2-4x+\sqrt{2-2m}+3=0\left(2\right)\end{matrix}\right.\)
\(\Rightarrow\left(1\right)\left(2\right)có-2-ngo-pb-và-không-có-nghiệm-chung\)
\(\Leftrightarrow\left\{{}\begin{matrix}\Delta'1>0\Leftrightarrow1+\sqrt{2-2m}>0\\\Delta'2>0\Leftrightarrow1-\sqrt{2-2m}>0\end{matrix}\right.\)\(\Leftrightarrow\dfrac{1}{2}< m\le1\left(m< 1\right)\Rightarrow\dfrac{1}{2}< m< 1\left(3\right)\)
\(giả-sử-có-ngo-chung\Rightarrow\left(1\right)-\left(2\right)=0\)
\(\Rightarrow-2\sqrt{2-2m}=0\Leftrightarrow m=1\Rightarrow m\ne1\left(4\right)\)
\(\left(3\right)\left(4\right)\Rightarrow\dfrac{1}{2}< m< 1\)
