\(\Delta'=m-2>0\Rightarrow m>2\)
Theo hệ thức Viet ta có: \(x_1+x_2=2\Rightarrow x_1=2-x_2\)
Do \(x_2\) là nghiệm nên \(x_2^2-2x_2+3-m=0\Leftrightarrow x_2^2-2x_2+4=m+1\)
Thay vào ta được:
\(2\left(2-x_2\right)^3+\left(x_2^2-2x_2+4\right)x_2^2=16\)
\(\Leftrightarrow2\left(8-12x_2+6x_2^2-x_2^3\right)+x_2^4-2x_2^3+4x_2^2=16\)
\(\Leftrightarrow x_2^4-4x_2^3+16x_2^2-24x_2=0\)
\(\Leftrightarrow x_2\left(x_2-2\right)\left(x_2^2-2x_2+12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x_2=0\Rightarrow x_1=2\\x_2=2\Rightarrow x_1=0\end{matrix}\right.\) \(\Rightarrow3-m=x_1x_2=0\Rightarrow m=3\)
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