để pt có 2 nghiệm x1,x2 => Δ'≥0
\(\Leftrightarrow\left(m+1\right)^2-m^2-2\ge0\)
\(\Leftrightarrow m^2+2m+1-m^2-2\ge0\Leftrightarrow2m\ge1\Leftrightarrow m\ge\frac{1}{2}\)
Theo viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1\cdot x_2=m^2+2\end{matrix}\right.\)
\(\left|x_1^4-x_2^4\right|=\left|\left(x_1^2-x_2^2\right)\left(x_1^2+x_2^2\right)\right|=\left|\left(x_1+x_2\right)\left(x_1-x_2\right)\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\right|=0\)
+) \(x_1+x_2=0\Leftrightarrow2\left(m+1\right)=0\Leftrightarrow m=-1\) (loại)
+) \(x_1-x_2=0\Leftrightarrow\left(x_1-x_2\right)^2=0\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=0\Leftrightarrow\left(2m+2\right)^2-4\left(m^2+2\right)=0\Leftrightarrow4m^2+8m+4-4m^2-8=0\Leftrightarrow8m=4\Leftrightarrow m=\frac{1}{2}\left(tm\right)\)
+) \(\left(x_1+x_2\right)^2-2x_1x_2=0\)
\(\Leftrightarrow\left(2m+2\right)^2-2\left(m^2+2\right)=0\)
\(\Leftrightarrow4m^2+8m+4-2m^2-4=0\)
\(\Leftrightarrow2m^2+8m=0\Leftrightarrow\left[{}\begin{matrix}m=0\\m=-4\end{matrix}\right.\)(ktm)
Vậy m = \(\frac{1}{2}\)
