\(2x^2-2x+m^2-2\)
\(\Delta=4m^2-4.2\left(m^2-2\right)=4m^2-8m^2+16=16-4m^2\)
\(\sqrt{\Delta}=\sqrt{16-4m^2}\)
=> \(\left\{{}\begin{matrix}x_1=\dfrac{2m+\sqrt{16-4m^2}}{4}\\x_2=\dfrac{2m-\sqrt{16-4m^2}}{4}\end{matrix}\right.\)
Ta có \(A=\left|2x_1x_2-x_1-x_2-4\right|\)
\(=\left|2\dfrac{2m+\sqrt{16-4m^2}}{4}\dfrac{2m-\sqrt{16-4m^2}}{4}-\dfrac{2m+\sqrt{16-4m}}{4}-\dfrac{2m-\sqrt{16-4m}}{4}-4\right|\)
\(=\left|\dfrac{4m^2-\left(16-4m^2\right)-8m-2\sqrt{16-4m^2}-8m+2\sqrt{16-4m^2}-32}{8}\right|\)
\(=\left|\dfrac{8m^2-16m-32}{8}\right|\)
\(\left|\dfrac{\left(m-1-\sqrt{5}\right)\left(m-1+\sqrt{5}\right)}{8}\right|\ge0\)
Dấu "=" xảy ra khi :\(m=1+\sqrt{5}\) hoặc \(m=1-\sqrt{5}\)
