Cho x1+x2=2m-2 x1x2= 2m-7 tìm m để biểu thức C =x1^2+ x2^2- x1x2 đạt GTNN

\(x_1^2+x_2^2-x_1x_2\\=(x_1^2+2x_1x_2+x_2^2)-3x_1x_2\\=(x_1+x_2)^2-3x_1x_2\\=(2m-2)^2-3(2m-7)\\=4m^2-8m+4-6m+21\\=4m^2-14m+25\\=4m^2-2.2m.\dfrac{7}{2}+\dfrac{49}{4}+\dfrac{51}{4}\\=\bigg(2m-\dfrac{7}{2}\bigg)^2+\dfrac{51}{4}\)

Vì \(\bigg(2m-\dfrac{7}{2}\bigg)^2\ge 0→C≥\dfrac{51}{4}\\→C_{min}=\dfrac{51}{4}\)

→ Dấu “=” xảy ra khi \(2m-\dfrac{7}{2}=0\\↔2m=\dfrac{7}{2}\\↔m=\dfrac{7}{4}\)

Vậy \(C_{min}=\dfrac{51}{4}\) khi \(m=\dfrac{7}{4}\)