\(\Delta\)' = m2 - m2 + m - 1 = m - 1
ta có phương trình có nghiệm x1 ; x2 \(\Leftrightarrow\) \(\Delta\)' \(\ge\) 0
\(\Leftrightarrow\) m - 1 \(\ge\) 0 \(\Leftrightarrow\) m \(\ge\) 1
ta có : A = x12 + x22 = (x1 + x2)2 - 2x1x2
áp dụng hệ thức vi ét ta có : \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=m^2-m+1\end{matrix}\right.\)
thay : (2m)2 - 2(m2 - m + 1) = 4m2 - 2m2 + 2m - 2
= 2m2 + 2m - 2 = 2 (m2 + m - 1) = 2 (m2 + 2.m.\(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{4}\) - 1) = 2 [(m + \(\dfrac{1}{2}\))2 - \(\dfrac{5}{4}\) ] = 2(m + \(\dfrac{1}{2}\))2 - \(\dfrac{5}{2}\) \(\ge\) \(-\dfrac{5}{2}\)
minA = \(\dfrac{-5}{2}\) khi m + \(\dfrac{1}{2}\) = 0 \(\Leftrightarrow\) m = \(-\dfrac{1}{2}\)
