a) Ta có : \(M=x^2-4x+7\)
\(=x^2-4x+4+3\)
\(=\left(x-2\right)^2+3\)\(\ge3\forall x\in R\)(vì \(\left(x-2\right)^2\ge0\))
Vậy Mmin = 3 khi x - 2 = 0 => x = 2
b)Ta có : N = x2 - x + 1
\(=x^2-x+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\in R\)
Vậy Nmin = \(\frac{3}{4}\) khi \(x=\frac{1}{2}\).