Bài 10:
1: Ta có: \(A=x^2-6x+11\)
\(=x^2-6x+9+2\)
\(=\left(x-3\right)^2+2\ge2\forall x\)
Dấu '=' xảy ra khi x=3
2: Ta có: \(B=3x^2-5x+7\)
\(=3\left(x^2-\dfrac{5}{3}x+\dfrac{7}{3}\right)\)
\(=3\left(x^2-2\cdot x\cdot\dfrac{5}{6}+\dfrac{25}{36}+\dfrac{59}{36}\right)\)
\(=3\left(x-\dfrac{5}{6}\right)^2+\dfrac{59}{12}\ge\dfrac{59}{12}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{5}{6}\)