Bài 6;
a: 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
b: Ta có: \(B=2x^2+10x-1\)
\(=2\left(x^2+5x-\dfrac{1}{2}\right)\)
\(=2\left(x^2+5x+\dfrac{25}{4}-\dfrac{27}{4}\right)\)
\(=2\left(x+\dfrac{5}{2}\right)^2-\dfrac{27}{2}\ge-\dfrac{27}{2}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{5}{2}\)