\(A=x^2+4x+5\)
\(=\left(x^2+4x+4\right)+1\)
\(=\left(x+2\right)^2+1\ge1\)
Vậy \(A_{min}=1\) khi x = -2
\(B=x^2+5x+7\)
\(=\left(x^2+2.x.\dfrac{5}{2}+\dfrac{25}{4}\right)+\dfrac{3}{4}\)
\(=\left(x+\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy \(B_{min}=\dfrac{3}{4}\) khi \(x=-\dfrac{5}{2}\)
\(C=\left(x-1\right)\left(x-3\right)+11\)
\(=x^2-3x-x+3+11\)
\(=x^2-4x+14\)
\(=\left(x^2-4x+4\right)+10\)
\(=\left(x-2\right)^2+10\ge10\)
Vậy \(C_{min}=10\) khi x = 2
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