*
\(A=x^2+2x-2=\left(x^2+2x+1\right)-3=\left(x+1\right)^2-3\ge-3\)
Vậy \(A_{min}=-3\Leftrightarrow x=-1\)
* Ta có:\(4x-x^2-5=-\left(x^2-2\cdot x\cdot2+4\right)-1=-\left(x-2\right)^2-1\le-1\)
\(\Rightarrow B\ge\dfrac{10}{-1}=-10\)
Vậy \(B_{min}=-10\Leftrightarrow x=2\)
* Ta có:
\(2x-x^2-5=-\left(x^2-2x+1\right)-4=-\left(x-1\right)^2-4\le-4\)
\(\Rightarrow C\ge\dfrac{-6}{-4}=\dfrac{3}{2}\)
Vậy \(C_{min}=\dfrac{3}{2}\Leftrightarrow x=1\)