\(A=\left(x+1\right)^2+\left(2-x\right)^2\ge\frac{1}{2}\left(x+1+2-x\right)^2=\frac{9}{2}\)
Hoặc nếu không thì làm như sau:
\(A=x^2+2x+1+x^2-4x+4\)
\(A=2x^2-2x+5=2x^2-2x+\frac{1}{2}+\frac{9}{2}\)
\(A=\frac{1}{2}\left(2x-1\right)^2+\frac{9}{2}\ge\frac{9}{2}\)
\(A_{min}=\frac{9}{2}\) khi \(x=\frac{1}{2}\)