A = \(x^2\) - 6x - 1
= (\(x^2\) - 2.x.3 + \(3^2\)) - \(3^2\) - 1
= \(\left(x+3\right)^2\) - 27 - 1
= \(\left(x+3\right)^2\) - 28
Ta có: \(\left(x+3\right)^2\) ≥ 0 ∀ x
⇒ \(\left(x+3\right)^2-28\) ≥ - 28
Hay A ≥ - 28
Dấu "=" xảy ra ↔ x + 3 = 0
x = - 3
Vậy min A = - 28 ↔ x = - 3
B = \(x^2\) + 3x + 7
= (\(x^2\) - 2.x.\(\frac{3}{2}\) + \(\frac{3}{2}^2\)) \(-\frac{3}{2}^2\) + 7
= \(\left(x+\frac{3}{2}\right)^2\) \(-\frac{9}{4}\) + 7
= \(\left(x+\frac{3}{2}\right)^2\) + \(\frac{19}{4}\)
Ta có: \(\left(x+\frac{3}{2}\right)^2\) ≥ 0 ∀ x
⇒ \(\left(x+\frac{3}{2}\right)^2+\frac{19}{4}\) ≥ \(\frac{19}{4}\)
Hay B ≥ \(\frac{19}{4}\)
Dấu "=" xảy ra ↔ \(x+\frac{3}{2}=0\)
\(x=-\frac{3}{2}\)
Vậy min B = \(\frac{19}{4}\) ↔ \(x=-\frac{3}{2}\)
\(\Leftrightarrow C=\left(x^2-4x+4\right)+\left(y^2-6y+9\right)+5\)
\(\Leftrightarrow\left(x-2\right)^2+\left(y-3\right)^2+5\ge5\forall x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
\(\Leftrightarrow D=\left(x^4-4x+3\right)+2017\)
\(\Leftrightarrow\)\(D=\left(x^4-2x^3+x^2\right)+\left(2x^3-4x^2+2x\right)+\left(3x^2-6x+3\right)+2017\)
\(\Leftrightarrow D=x^2\left(x^2-2x+1\right)+2x\left(x^2-2x+1\right)+3\left(x^2-2x+1\right)+2017\)
\(\Leftrightarrow D=\left(x^2+2x+3\right)\left(x-1\right)^2+2017\ge2017\)
Dấu "=" xảy ra ⇔ x = 1