\(Sửa:A=x^4-6x^3+13x^2-12x+2021\\ A=\left(x^4-6x^3+9x^2\right)+4\left(x^2-3x\right)+4+2017\\ A=\left(x^2-3x\right)^2+4\left(x^2-3x\right)+4+2017\\ A=\left(x^2-3x+2\right)^2+2017\ge2017\\ A_{min}=2017\Leftrightarrow x^2-3x+2=0\Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Có điều kiện không bạn.

câu 1

x^2 -5x +y^2+xy -4y +2014 

=(y^2+xy +1/4x^2) -4(y+1/2x)+4 +3/4x^2-3x+2010

=(y+1/2x-2)^2 +3/4(x^2-4x+4)+2007

=(y+1/2x-2)^2 +3/4(x-2)^2 +2007

GTNN là 2007<=> x=2 và y=1

M= \(x^2-3x+5=x^2-2\times\frac{3}{2}\times x+\frac{9}{4}-\frac{9}{4}+5\)

M =   \(\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\)

Vì \(\left(x-\frac{3}{3}\right)^2\ge0\)

=> \(\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)

Vậy MIN  M = \(\frac{11}{4}\)dấu bằng xảy ra khi và chỉ khi \(x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)

\(M=x^2-3x+5=\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}\right)+5-\frac{9}{4}\)

\(=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)

Vậy \(MinM=\frac{11}{4}\Leftrightarrow\left(x-\frac{3}{2}\right)^2=0\Leftrightarrow x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)

Max:

\(M=\frac{x^2+xy+y^2}{x^2+y^2}=1+\frac{xy}{x^2+y^2}\le1+\frac{xy}{2\left|xy\right|}\le1+\frac{xy}{2xy}=1+\frac{1}{2}=\frac{3}{2}\)

Dấu "=" xảy ra tại x=y