bài 1:= \(2x\left(x-3\right)-6\left(x-3\right)+2y\left(x-3\right)\)

         =\(2\left(x-3\right)\left(x+y-3\right)\)

bài 2:P=\(x^2-2x+1+y^2+6y+9+2\)

         P=\(\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\)

vậy Pmin=2 khi x=1 và y=-3

Ta có : A = x² + 2x + y² + 6y + 10

=> A = (x² + 2x + 1) + (y² + 6y + 9)

=> A = (x + 1)² + (y + 3)²

Mà : (x + 1)² và (y + 3)² \(\ge0\forall x,y\)

Nên : A = (x + 1)² + (y + 3)² \(\ge0\forall x,y\)

Vậy A_min = 0 tại x = -1 và y = -3

\(A=x^2+2x+y^2+6y+10\)

\(=x^2+2x+y^2+6y+1+9\)

\(=\left(x^2+2x+1\right)+\left(y^2+6y+9\right)\)

\(=\left(x+1\right)^2+\left(y+3\right)^2\)

vì \(\left(x+1\right)^2\ge0\forall x;\left(y+3\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+1\right)^2+\left(y+3\right)^2\ge0\forall x\)

vậy \(MinA=0\Leftrightarrow\orbr{\begin{cases}x=-1\\y=-3\end{cases}}\)

A= \(x^2+2x+1+y^2+6y+9\) \(=\left(x+1\right)^2+\left(y+3\right)^2\ge0\)

dau "=" xay ra \(\Leftrightarrow x=-1,y=-3\)

min A= 0 khi x=-1,y=-3

Ta có : 2x² - 6x 

= \(\left(\sqrt{2}x\right)^2-2.\sqrt{2}x.6+36-36\)

Q\(=\left(\sqrt{2}x-6\right)^2-36\)

Vì \(\left(\sqrt{2}x-6\right)^2\ge0\forall x\)

Nên : Q = \(=\left(\sqrt{2}x-6\right)^2-36\) \(\ge-36\forall x\)

Vậy \(Q_{min}=-36\) khi \(\sqrt{2}x-6=0\) => \(\sqrt{2}x=6\) => \(x=6:\sqrt{2}=3\sqrt{2}\)

\(B=\left(x-2y\right)^2+y^2+2x+6y+2046=\left[\left(x-2y\right)^2+2\left(x-2y\right)+1\right]+\left(y^2+10y+25\right)+2020=\left(x-2y+1\right)^2+\left(y+5\right)^2+2020\ge2020\)

\(minB=2020\Leftrightarrow\)\(\left\{{}\begin{matrix}x=-11\\y=-5\end{matrix}\right.\)

ta có \(A=x^2+y^2+9-2xy-6x+6y+x^2-4x+4+2004\)

             \(=\left(x-y-3\right)^2+\left(x-2\right)^2+2004\)

vì \(\left(x-y-3\right)^2+\left(x-2\right)^2\ge0\)

=> \(A\ge2004\)

dấu = xảy ra <=> x=2 và y=-1

a.

\(A=\left(x^4+y^2+1-2x^2y+2x^2-2y\right)+2\left(y^2-2y+1\right)+2026\)

\(A=\left(x^2-y+1\right)^2+2\left(y-1\right)^2+2026\ge2026\)

\(A_{min}=2026\) khi \(\left(x;y\right)=\left(0;1\right)\)

b.

Đặt \(x-1=t\Rightarrow x=t+1\)

\(\Rightarrow A=\dfrac{3\left(t+1\right)^2-8\left(t+1\right)+6}{t^2}=\dfrac{3t^2-2t+1}{t^2}=\dfrac{1}{t^2}-\dfrac{2}{t}+3=\left(\dfrac{1}{t}-1\right)^2+2\ge2\)

\(A_{min}=2\) khi \(t=1\Rightarrow x=2\)

\(A=\dfrac{3x^2-8x+6}{x^2-2x+1}=\dfrac{3x^2-8x+6}{\left(x-1\right)^2}=\dfrac{2\left(x-1\right)^2+\left(x-2\right)^2}{\left(x-1\right)^2}=2+\dfrac{\left(x-2\right)^2}{\left(x-1\right)^2}\ge2\)

Dấu \("="\Leftrightarrow x=2\)