a) -( x-y)² - (x-1)² -2 

GTLN = -2

\(A=\left(x^2-4x+4\right)+4=\left(x-2\right)^2+4\ge4\)

\(minA=4\Leftrightarrow x=2\)

\(B=\left(4x^2-12x+9\right)+2=\left(2x-3\right)^2+2\ge2\)

\(minB=2\Leftrightarrow x=\dfrac{3}{2}\)

\(C=3\left(x^2+2x+1\right)-8=3\left(x+1\right)^2-8\ge-8\)

\(minC=-8\Leftrightarrow x=-1\)

\(D=-\left(x^2-2x+1\right)-4=-\left(x-1\right)^2-4\le-4\)

\(maxD=-4\Leftrightarrow x=1\)

\(E=-\left(4x^2-6x+\dfrac{9}{4}\right)-\dfrac{11}{4}=-\left(2x-\dfrac{3}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\)

\(maxA=-\dfrac{11}{4}\Leftrightarrow x=\dfrac{3}{4}\)

\(F=-2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{55}{8}=-2\left(x-\dfrac{1}{4}\right)^2-\dfrac{55}{8}\le-\dfrac{55}{8}\)

\(maxF=-\dfrac{55}{8}\Leftrightarrow x=\dfrac{1}{4}\)

\(G=\left(x^2-4xy+4y^2\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-2y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(maxG=\dfrac{3}{4}\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-\dfrac{1}{2}\end{matrix}\right.\)

\(H=-\left(x^2-2x+1\right)-\left(y^2+4y+4\right)+16=-\left(x-1\right)^2-\left(y+2\right)^2+16\le16\)

\(maxH=16\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(A=\left(x^2+2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{5}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\\ A_{min}=-\dfrac{5}{4}\Leftrightarrow x=-\dfrac{3}{2}\\ B=\left(x^2+2xy+y^2\right)+\left(x^2+6x+9\right)+3\\ B=\left(x+y\right)^2+\left(x+3\right)^2+3\ge3\\ B_{min}=3\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-3\end{matrix}\right.\\ C=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1\le1\\ C_{max}=1\Leftrightarrow x=1\)

1. \(A=2x^2-6x-2xy+y^2+10\)

\(\Leftrightarrow A=\left(x^2-2xy+y^2\right)+\left(x^2-6x+9\right)+1\)

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

Vì \(\left(x-y\right)^2\ge0\) ; \(\left(x-3\right)^2\ge0\)\(\forall x;y\)

\(\Rightarrow A=\left(x-y\right)^2+\left(x-3\right)^2+1\ge1\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow x=y=3\)

Vậy minA = 1 \(\Leftrightarrow x=y=3\)

2. \(A=5+2xy+14y-x^2-5y^2-2x\)

\(\Leftrightarrow A=-\left(x^2-2xy+y^2+2x-2y+1\right)-\left(4y^2-12y+9\right)+15\)

\(\Leftrightarrow A=-\left(x-y+1\right)^2-\left(2y-3\right)^2+15\)

Vì \(\left\{{}\begin{matrix}\left(x-y+1\right)^2\ge0\\\left(2y-3\right)^2\ge0\end{matrix}\right.\)\(\forall x;y\)

\(\Rightarrow A=-\left(x-y+1\right)^2-\left(2y-3\right)^2+15\le15\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y+1\right)^2=0\\\left(2y-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\y=\frac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{3}{2}\end{matrix}\right.\)

Vậy maxA = 15 \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{3}{2}\end{matrix}\right.\)

1.

Vì ;

Dấu "=" xảy ra

Vậy minA = 1

2.

Vì

Dấu "=" xảy ra

Vậy maxA = 15