ta có \(2B=2x^2-4xy+4y^2+10x\) 

                \(=\left(x^2-4xy+4y^2\right)+\left(x^2+10x+25\right)-25\)

                 \(=\left(x-2y\right)^2+\left(x+5\right)^2-25\)

vì \(\left(x-2y\right)^2>=0;\left(x+5\right)^2>=0\)

=>\(2B>=-25=>b>=-\frac{25}{2}\)

dấu = xảy ra <=> \(\hept{\begin{cases}x=-5\\y=-10\end{cases}}\)

b)   ta có 

\(Q=x^2-6xy+9y^2+x^2-x+\frac{1}{4}+\frac{3}{4}\)

     \(=\left(x-3y\right)^2+\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)

=> Q>=3/4

dấu = xảy ra <=> \(\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{3}{2}\end{cases}}\)

x^2 - 2xy + 6y^2 - 12x + 2y +45 
= x^2 - 2x(y+6) + (y+6)^2 - (y+6)^2 + 6y^2 +2y + 45 
= (x - y - 6)^2 - y^2 - 12y - 36 + 6y^2 + 2y + 45 
= (x - y - 6)^2 + 5y^2 - 10y + 9 
= (x - y - 6)^2 + 5.(y^2 - 2y +1) + 4 
= (x - y - 6)^2 + 5.(y-1)^2 + 4 
=>> MIN = 4 khi (x;y) = {(7;1)}

\(A=x^2-2xy+6y^2-12x+2y+45\)

\(=x^2+y^2+36-2xy-12x+12y+5y^2-10y+5+4\)

\(=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\ge4\)

GTNN A = 4 Khi: \(\hept{\begin{cases}y-1=0\\x-y-6=0\end{cases}\Rightarrow\hept{\begin{cases}y=1\\x=7\end{cases}}}\)

\(A=x^2-2xy+6y^2-12x+2y\)\(+45\)

 \(=x^2+y^2+36-2xy-12x\)\(+12y+5y^2-10y+5+4\)

 \(=\left(x-y-6\right)^2+5\left(y-1\right)^2\)\(+4\ge4\)

GTNN của A là 4 khi \(\hept{\begin{cases}y-1=0\\x-y-6=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}y=1\\x-y=6\end{cases}\Rightarrow\hept{\begin{cases}y=1\\x=7\end{cases}}}\)

Vậy BT A đạt giá trị nhỏ nhất là 4 tại x = 7 và y = 1

\(C=\left(x^2-6xy+9y^2\right)+\left(x^2-2x+1\right)+2017=\left(x-3y\right)^2+\left(x-1\right)^2+2017\)

\(\ge0+0+2017=2017.\Rightarrow C_{min}=2017\Leftrightarrow\hept{\begin{cases}x-1=0\\x-3y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{1}{3}\end{cases}}\)

C= (x-3y)²+(x-1)²+2017 \(\ge2017\)

Min C = 2017

\(C=2x^2+9y^2-6xy-2x+2018\)

\(=x^2-6xy+9y^2+x^2-2x+1+2017\)

\(=\left(x-3y\right)^2+\left(x-1\right)^2+2017\ge2017\forall x;y\)

Dấu"=" xảy ra<=> \(\hept{\begin{cases}\left(x-3y\right)^2=0\\\left(x-1\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x-3y=0\\x=1\end{cases}\Rightarrow y=\frac{1}{3}}\)

\(A=x^2-2xy+6y^2-12x+2y+54\)

\(A=x^2-2xy+y^2-12x+12y+36+5y^2-10y+5+4\)

\(A=\left(x-y\right)^2-2.6\left(x-y\right)+36+5\left(y^2-2y+1\right)+4\)

\(A=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\)

Do: \(\left(x-y-6\right)^2\ge0\forall xy\); \(5\left(y-1\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-y-6\right)^2+5\left(y-1\right)^2\ge0\)

\(\Leftrightarrow A=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\ge4\)

\(\Rightarrow A_{Min}=4\)

Dấu "=" xảy ra khi \(x=7;y=1\)

\(A=x^2-2xy+6y^2-12x+2y+45\)

\(=x^2+y^2+36-2xy-12x+12y+5y^2-10y+5+4\)

\(=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\ge4\)

Gía trị nhỏ nhất : \(A=4\)Khi \(\hept{\begin{cases}y-1=0\\x-y-6=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\x=7\end{cases}}\)

\(A=\left(x-y-6\right)^2+6y^2+2y+45-\left(y^2+12y+36\right)\\ \)

\(A=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\)\(\ge4\)

Amin=4 khi y=1; x=7

A = (x - y - 6)² - 6y² - 2y - 45 - (y² - 12y - 36)

A = (x - y -6)² + 5(y-1)² +4 \(\ge\)4

Amin = 4 khi y = 1; x = 7

#chanh

\(A=\left(x-y-6\right)^2+6y^2+2y+45-\left(y^2+12y+36\right) \)

\(A=\left(x-7-6\right)^2+5\left(y-1^2\right)+4\ge4\)

\(Amin=4\)\(khi\)\(y=1;x=7\)