\(C=1-6y-5y^2-12xy-9x^2\)

\(\Rightarrow C=-4y^2-12xy-9x^2-y^2-6y+1\)

\(\Rightarrow C=-\left(4y^2+12xy+9x^2\right)-\left(y^2+6y+9\right)+1+9\)

\(\Rightarrow C=-\left(2y-3x\right)^2-\left(y+3\right)^2+10\)

mà \(\left\{{}\begin{matrix}-\left(2y-3x\right)^2\le0,\forall x;y\\-\left(y+3\right)^2\le0,\forall y\end{matrix}\right.\)

\(\Rightarrow C=-\left(2y-3x\right)^2-\left(y+3\right)^2+10\le10\)

\(\Rightarrow GTLN\left(C\right)=10\left(tạix=-2;y=-3\right)\)

\(9x^2+5y^2-6xy-6x-6y+20\)

\(=9x^2+y^2+1-6x+2y-6xy+4y^2-8y+4+15\)

\(=\left(3x-y-1\right)^2+4\left(y-1\right)^2+15\ge15\)

Dấu \(=\)khi \(\hept{\begin{cases}3x-y-1=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{2}{3}\\y=1\end{cases}}\).

Answer:

\(B=-5x^2-5y^2+8x-6y-1\)

\(\Rightarrow B=\left(-5x^2+8x-\frac{16}{5}\right)+\left(-5y^2-6y-\frac{9}{5}\right)+4\)

\(\Rightarrow B=-5\left(x-\frac{4}{5}\right)^2-5\left(y+\frac{3}{5}\right)^2+4\)

Có:

\(\hept{\begin{cases}\left(x-\frac{4}{5}\right)^2\ge0\forall x\Rightarrow-5\left(x-\frac{4}{5}\right)^2\le0\\\left(y+\frac{3}{5}\right)^2\ge0\forall y\Rightarrow-5\left(y+\frac{3}{5}\right)^2\le0\end{cases}}\)

Do vậy:

\(-5\left(x-\frac{4}{5}\right)^2-5\left(y+\frac{3}{5}\right)^2+4\le4\forall x;y\) hay \(B\le4\)

Vậy "=" xảy ra khi:

\(\hept{\begin{cases}x-\frac{4}{5}=0\\y+\frac{3}{5}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{4}{5}\\y=\frac{-3}{5}\end{cases}}\)

Vậy giá trị lớn nhất của biểu thức \(B=4\) khi \(\hept{\begin{cases}x=\frac{4}{5}\\y=\frac{-3}{5}\end{cases}}\)

\(C=-5x^2-2xy-2y^2+14x+10y-1\)

\(\Rightarrow5C=\left(-25x^2-10xy-y^2+70x+14y-49\right)+\left(-9y^2+36y-36\right)+80\)

\(\Rightarrow5C=-\left(5x+y-7\right)^2-9\left(y-2\right)^2+80\)

\(\Rightarrow C=-\frac{1}{5}\left(5x+y-7\right)^2-\frac{9}{2}\left(y-2\right)^2+16\)

Có:

\(\hept{\begin{cases}\left(5x+y-7\right)^2\ge0\forall x;y\Rightarrow-\frac{1}{5}\left(5x+y-7\right)^2\le0\\\left(y-2\right)^2\ge0\forall y\Rightarrow-\frac{9}{5}\left(y-2\right)^2\le0\end{cases}}\)

Do vậy:

\(-\frac{1}{5}\left(5x+y-7\right)^2-\frac{9}{5}\left(y-2\right)^2+16\le16\) hay \(C\le16\)

Dấu "=" xảy ra khi: 

\(\hept{\begin{cases}5x+y-7=0\\y-2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}\)

Vậy giá trị lớn nhất của biểu thức \(C=16\) khi \(\hept{\begin{cases}x=1\\y=2\end{cases}}\)

a) A = x² - 6x + 13 = x² - 2.x.3 + 3³ +4 = (x-3)^{2 +} 4 >= 4 suy ra minA=4 
mấy câu kia giải tương tự

C= \(\frac{1}{2}\)

Lời giải:
$A=(9x^2-6xy+y^2)+5y^2-6x-6y+20$

$=(3x-y)^2-2(3x-y)+4y^2-8y+20$

$=(3x-y)^2-2(3x-y)+1+(4y^2-8y+4)+15$

$=(3x-y-1)^2+(2y-2)^2+15\geq 15$

Vậy $A_{\min}=15$.

Giá trị này đạt tại $3x-y-1=2y-2=0$

$\Leftrightarrow (x,y)=(\frac{2}{3},1)$