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

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

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

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

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 $3x-y-1=2y-2=0$

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

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

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

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

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

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

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

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

\(A=\sqrt{1-6x+9x^2}+\sqrt{9x^2-12x+4}\)

\(A=\sqrt{1^2-2\cdot3x\cdot1+\left(3x\right)^2}+\sqrt{\left(3x\right)^2-2\cdot2\cdot3x+2^2}\)

\(A=\sqrt{\left(1-3x\right)^2}+\sqrt{\left(3x-2\right)^2}\)

\(A=\left|1-3x\right|+\left|3x-2\right|\)

\(A=\left|1-3x+3x-2\right|\)

\(A=\left|-1\right|=1\)

Dấu "=" xảy ra \(\left(1-3x\right)\left(3x-2\right)\ge0\)

\(\Rightarrow\dfrac{1}{3}\le x\le\dfrac{2}{3}\)

Vậy: \(A_{min}=1\) khi \(\dfrac{1}{3}\le x\le\dfrac{2}{3}\)

A=9x^2+18xy-12x+13y^2-24y+5

\(=\left(3x\right)^2+2.3.3xy-2.3x.2+9y^2+4y^2-12y-12y+4+9-8\)

\(=\left[\left(3x\right)^2+\left(3y\right)^2+2^2+2.3x.3y+2.3x.2+2.3y.2\right]+\left[\left(2y\right)^2-2.2y.3+9\right]-8\)

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

Vậy \(MinA=-8\Leftrightarrow\hept{\begin{cases}\left(3x+3y+2\right)^2=0\\\left(2y-3\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}3x+3y+2=0\\2y-3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-6,5\\y=1,5\end{cases}}}\)

5

5

\(\text{A=9x^2+18xy-12x+13y^2-24y+5}\)

\(=\left[\left(3x\right)^2+\left(3y\right)^2+2^2-12x+18xy-12y\right]+\left[\left(2y\right)^2-2.2y.3+9\right]-8\)

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

Vậy \(MinA=-8\Leftrightarrow\hept{\begin{cases}3x+3y-2=0\\2y-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1,5\\y=1,5\end{cases}}}\)

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

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

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

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

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

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