x² -2x+9y²-6y+2=0

=> x² - 2x.1 + 1² + (3y)² - 2.3y.1 + 1² = 0

=> ( x - 1 )² + ( 3y - 1 )² = 0

Vì ( x -1 )² \(\ge\)0

( 3y - 1 )² \(\ge\)0

=> ( x - 1 )² + ( 3y - 1 ) ² \(\ge\)0

Dấu " = "  xảy ra khi :

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

Vậy \(x=1\) và \(y=\frac{1}{3}\)

Study well 

\(x^2-2x+9y^2-6y+2=0\)

\(\Rightarrow x^2-2x+1+\left(3y\right)^2-6y+1=0\)

\(\Rightarrow\left(x-1\right)^2+\left(3y-1\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(3y-1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=\frac{1}{3}\end{cases}}}\)

Vậy.......

x² -2x+9y²-6y+2=0

( x² -2x + 1 ) + ( 3y )² - 6y + 1 = 0

( x-1 )²  + ( 3y-1)² = 0

Mà ( x-1 )²  \(\ge\)0với mọi x

     (  3y-1)²  \(\ge\) 0 với mọi y 

nên : \(\hept{\begin{cases}x-1=0\\3y-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=\frac{1}{3}\end{cases}}}\)

                       Vậy x=1 ; y = 1/3 

      a)     x² + y² - 2x + 4y + 5 = 0

\(\Leftrightarrow\)( x² - 2x + 1 ) + ( y² + 4y + 4 ) = 0

\(\Leftrightarrow\)( x - 1 )² + ( y + 2 )² = 0

\(\Rightarrow\)x - 1 = 0 và y + 2 = 0

\(\Rightarrow\)x = 1 và y = - 2

Vậy : x = 1 và y = - 2

b) 4x² + 9y² - 4x - 6y + 2 = 0

\(\Leftrightarrow\)[ ( 2x )² - 4x + 1 ] + [ ( 3y )² - 6y + 1 ] = 0

\(\Leftrightarrow\)( 2x - 1 )² + ( 3y - 1 )² = 0

\(\Rightarrow\)2x - 1 = 0 và 3y - 1 = 0

\(\Rightarrow\)x = 1 / 2 và y = 1 / 3

Vậy : x = 1 / 2 và y = 1 / 3

a) \(x^2+y^2-2x+4y+5=0\)

    \(x^2+y^2-2x+4y+1+4=0\)

    \(\left(x^2-2x+1\right)\left(y^2+4y+4\right)=0\)

     \(\left(x-1\right)^2\left(y+2\right)^2=0\)

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

b) \(4x^2+9y^2-4x-6y+2=0\)

    \(\left(4x^2-4x+1\right)\left(9y^2-6y+1\right)=0\)

    \(\left(2x-1\right)^2\left(3y-1\right)^2=0\)

    \(\Rightarrow\orbr{\begin{cases}2x-1=0\\3y-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{3}\end{cases}}}\)

\(a,4x^2+9y^2+4x-24y+17=0\)

\(\Rightarrow\left(4x^2+4x+1\right)+\left(9y^2-24y+16\right)=0\)

\(\Rightarrow\left(2x+1\right)^2+\left(3y-4\right)^2=0\)

\(\left(2x+1\right)^2\ge0;\left(3y-4\right)^2\ge0\)

\(\Rightarrow\hept{\begin{cases}\left(2x+1\right)^2=0\\\left(3y-4\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}2x+1=0\\3y-4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{4}{3}\end{cases}}}\)

a) Ta có: \(Q=-x^2-y^2+4x-4y+2=-\left(x^2+y^2-4x+4y-2\right)\)

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

\(=-\left[\left(x-2\right)^2+\left(y+2\right)^2\right]+10\le10\forall x,y\)

Vậy Max_Q=10 khi x=2, y=-2

b) +Ta có: \(A=-x^2-6x+5=-\left(x^2+6x-5\right)=-\left(x^2+6x+9-14\right)\)

\(=-\left(x^2+6x+9\right)+14=-\left(x+3\right)^2+14\le14\forall x\)

Vậy Max_A=14 khi x=-3

+Ta có: \(B=-4x^2-9y^2-4x+6y+3=-\left(4x^2+9y^2+4x-6y-3\right)\)

\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)

\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2\right]+5\le5\forall x,y\)

Vậy Max_B=5 khi x=-1/2, y=1/3

c) Ta có: \(P=x^2+y^2-2x+6y+12=x^2-2x+1+y^2+6y+9+2\)

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

Vậy Min_P=2 khi x=1, y=-3

\(x^2+2x+9y^2+6y+15\)

\(=\left(x^2+2x+1\right)+\left(9y^2+6y+1\right)+13\)

\(=\left(x+1\right)^2+\left(3y+1\right)^2+13\ge13>0\)

Ta có:

\(x^2+2x+9y^2-6y+3=0\)

\(\Leftrightarrow\left(x^2+2x+1\right)+\left(9y^2-6y+1\right)+1=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(3y-1\right)^2+1=0\)

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

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

\(\Rightarrow\left(x+1\right)^2+\left(3y-1\right)^2+1\ge1>0\)

Vậy không tồn tại x và y để thỏa mãn đề bài...!