Ta sẽ tạo các tổng bình phương như sau:

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

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

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

Từ (1) và (2) \(\Rightarrow\hept{\begin{cases}2x-1=0\\3y-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{3}\end{cases}}}\)

Chu Văn Long thiếu câu dấu = xảy ra trước cái từ 1 và 2

Bài 1 :

Câu a : \(A=x^2-3x+5=\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{11}{4}=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}>0\)

Câu b : \(A=x^2-3x+5=\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{11}{4}=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)

Vậy \(GTNN\) của \(A\) là \(\dfrac{11}{4}\) . Dấu \("="\) xảy ra khi \(\left(x-\dfrac{3}{2}\right)^2=0\Leftrightarrow x=\dfrac{3}{2}\)

Bài 2 :

Câu a : \(x^2-6x+y^2-4y+13=0\)

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

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

Do : \(\left(x-3\right)^2\ge0\) and \(\left(y-2\right)^2\ge0\)

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

Vậy \(x=3\) and \(y=2\)

Câu b : \(4x^2-4x+y^2+6y+10=0\)

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

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

Because the : \(\left(2x-1\right)^2\ge0\) and \(\left(y+3\right)^2\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(2x-1\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\)

Vậy \(x=\dfrac{1}{2}\) và \(y=-3\)

4x^{2 -} 4x + 9y² - 6y + 2 = 0
4x^{2 -} 4x + 9y² - 6y + 1 + 1 = 0
(4x^{2 -} 4x + 1) + (9y² - 6y + 1) = 0
(2x - 1)² + (3y - 1)² =0   

 =>   (2x - 1)² = 0          2x - 1 = 0                   2x   = 1              x = 1/2

                           <=>                          <=>                     <=>

        (3y - 1)² = 0         3y - 1 = 0                   3y   = 1              y = 1/3
              Vậy x = 1/2 và y = 1/3

x²+y²-4x+6y+13=0

(x²-4x+4)+(y²+6y+9)=0

(x-2)²+(y+3)²=0

suy ra x-2=0 hoặc y+3=0

*x-2=0=>x=2      *y+3 =0=> y=-3

vậy x=2,y=-3

Ta co pt \(\Leftrightarrow x^2-4x+4+y^2+6y+9=0\)

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

mà \(\hept{\begin{cases}\left(x-2\right)^2\ge0\\\left(y+3\right)^2\ge0\end{cases}}\)

Nên dấu \(=\)xảy ra khi \(\hept{\begin{cases}\left(x-2\right)^2=0\\\left(y+3\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)

Vậy \(x=2;y=-3\)

\(^{x^2-4x+4+y^2+6y+9=0}\)0

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

x=2 va y=-3

ngu cho

\(x^2+y^2-4x+6y+13=0\)

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

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

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

=> ''='' xảy ra khi x = 2; y = -3

Vậy.........

Lời giải:
\(x^2+y^2-4x+6y+13=0\)

\(\Leftrightarrow (x^2-4x+4)+(y^2+6y+9)=0\)

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

Vì \((x-2)^2; (y+3)^2\ge 0, \forall x,y\Rightarrow (x-2)^2+(y+3)^2\geq 0\)

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

\(x^2+y^2-4x+6y+13=0\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2+6y+9\right)=0\)

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

Mà ta lại có: \(\left(x-2\right)^2+\left(y+3\right)^2\ge0\left(\forall x;y\right)\)

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

x² + y² - 4x + 6y + 13 = 0

=> x²+y²-4x+6y+9+4=0

=> (x²-4x+4)+(y²+6y+9)=0

=> (x-2)²+(y+3)²=0

=> \(\left[{}\begin{matrix}x-2=0\\y+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

vậy x=2,y=-3

x² + y² - 4x + 6y + 13 = 0

=> y² + 2.3y + 3² + x² - 2.2x + 2² = 0

=> ( y + 3)² + ( x - 2)² = 0

=> y = -3 ; x = 2