4x^2+y^2-4x+10y+26=0

<=>4x²-4x+1+y²+10x+25=0

<=>(2x-1)²+(y+5)²=0

<=>2x-1=0 và y+5=0

<=>x=1/2 và y=-5

          \(4x^2+y^2-4x+10y+26=0\)

\(\Leftrightarrow\)\(\left(4x^2-4x+1\right)+\left(y^2+10y+25\right)=0\)

\(\Leftrightarrow\)\(\left(2x-1\right)^2+\left(y+5\right)^2=0\)

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

\(\Leftrightarrow\)\(\hept{\begin{cases}x=\frac{1}{2}\\y=-5\end{cases}}\)

Vậy..

4x^2 +y^2 -4x+10y+26=0 

4x^2-4x+1 +y^2+10y+25 =0

(2x-1)^2+(y+5)^2=0 

suy ra 2x-1=0 và y+5=0 

 x=1/2,y=-5

4x² + y² - 4x + 10y + 26 = 0

<=> ( 4x² - 4x + 1 ) + ( y² + 10y + 25 ) = 0

<=> ( 2x - 1 )² + ( y + 5 )² = 0

<=> \(\hept{\begin{cases}2x-1=0\\y+5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-5\end{cases}}\)

4x² - 4x + y² + 10y + 26 = 0

<=> [(2x)² - 2.2x + 1] + (y² + 2.5y + 5²) = 0

<=> (2x - 1)² + (y + 5)² = 0

Mà \(\left(2x-1\right)^2\ge0\forall x;\left(y+5\right)^2\ge0\forall y\)

nên \(\left\{{}\begin{matrix}\left(2x-1\right)^2=0\\\left(y+5\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-5\end{matrix}\right.\)

\(4x^2-4x+y^2+10y+26=0\)

=> \(4x^2-4x+y^2+10y+25+1=0\)

=> \(\left(4x^2-4x+1\right)+\left(y^2+10y+25\right)=0\)

=> \(\left(2x-1\right)^2+\left(y+5\right)^2=0\)

Ta thấy:

\(\left(2x-1\right)^2\ge0\)

\(\left(y+5\right)^2\ge0\)

=>\(\left(2x-1\right)^2+\left(y+5\right)^2\ge0\)

Mà \(\left(2x-1\right)^2+\left(y+5\right)^2=0\)

=>\(\left\{{}\begin{matrix}2x-1=0\\y+5=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-5\end{matrix}\right.\)

Vậy x = \(\dfrac{1}{2}\); y = -5

a/

\(\Leftrightarrow x^2-2xy+y^2+2x^2+10x+26=0\)

\(\Leftrightarrow\left(x-y\right)^2+2\left(x-\frac{5}{2}\right)^2+\frac{27}{2}=0\)

\(VT>0\Rightarrow\) ko tồn tại x; y thỏa mãn

b/

\(\Leftrightarrow4x^2-4x+1+3\left(y^2+10y+25\right)+2=0\)

\(\Leftrightarrow\left(2x-1\right)^2+3\left(y+5\right)^2+2=0\)

\(\Rightarrow\) Không tồn tại x; y thỏa mãn

c/

\(3\left(x^2-4x+4\right)+6\left(y^2-\frac{10}{3}y+\frac{25}{9}\right)+\frac{34}{3}=0\)

\(\Leftrightarrow3\left(x-2\right)+6\left(y-\frac{5}{3}\right)^2+\frac{34}{3}=0\)

Không tồn tại x; y thỏa mãn

\(4x^2+y^2-12x+10y+34=0\)

\(\Leftrightarrow4x^2-12x+9+y^2+10y+25=0\)

\(\Leftrightarrow\left(2x-3\right)^2+\left(y+5\right)^2=0\left(1\right)\)

mà \(\left\{{}\begin{matrix}\left(2x-3\right)^2\ge0,\forall x\\\left(y+5\right)^2\ge0,\forall y\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\left(2x-3\right)^2=0\\\left(y+5\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\y+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-5\end{matrix}\right.\)

Ta có : \(4x^2+y^2-12x+10y+34=0\)

\(\Leftrightarrow4x^2-12x+9+y^2+10y+25=0\)

\(\Leftrightarrow\left(2x-3\right)^2+\left(y+5\right)^2=0\left(1\right)\)

Ta thấy : \(\left(2x-3\right)^2;\left(y+5\right)^2\ge0\)

Nên để (1) thoả mãn : 

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\y+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-5\end{matrix}\right.\)

Vậy........

4\(x^2\) + y² - 12\(x\) + 10y + 34 = 0

(4\(x^2\) - 12\(x\) + 9) + (y² + 10y + 25) = 0

(2\(x\) - 3)² + (y + 5)² = 0

(2\(x\) - 3)² ≥ 0 ∀ \(x\); (y + 5)² ≥ 0 ∀ y

(2\(x-3\))² + (y + 5)² = 0 ⇔ \(\left\{{}\begin{matrix}2x-3=0\\y+5=0\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-5\end{matrix}\right.\)

Kl: (\(x;y\)) = ( \(\dfrac{3}{2}\); -5)

\(\Leftrightarrow4x^2-12x+9+y^2+10y+25=0\)

\(\Leftrightarrow\left(2x-3\right)^2+\left(y+5\right)^2=0\) (1)

Do \(\left(2x-3\right)^2\ge0\) và \(\left(y+5\right)^2\ge0\)

\(\Rightarrow\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\left(2x-3\right)=0\\y+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-5\end{matrix}\right.\)