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

a: \(x^2+3y^2-4x+6y+7=0\)

\(\Leftrightarrow x^2-4x+4+3y^2+6y+3=0\)

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

\(\Leftrightarrow\left(x,y\right)=\left(-2;1\right)\)

câu 2 : x^2-6x+9+y^2+10y+25+(4z-1)^2=0

(y+5)^2=0 => y=-5

câu này căng đấy nhưng tớ sẽ cố giúp

thế này: 

4x² +y²-4x+10y+26=0.

= 4x\(^2\)- 4x+1+y\(^2\)+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..

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