Ta có : 

\(x^2+4y^2-4x-4y+5=0\)

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

\(\Leftrightarrow\)\(\left[x^2-2.x.2+2^2\right]+\left[\left(2y\right)^2-2.2y.1+1^2\right]=0\)

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

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

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

Vậy \(x=2\) và \(y=\frac{1}{2}\)

Chúc bạn học tốt ~ 

        \(x^2+4y^2-4x-4y+5=0\)

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

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

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

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

Vậy

\(x^2+4y^2-4x-4y+5=0\)( 1 )

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

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

Mà  \(\left(x-2\right)^2\ge0\forall x\)

       \(\left(2y-1\right)^2\ge0\forall y\)

\(\Rightarrow\left(1\right)\ge0\forall x;y\)

Dấu "=" xảy ra khi :

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

Vậy ...

a) \(x^2-10x+4y^2-4y+26=0\)

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

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

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

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

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

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

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

a) \(x^2+4y^2-6x-4y+10=0\)

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

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

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

b) \(2x^2+y^2+2xy-10x+25=0\)

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

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

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

c) \(x^2+2xy+4x-4y-2xy+5=0\)

\(\Leftrightarrow x^2-4x-4y+5=0\)

Xem lại đề câu c).

a) x² + 4y² - 6x - 4y + 10 = 0

<=> x² - 6x + 9 + 4y² - 4y + 1 = 0

<=> ( x - 3 )² + ( 4y - 1 )² = 0

<=> \(\hept{\begin{cases}x-3=0\\4y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=\frac{1}{4}\end{cases}}\)

b) 2x² + y² + 2xy - 10x + 25 = 0

<=> x² + 2xy + y² + x² - 10x + 25 = 0

<=> ( x + y )² + ( x - 5 )² = 0

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

c) Xem lại đề 

\(x^2+4y^2-4x-4y+5=0\)

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

\(\Leftrightarrow\left(x^2-2\cdot x\cdot2+2^2\right)+\left[\left(2y\right)^2-2\cdot2y\cdot1+1^2\right]=0\)

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

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

Vậy....

nhom (x2-4x+5)+(4y2-4y)

\(x^2+4y^2-4x-4y+5=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}\left(x-2\right)^2=0\\\left(2y-1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-2=0\\2y-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\y=\frac{1}{2}\end{cases}}\)

Vậy x = 2 ; y = \(\frac{1}{2}\)

b) 4x^2+y^2-20x-2y+26=0;
(4x^2-20x+25)+(y^2-2y+1)=(2x-5)^2+(y-1)^2=0
<=>x=5/2; y=1