\(x^2+4y^2-4x-4y+5=0\)
\(\Leftrightarrow x^2-4x+4+4y^2-4y+1=0\)
\(\Leftrightarrow\left(x^2-2\cdot x\cdot2+2^2\right)+\left[\left(2y\right)^2-2\cdot2y\cdot1+1^2\right]=0\)
\(\Leftrightarrow\left(x-2\right)^2+\left(2y-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-2=0\\2y-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=\frac{1}{2}\end{cases}}}\)
Vậy....
\(x^2+4y^2-4x-4y+5=0\)
\(\Leftrightarrow\)\(\left(x^2-4x+4\right)+\left(4y^2-4y+1\right)=0\)
\(\Leftrightarrow\)\(\left(x-2\right)^2+\left(2y-1\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-2\right)^2=0\\\left(2y-1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-2=0\\2y-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\y=\frac{1}{2}\end{cases}}\)
Vậy x = 2 ; y = \(\frac{1}{2}\)
\(x^2+4y^2-4x-4y+5=0\)
\(\Leftrightarrow\left(x^2-4x+4\right)+\left(4y^2-4y+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)^2+\left(2y-1\right)^2=0\)
Mà \(\hept{\begin{cases}\left(x-2\right)^2\ge0\\\left(2y-1\right)^2\ge0\end{cases}}\Rightarrow\left(x-2\right)^2+\left(2y-1\right)^2\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\\left(2y-1\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=\frac{1}{2}\end{cases}}\)
Vậy x = 2 và \(y=\frac{1}{2}\)
