Question

Tìm x, y : \(4x^2-4x+y^2+2y+2=0\)

Answer 1

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

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

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

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

Lại có: \(\left\{\begin{matrix}\left(2x-1\right)^2\ge0\\\left(y+1\right)^2\ge0\end{matrix}\right.\)\(\forall x;y\)

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

Vậy x = 1/2 ; y = -1