\(\Leftrightarrow\left(x^2+\dfrac{y^2}{4}+\dfrac{9}{4}+xy-3x-\dfrac{3y}{2}\right)+\dfrac{3}{4}\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow\left(x+\dfrac{y}{2}-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\left(y-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{y}{2}-\dfrac{3}{2}=0\\y-1=0\end{matrix}\right.\)
\(\Rightarrow x=y=1\)
<=> \(x^2+y^2+\left(xy-3x\right)-\left(3y-3\right)=0\)
\(\Leftrightarrow x^2+y^2+3x^2\left(y-1\right)-3\left(y-1\right)=0\)
<=> \(x^2+y^2\left(3x^2-3\right)\left(y-1\right)=0\)
\(\Leftrightarrow x^2+y^2.3.\left(x-1\right)\left(x+1\right)\left(y-1\right)=0\)
=> x = 0 ; 1 ; -1 . y =0 ; 1
P/s : ngủ được ròi:3
\(x^2 + y^2 + xy - 3x - 3y + 3 =0\)
⇔ \(2x^2 + 2y^2 + 2xy-6x-6y+6=0\)
⇔ \((x^2+y^2 +4+2xy-4x-4y) + (x^2-2x+1) + (y^ -2y+1)=0\)
⇔ \((x+y-2)^2 + (x-1)^2 + (y-1)^2 =0\)
⇔ \(\begin{cases} x+y-2=0\\ x-1=0\\ y-1=0 \end{cases} \) ⇔ \(\begin{cases} x+y=2\\ x=1\\ y=1 \end{cases} \) ⇔\(\begin{cases} x=1\\ y=1 \end{cases} \)
Vậy \((x;y)=(1;1)\)
