Giải pt: \(x^2-4x+3=0\\↔x^2-3x-x+3=0\\↔x(x-3)-(x-3)=0\\↔(x-1)(x-3)=0\\↔\left[\begin{array}{1}x-1=0\\x-3=0\end{array}\right.\\↔\left[\begin{array}{1}x=1\\x=3\end{array}\right.\)
Thế \(x=1\) vào pt \(x^2+xy+y^2=1\)
\(→1+y+y^2=1\\↔y(y+1)=0\\↔\left[\begin{array}{1}y=0\\y+1=0\end{array}\right.\\↔\left[\begin{array}{1}y=0\\y=-1\end{array}\right.\)
Thế \(x=3\) vào pt \(x^2+xy+y^2=1\)
\(→y^2+3y+9=1\\↔y^2+3y+8=0\\↔y^2+2.\dfrac{3}{2}y+\dfrac{9}{4}+\dfrac{23}{4}=0\\↔(y+\dfrac{3}{2})^2+\dfrac{23}{4}=0\)
Vì \( (y+\dfrac{3}{2})^2\ge 0→(y+\dfrac{3}{2})^2+\dfrac{23}{4}>0\)
\(→x=3\) thì pt \(x^2+xy+y^2\) vô nghiệm
Suy ra: \( (x;y)=\{(1;0);(1;-1)\}\)
Vậy hệ pt có nghiệm \( (x;y)=\{(1;0);(1;-1)\}\)
