- Trừ hai pt ta được :\(x^3-y^3-x^2+y^2+x-y+1-1=2y-2x\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x^2+xy+y^2-x-y+3=0\end{matrix}\right.\)

TH1 : x = y

PT ( I ) TT : \(x^3-x^2+x+1-2x=x^3-x^2-x+1=0\)

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

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

\(\Leftrightarrow x=y=\pm1\)

TH₂ : \(x^2+xy+y^2-x-y+3=0\)

\(\Leftrightarrow x^2+\dfrac{y^2}{4}+\dfrac{1}{4}+xy-x-\dfrac{1}{2}y+\dfrac{3}{4}y^2-\dfrac{1}{2}y+\dfrac{11}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}y-\dfrac{1}{2}\right)^2+\left(\dfrac{y\sqrt{3}}{2}-\dfrac{1}{2\sqrt{3}}\right)^2+\dfrac{8}{3}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}y-\dfrac{1}{2}\right)^2+\left(\dfrac{y\sqrt{3}}{2}-\dfrac{1}{2\sqrt{3}}\right)^2=-\dfrac{8}{3}\left(VL\right)\)

Vậy ....

2x + y = 1 <=> y = 1 - 2x

Thế vào pt còn lại thì:

x^2 + (1 - 2x)^2 - x(1 - 2x) = 3

<=> x^2 + 4x^2 - 4x + 1 - x + 2x^2 - 3 = 0

<=> 7x^2 - 5x - 2 = 0

<=> (x - 1)(7x + 2) = 0

<=> x = 1 hoặc x = -2/7

Với x = 1 <=> y = 1 - 2.1 = -1

Với x = -2/7 <=> y = 1 - 2.(-2/7) = 11/7