Question

giải hệ phương trình sau

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

Answer 1

Lời giải:
HPT \(\Leftrightarrow \left\{\begin{matrix} x^2+y^2+2xy-(x+y)=3xy\\ 2x^3=x^2+y^2+2xy\end{matrix}\right.\)

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

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

\(\Rightarrow (x+y)^3-2x^3=3xy(x+y)\)

\(\Leftrightarrow x^3-y^3=0\)

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

TH1: $x-y=0\Rightarrow x=y$. Thay vào PT(1):

\(2x^2-2x=x^2\Leftrightarrow x^2-2x=0\Rightarrow x=0\) hoặc $x=2$

Nếu $x=0\Rightarrow y=0$

Nếu $x=2\Rightarrow y=2$

TH2: \(x^2+xy+y^2=0\Leftrightarrow (x+\frac{y}{2})^2+\frac{3}{4}y^2=0\)

\(\Rightarrow x+\frac{y}{2}=y=0\Rightarrow x=y=0\)

Vậy $(x,y)=(0,0); (2,2)$