Giải hệ phương trình $\begin{cases}x^2+y^2=xy+x+y\\ x^2-y^2=3\end{cases} $
Đặt $x-y = a, x+y=b$. Ta biết rằng
$\begin{cases}x^2+y^2= \frac{a^2+b^2}{2}\\ xy= \frac{b^2-a^2}{4} \end{cases}$
Lúc đó HPT
$\Leftrightarrow \begin{cases} \frac{a^2+b^2}{2}=\frac{b^2-a^2}{4}+b \\ ab=3 \end{cases}$
$\Leftrightarrow \begin{cases} 3a^2+b^2-4b=0\\ a=\frac{3}{b} \end{cases}$
$\Leftrightarrow \begin{cases} \frac{27}{b^2} +b^2-4b=0\\ a=\frac{3}{b} \end{cases}$
$\Leftrightarrow \begin{cases} b^4-4b^3+27=0\\ a=\frac{3}{b} \end{cases}$
$\Leftrightarrow \begin{cases} (b-3)^2(b^2+2b+3)=0\\ a=\frac{3}{b} \end{cases}$
$\Leftrightarrow \begin{cases} b=3\\ a=1 \end{cases}$
$\Leftrightarrow \begin{cases} x=2\\ y=1 \end{cases}$
Cảm ơn bạn vì lời giải. Chi tiết rất dễ hiểu –  motthoitraique97 27-09-12 12:01 AM
Hãy ấn nút tam giác màu xanh bên cạnh đáp án để chấp nhận nếu như bạn thấy lời giải này chính xác nhé. Thanks! –  Trần Nhật Tân 26-09-12 11:48 PM

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