can gap dap an ve he pt

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N

ngocthao1995

PT{(xy)2=1(xy)2x2y+xxy2y1=0{(xy)2=1(xy)2xy(xy)+(xy)1=0PT \Leftrightarrow \left\{ \begin{array}{l} (x-y)^2=1-(xy)^2\\ x^2y+x-xy^2-y-1=0 \end{array} \right.\\\left\{ \begin{array}{l} (x-y)^2=1-(xy)^2 \\ xy(x-y)+(x-y)-1=0 \end{array} \right.
Đặt {xy=uxy=v\left\{ \begin{array}{l} x-y=u \\ xy=v \end{array} \right.
{u2=1v2uv+u1=0\Rightarrow \left\{ \begin{array}{l} u^2=1-v^2 \\ uv+u-1=0 \end{array} \right.
Giải tìm u,v --> nghiệm
 
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