от math10.com » 15 Яну 2015, 23:43
[tex]\begin{tabular}{|l}xy-3x=x^2+x-12 \Rightarrow x(y-1)=x^2+3x-12\\y^2-xy+x+y-2=0 \Rightarrow x(y-1)=(y-1)(y+2)\end{tabular}[/tex]
За [tex]y=1\Rightarrow \begin{tabular}{|l}0.x=x^2+3x-12 \\0.x=0.3 \end{tabular} \Rightarrow x_1=\frac{-3+\sqrt{57}}{2}; x_2=\frac{-3-\sqrt{57}}{2}[/tex]
За [tex]y\ne 1 \Rightarrow \begin{tabular}{|l}x(y-1)=x^2+3x-12\\x=\frac{\cancel{(y-1)}(y+2)}{\cancel{(y-1)}}=y+2\Rightarrow y=x-2\end{tabular}[/tex]
[tex]\Rightarrow x(x-3)=x^2+3x-12 \Rightarrow 6x=12 \Rightarrow x=2 \Rightarrow y=0[/tex]