Помогите ж) з) пожалуйста

[latex]1)\quad \left \{ {{x+y=xy} \atop {x^2+y^2=4xy}} \right. \; \left \{ {x+{y=xy} \atop {(x+y)^2-2xy=4xy}} \right. \; \left \{ {{x+y=xy} \atop {(xy)^2-6xy=0}} \right. \\\\t=xy\; ,\; \; t^2-6t=0\; \; \to \; \; t(t-6)=0\; ,\; t_1=0\; ,\; t_2=6\\\\ a)\; \; \left \{ {{x+y=0} \atop {xy=0}} \right. \; \; ili\; \; b)\; \; \left \{ {{x+y=6} \atop {xy=6}} \right. \\\\a)\quad x=0,\; y=0[/latex] [latex]b)\quad \left \{ {{y=6-x} \atop {x(6-x)=0}} \right. \; \left \{ {{y=6-x} \atop {x_1=0,\; x_2=6}} \right.\Rightarrow \quad \left \{ {{y_1=6,\; y_2=0} \atop {x_1=0,\; x_2=6}} \right. \\\\Proverka:[/latex] [latex]Otvet:\; \; (0,0)\; .\\\\2)\quad \left \{ {{x-y=0,25xy} \atop {x^2+y^2=2,5xy}} \right. \; \left \{ {{x-y=\frac{1}{4}xy} \atop {(x-y)^2+2xy=2,5xy}} \right. \; \left \{ {{x-y=\frac{1}{4}xy} \atop {(\frac{1}{4}xy)^2-\frac{1}{2}xy=0}} \right. \\\\ \left \{ {{x-y=\frac{1}{4}xy} \atop {\frac{1}{16}(xy)^2-\frac{1}{2}xy=0}} \right. [/latex] [latex]t=xy\; ,\; \; \frac{t^2}{16}-\frac{t}{2}=0\, |\cdot 16\\\\t^2-8t=0\; ,\; \; t(t-8)=0\; ,\; \; t_1=0\; ,\; \; t_2=8\\\\xy=0\; \; ili\; \; xy=8\\\\a)\; xy=0,x-y=\frac{1}{4}xy\; \; \to \; \; x-y=0,\; \; x=y\; \left \{ {{x=y} \atop {xy=0}} \right. \to \; x=0\; ,\; y=0\\\\b)\; xy=8,x-y=\frac{8}{4}\; \to \; x-y=2,\; y=x-2\; \left \{ {{y=x-2} \atop {xy=8}} \right. \\\\x(x-2)=8\; ,\; \; x^2-2x-8=0\; ,\\\\x_1=-2\; ,\; \; x_2=4\\\\y_1=x_1-2=-4\; ,\; \; y_2=x_2-2=2[/latex] [latex]Otvet:\; \; (0,0)\; ,\; (-2,-4)\; ,\; (4,2)\; .[/latex] P.S.  Ответ пишем только после прверки.