Решите систему уравнений ху + х - у = 7, х^2у - ху^2=6

Question 1

Question 2

$\left \{ {{xy+x-y=7} \atop {x^{2}y-xy^{2}=6}} \right.\\\\\left \{ {{xy+(x-y)=7} \atop {xy(x-y)=6}} \right.$

Сделаем замену :

xy = m , x - y = n

$\left \{ {{m+n=7} \atop {m*n=6}} \right. \\\\\left \{ {{m=7-n} \atop {(7-n)*n=6}} \right.\\\\\left \{ {{m=7-n} \atop {-n^{2}+7n-6=0 }} \right.\\\\\left \{ {{m=7-n} \atop {n^{2}-7n+6=0 }} \right.\\\\\left \{ {{m=7-n} \atop {\left[\begin{array}{ccc}n_{1}=1 \\n_{2} =6\end{array}\right }} \right.$

$\left \{ {{\left[\begin{array}{ccc}m_{1}=6 \\m_{2}=1 \end{array}\right } \atop {\left[\begin{array}{ccc}n_{1}=1 \\n_{2}=6 \end{array}\right }} \right.$

$1)\left \{ {{xy=6} \atop {x-y=1}} \right.\\\\\left \{ {{x=y+1} \atop {(y+1)*y=6}} \right.\\\\\left \{ {{x=y+1} \atop {y^{2}+y-6=0 }} \right.\\\\\left \{ {{x=y+1} \atop {\left[\begin{array}{ccc}y_{1}=-3 \\y_{2}=2 \end{array}\right }} \right. \\\\\left \{ {{\left[\begin{array}{ccc}x_{1}=-2 \\x_{2}=3 \end{array}\right } \atop {\left[\begin{array}{ccc}y_{1}=-3 \\y_{2}=2 \end{array}\right}} \right.$

$2)\left \{ {{xy=1} \atop {x-y=6}} \right.\\\\\left \{ {{x=y+6} \atop {xy=1}} \right.\\\\\left \{ {{x=y+6} \atop {(y+6)*y=1}} \right. \\\\\left \{ {{x=y+6} \atop {y^{2}+6y-1=0 }} \right.\\\\y^{2} +6y-1=0\\\\D=6^{2}-4*(-1)=36+4=40=(2\sqrt{10})^{2}\\\\y_{1}=\frac{-6-2\sqrt{10}}{2}=-3-\sqrt{10}\\\\y_{2}==\frac{-6+2\sqrt{10}}{2}=\sqrt{10}-3\\\\x_{1} =-3-\sqrt{10}+6=3-\sqrt{10}\\\\x_{2}=\sqrt{10}-3+6=\sqrt{10}+3$

$Otvet:\boxed{(-2;-3),(3;2),(3-\sqrt{10} ;-3-\sqrt{10} ),(\sqrt{10}+3;\sqrt{10}-3)}$