Question

поможітььььь розв'яжіть систему рівнянь!!!!​

Answer 1

$\displaystyle\bf\\\left \{ {{\dfrac{x}{y} +\dfrac{6y}{x}=5 } \atop {x^{2} +4xy-3y^{2} =18}} \right. \\\\\\\frac{x}{y} +6\cdot\frac{y}{x} =5\\\\\\\frac{x}{y} =m \ \ \ \Rightarrow \ \ \ \frac{y}{x} =\frac{1}{m} \\\\\\m+\frac{6}{m}-5=0\\\\\\m^{2} -5m+6=0 \ \ ; \ \ m\neq 0\\\\Teorema \ Vieta \ :\\\\m_{1} +m_{2} =5\\\\m_{1} \cdot m_{2} =6\\\\m_{1} =2 \ \ ; \ \ m_{2} =3$

$\displaystyle\bf\\1)\\\\\frac{x}{y} =2 \ \ \ \Rightarrow \ \ x=2y\\\\x^{2} +4xy-3y^{2} =18\\\\(2y)^{2} +4y\cdot 2y-3y^{2} =18\\\\4y^{2} +8y^{2} -3y^{2} =18\\\\9y^{2} =18\\\\y^{2} =2\\\\y_{1} =-\sqrt{2} \ \ \ ; \ \ \ y_{2} =\sqrt{2} \\\\x_{1} =-2\sqrt{2} \ \ \ ; \ \ \ x_{2} =2\sqrt{2} \\\\2)\\\\\frac{x}{y} =3 \ \ \ \Rightarrow \ \ x=3y\\\\x^{2} +4xy-3y^{2} =18\\\\(3y)^{2} +4y\cdot 3y-3y^{2} =18\\\\9y^{2} +12y^{2} -3y^{2} =18\\\\18y^{2} =18\\\\y^{2} =1$

$\displaystyle\bf\\y_{1} =-1 \ \ \ ; \ \ \ y_{2} =1 \\\\x_{1} =-3 \ \ \ ; \ \ \ x_{2} =3 \\\\\\Otvet \ : \ (-2\sqrt{2} \ ; \ -\sqrt{2} ) \ , \ (2\sqrt{2} \ ; \ \sqrt{2} ) \ , \ (-3 \ ; \ -1) \ , \ (3 \ ; \ 1)$