Question

Найти кол-во решений системы уравнений { x^2+y^2=5 { 3xy+2x=2y

Answer 1

$\left \{ {{x^2+y^2=5} \atop {3xy+2x=2y}} \right. \; \left \{ {x^2+y^2=5} \atop {3xy=2y-2x}} \right. \; \left \{ {{x^2+y^2=5} \atop {3xy=2\cdot (y-x)}} \right. \; \left \{ {{x^2+y^2=5} \atop {9x^2y^2=4(y-x)^2}} \right. \\\\9x^2y^2=4y^2+4x^2-8xy\\\\9x^2y^2=4(x^2+y^2)-8xy\\\\9x^2y^2=4\cdot 5-8xy\\\\t=xy\; ,\; \; 9t^2+8t-20=0\; ,\; D/4=196\; ,\; t_1=-2,\; t_2=\frac{10}{9}\\\\ 1)\; \left \{ {{x^2+y^2=5} \atop {xy=-2}} \right. \quad ili \quad 2)\;\left \{ {{x^2+y^2=5} \atop {xy=\frac{10}{9}}} \right.$

$1)\; \; \left \{ {{x^2+2xy+y^2=5+2\cdot (-2)} \atop {xy=-2}} \right. \; \left \{ {{(x+y)^2=1} \atop {xy=-2}} \right. \; \left \{ {{x+y=\pm 1} \atop {xy=-2}} \right. \\\\a) \left \{ {x+{y=1} \atop {xy=-2}} \right. \; \left \{ {{y=1-x} \atop {x(1-x)=-2}} \right. \; \left \{ {{y=1-x} \atop {x^2-x-2=0}} \right. \; \left \{ {{y_1=2,\; y_2=-1} \atop {x_1=-1},\; x_2=2} \right. \\\\b)\; \left \{ {{x+y=-1} \atop {xy=-2}} \right. \; \left \{ {{y=-1-x} \atop {x(-1-x)=-2}} \right.$

$\left \{ {{y=-1-x} \atop {x^2+x-2=0}} \right. \; \left \{ {{y_1=1,\; y_2=-2} \atop {x_1=-2,\; x_2=1}} \right. \\\\2)\; \; \left \{ {{x^2+2xy+y^2=5+2\cdot \frac{10}{9}} \atop {xy=\frac{10}{9}}} \right. \; \left \{ {{(x+y)^2=\frac{65}{9}} \atop {xy=\frac{10}{9}}} \right. \; \left \{ {{x+y=\pm \frac{\sqrt{65}}{3}} \atop {xy=\frac{10}{9}}} \right.$

$a)\; \left \{ {{x+y=\frac{\sqrt{65}}{3}} \atop {xy=\frac{10}{9}}} \right. \; \left \{ {{y=\frac{\sqrt{65}}{3}-x} \atop {x(\frac{\sqrt{65}}{3}-x)=\frac{10}{9}}} \right.$

$x^2-\frac{\sqrt{65}}{3}x+\frac{10}{9}=0\, |\cdot 9\\\\9x^2-3\sqrt{65}x+10=0\\\\D=9\cdot 65-4\cdot 9\cdot 10=225\; ,\; \sqrt{D}=15\\\\x_1=\frac{3\sqrt{65}-15}{18}= \frac{\sqrt{65}-5}{6} \; ,\; \; x_2= \frac{\sqrt{65}+5}{6} \\\\y_1=\frac{\sqrt{65}}{3}- \frac{\sqrt{65}-5}{6} = \frac{\sqrt{65}+5}{6} \; ,\; \; y_2=\frac{\sqrt{65}}{3}- \frac{\sqrt{65}+5}{6}= \frac{\sqrt{65}-5}{6}$

$b)\; \left \{ {{x+y=-\frac{\sqrt{65}}{3}} \atop {xy=\frac{10}{9}}} \right. \left \{ {{y=-\frac{\sqrt{65}}{3}-x} \atop {xy=\frac{10}{9}} \right.$ 

$x(-\frac{\sqrt{65}}{3}-x)=\frac{10}{9}\\\\x^2+\frac{\sqrt{65}}{3}x+\frac{10}{9}=0\, |\cdot 9\\\\9x^2+3\sqrt{65}x+10=0\; ,\; D=9\cdot 65-4\cdot 9\cdot 10=225\\\\x_1= \frac{-3\sqrt{65}-15}{18}= \frac{-\sqrt{65}-5}{6}\; ,\; x_2= \frac{-\sqrt{65}+5}{6} \\\\y_1=-\frac{\sqrt{65}}{3}-\frac{-\sqrt{65}-5}{6}= \frac{-\sqrt{65}+5}{6} \; ,\; y_2=-\frac{\sqrt{65}}{3}- \frac{-\sqrt{65}+5}{6}= \frac{-\sqrt{65}-5}{6}$

$Otvet:\; (-1,2)\; ,\; (2,-1)\; ,\; (-2,1)\; ,\; (1,-2)\; ,\\\\\left( \frac{\sqrt{65}-5}{6} , \frac{\sqrt{65}+5}6} \right)\; ,\; \left ( \frac{\sqrt{65}+5}{6} , \frac{\sqrt{65}-5}{6} \right )\; ,$

$\left ( \frac{-\sqrt{65}-5}{6} , \frac{-\sqrt{65}+5}{6} \right )\; ,\; \left ( \frac{-\sqrt{65}+5}{6} , \frac{-\sqrt{65}-5}{6}\right )$