Question

Решить систему
X^2+y^2=37
Xy=6

Answer 1

$\left \{ {{x^2+y^2=37} \atop {xy=6}} \right. \left \{ {{x^2+y^2=37} \atop {y= \frac{6}{x} }} \right. \left \{ {{x^2+( \frac{6}{x})^2=37} \atop {y= \frac{6}{x} }} \right. \left \{ {{x^4-37x^2+36=0} \atop {y= \frac{6}{x}}} \right. \\ \\ t=x^2,t \geq 0 \\ \\ t^2-37t+36=0;D=1369-144=1225=35^2 \\ \\ 1) t_{1} =(37+35)/2=36=6^2; \left \{ {{ x_{1}=6 } \atop { y_{1} =1}} \right. ; \left \{ {{ x_{2}=-6 } \atop { y_{2} =-1}} \right.$
$2) t_{2} =(37-35)/2=1; \left \{ {{ x_{3}=1 } \atop { y_{3} =6}} \right. ; \left \{ {{ x_{4}=-1 } \atop { y_{4} =-6}} \right. \\ \\ otvet:(6;1)(-6;-1)(1;6)(-1;-6)$



$\left \{ {{x^2+y^2=37} \atop {xy=6}} \right. \left \{ {{x^2+2xy+y^2-2xy=37} \atop {xy=6}} \right. \left \{ {{(x+y)^2-2xy=37} \atop {xy=6}} \right. \left \{ {{(x+y)^2-12=37} \atop {xy=6}} \right. \\ \\ \left \{ {{(x+y)^2-49=0} \atop {xy=6}} \right. \left \{ {{(x+y+7)(x+y-7)=0} \atop {xy=6}} \right. \\ \\ 1)\left \{ {{x+y+7=0} \atop {xy=6}} \right. \left \{ {{y=-x-7} \atop {xy=6}} \right. \left \{ {{y=-x-7} \atop {x(-x-7)=6}} \right.\left \{ {{y=-x-7} \atop {x^2+7x+6=0}} \right.$
D=49-24=25
$\left \{ {{ x_{1}=(-7-5)/2=-6 } \atop { y_{1}=6-7=-1 }} \right. \left \{ {{ x_{2}=(-7+5)/2=-1 } \atop { y_{1}=1-7=-6 }} \right. \\ \\ 2)\left \{ {{x+y-7=0} \atop {xy=6}} \right. \left \{ {{y=7-x} \atop {xy=6}} \right. \left \{ {{y=7-x} \atop {x(7-x)=6}} \right.\left \{ {{y=7-x} \atop {x^2-7x+6=0}} \right. \\ D=25 \\ \\ \left \{ {{ x_{3}=(7-5)/2=1 } \atop { y_{3}=7-1=6 }} \right. \left \{ {{ x_{4}=(7+5)/2=6} \atop { y_{4}=7-6=1 }} \right.$