Question

Решите подробно систему уравнений:
x^2 +y^2 = 27
xy=6

Answer 1

$\left \{ {{x^2+y^2=27} \atop {xy=6}} \right.\; \left \{ {{x^2+2xy+y^2=27+2\cdot 6} \atop {y=6}} \right. \; \left \{ {{(x+y)^2=39} \atop {xy=6}} \right. \; \left \{ {{x+y=\pm \sqrt{39}} \atop {xy=6\; \; \; }} \right.\\\\1)\; \; x+y=\sqrt{39}\; ,\; \; y=x-\sqrt{39}\; ,\\\\x(x-\sqrt{39})=6\; ,\; \; x^2-x\sqrt{39}-6=0\; ,\; \; D=39+24=63\; ,\\\\x_1=\frac{\sqrt{39}-\sqrt{63}}{2}=\frac{\sqrt3\cdot (\sqrt{13}-\sqrt{21})}{2}\; ,\; \; x_2=\frac{\sqrt{39}+\sqrt{63}}{2}=\frac{\sqrt3\cdot (\sqrt{13}+\sqrt{21})}{2}$

$y_1=x_1-\sqrt{39}=\frac{\sqrt{39}-\sqrt{63}}{2}-\sqrt{39}=-\frac{\sqrt{39}+\sqrt{63}}{2}=-\frac{\sqrt3\cdot (\sqrt{13}+\sqrt{21})}{2}\\\\y_2=x_2-\sqrt{39}=\frac{\sqrt{39}+\sqrt{63}}{2}-\sqrt{39}=\frac{-\sqrt{39}+\sqrt{63}}{2}=\frac{\sqrt3\cdot (\sqrt{21}-\sqrt{13})}{2}\\\\2)\; \; x+y=-\sqrt{39}\; \; ,\; \; y=-x-\sqrt{39}\\\\x(-x-\sqrt{39})=6\; \; ,\; \; x^2+x\sqrt{39}+6=0\; ,\; \; D=39-24=15\; ,\\\\x_1=\frac{-\sqrt{39}-\sqrt{15}}{2}=-\frac{\sqrt3\cdot (\sqrt{13}+\sqrt5)}{2}\; ,\; \; x_2=\frac{-\sqrt{39}+\sqrt{15}}{2}=\frac{\sqrt3\cdot (\sqrt5-\sqrt{13})}{2}$

$y_1=-x_1-\sqrt{39}=\frac{\sqrt{39}+\sqrt{15}}{2}-\sqrt{39}=\frac{-\sqrt{39}+\sqrt{15}}{2}=\frac{\sqrt3\cdot (\sqrt5-\sqrt{13})}{2}\\\\y_2=-x_2-\sqrt{39}=\frac{\sqrt{39}-\sqrt{15}}{2}-\sqrt{39}=\frac{-\sqrt{39}-\sqrt{15}}{2}=-\frac{\sqrt3\cdot (\sqrt5+\sqrt{13})}{2}\\\\Otvet:\; \; \Big (\frac{\sqrt3\cdot (\sqrt{13}-\sqrt{21})}{2}\, ;\, -\frac{\sqrt3\cdot (\sqrt{13}+\sqrt{21})}{2}\Big )\; ,\; \Big (\frac{\sqrt3\cdot (\sqrt{13}+\sqrt{21})}{2}\, ,\, \frac{\sqrt3\cdot (\sqrt{21}-\sqrt{13})}{2}\Big )\; ,$

$\Big (-\frac{\sqrt3\cdot (\sqrt{5}+\sqrt{13})}{2}\, ,\, \frac{\sqrt3\cdot (\sqrt5-\sqrt{13})}{2}\Big )\; ,\; \Big (\frac{\sqrt3\cdot (\sqrt5-\sqrt{13})}{2}\, ,\, -\frac{\sqrt3\cdot (\sqrt5+\sqrt{13})}{2}\Big )\, .$