Question

решите систему уравнений
$x {}^{2} + y {}^{2} = 4 \\ xy = 1$

Answer 1

$\left \{ {{x^2+y^2=4} \atop {xy=1}} \right. \\\\(x+y)^2=\underbrace {x^2+y^2}_{4}+\underbrace {2xy}_{2\cdot 1}=4+2\cdot 1=6\; \; \to \; \; \left \{ {{x+y=\pm \sqrt6} \atop {xy=1}} \right. \\\\a)\; \; \left \{ {{x+y=\sqrt6} \atop {xy=1}} \right. \; \left \{ {{y=\sqrt6-x} \atop {x(\sqrt6-x)=1}} \right. \; \left \{ {{y=\sqrt6-x} \atop {x^2-x\sqrt6+1=0}} \right. \\\\x^2-\sqrt6\cdot x+1=0\; ,\; \; D=6-4=2\; ,\\\\x_1= \frac{\sqrt6-\sqrt2}{2}\; ,\; \; x_2= \frac{\sqrt6+\sqrt2}{2}$

$y_1=\sqrt6-\frac{\sqrt6-\sqrt2}{2}=\frac{\sqrt6+\sqrt2}{2}\\\\y_2=\sqrt6-\frac{\sqrt6+\sqrt2}{2}=\frac{\sqrt6-\sqrt2}{2}\\\\b)\; \; \left \{ {{x+y=-\sqrt6} \atop {xy=1}} \right. \; \left \{ {{y=-\sqrt6-x} \atop {x(-\sqrt6-x)=1}} \right. \; \left \{ {{y=-\sqrt6-x} \atop {x^2+\sqrt6\cdot x+1=0}} \right. \\\\x^2+\sqrt6\cdot x+1=0\; ,\; \; D=6-4=2\\\\x_1=\frac{-\sqrt6-\sqrt2}{2}\; ,\; \; x_2=\frac{-\sqrt6+\sqrt2}{2}\\\\y_1=-\sqrt6-\frac{-\sqrt6-\sqrt2}{2}=\frac{-\sqrt6+\sqrt2}{2} \\\\y_2=-\sqrt6-\frac{-\sqrt6+\sqrt2}{2}=\frac{-\sqrt6+\sqrt2}{2}$

$Otvet:\; \; ( \frac{\sqrt6-\sqrt2}{2};\frac{\sqrt6+\sqrt2}{2})\; ,\; (\frac{\sqrt6+\sqrt2}{2};\frac{\sqrt6-\sqrt2}{2})\; ,\\\\(\frac{-\sqrt6-\sqrt2}{2}; \frac{-\sqrt6+\sqrt2}{2})\; ,\; (\frac{-\sqrt6+\sqrt2}{2}; \frac{-\sqrt6+\sqrt2}{2})\; .$