Question

решите систему уравнений под номером 11

Answer 1

$\left \{ {{ \frac{x^2+4y^2}{xy}=5} \atop {2x^2-y^2=31}} \right. \; \; ODZ:\; x\ne 0\; ,\; y\ne 0\\\\\left \{ {{x^2+4y^2=5xy} \atop {2x^2-y^2=31}} \right. \; \; \left \{ {{x^2+4y^2-5xy=0} \atop {2x^2-y^2=31}} \right. \\\\x^2+4y^2-5xy=(x^2+y^2-2xy)+3y^2-3xy=(y-x)^2+3y(y-x)\\=(y-x)\cdot (y-x+3y)=(y-x)(4y-x)=0\; \; \Rightarrow \\\\y-x=0\; \; ili\; \; \; 4y-x=0\; \; \Rightarrow \; \; x=y\; \; ili\; \; x=4y\\\\a)\; \left \{ {{x=y} \atop {2x^2-y^2=31}} \right. \; \left \{ {{x=y} \atop {2y^2-y^2=31}} \right. \left \{ {{x=y} \atop {y^2=31}} \right. \; \left \{ {{x=y} \atop {y=\pm \sqrt{31}}} \right.$

$\underline {(-\sqrt{31},-\sqrt{31})\; ,\; \; (\sqrt{31},\sqrt{31})}\\\\b)\; \; \left \{ {{x=4y} \atop {2x^2-y^2=31}} \right. \; \left \{ {{x=4y} \atop {2\cdot 16y^2-y^2=31}} \right.\; \left \{ {{x=4y} \atop {31y^2=31}} \right. \; \left \{ {{x=4y} \atop {y^2=1}} \right. \; \left \{ {{x=\pm 4} \atop {y=\pm 1}} \right.\\\\\underline {(-1,-4)\; ,\; \; (1,4)}\\\\Otvet:\; \; (-\sqrt{31},-\sqrt{31})\; ,\; (\sqrt{31}\; ,\; \sqrt{31})\; ,(-1,-4)\; ,\; (1,4)\; .$