решите систему уравнений под номером 11

$\left \ \fracx^2+4y^2xy=5 \atop 2x^2-y^2=31 \right. \; \; ODZ:\; x\ne 0\; ,\; y\ne 0\\\\\left \ x^2+4y^2=5xy \atop 2x^2-y^2=31 \right. \; \; \left \ x^2+4y^2-5xy=0 \atop 2x^2-y^2=31 \right. \\\\x^2+4y^2-5xy=(x^2+y^2-2xy)+3y^2-3xy=(y-x)^2+3y(y-x)\\=(y-x)\cdot (y-x+3y)=(y-x)(4y-x)=0\; \; \Rightarrow \\\\y-x=0\; \; ili\; \; \; 4y-x=0\; \; \Rightarrow \; \; x=y\; \; ili\; \; x=4y\\\\a)\; \left \ x=y \atop 2x^2-y^2=31 \right. \; \left \ x=y \atop 2y^2-y^2=31 \right. \left \ x=y \atop y^2=31 \right. \; \left \ x=y \atop y=\pm \sqrt31 \right.$

$\underline (-\sqrt31,-\sqrt31)\; ,\; \; (\sqrt31,\sqrt31)\\\\b)\; \; \left \ x=4y \atop 2x^2-y^2=31 \right. \; \left \ x=4y \atop 2\cdot 16y^2-y^2=31 \right.\; \left \ x=4y \atop 31y^2=31 \right. \; \left \ x=4y \atop y^2=1 \right. \; \left \ x=\pm 4 \atop y=\pm 1 \right.\\\\\underline (-1,-4)\; ,\; \; (1,4)\\\\Otvet:\; \; (-\sqrt31,-\sqrt31)\; ,\; (\sqrt31\; ,\; \sqrt31)\; ,(-1,-4)\; ,\; (1,4)\; .$

О проекте

решите систему уравнений под номером 11

Добро пожаловать!