Question

ПОМОгИТЕ ПЛИЗ $\left \{ {{\frac{5}{x^{2}+xy }+\frac{4}{y^2+xy}=\frac{13}{6} } \atop {\frac{8}{x^{2}+xy}-\frac{1}{y^2+xy} =1}} \right.$

Answer 1

$\left \{ {{\frac{5}{x^2+xy}+\frac{4}{y^2+xy}=\frac{13}{6} } \atop {{{\frac{8}{x^2+xy}-\frac{1}{y^2+xy}=1\\ }} \right. \\ \\ a={\frac{1\\ }{x^2+xy};b=\frac{1}{y^2+xy};\\ \\ odz: x\neq 0;y\neq 0;x\neq -y$

$\left \{ {{\frac{5}{a} +\frac{4}{b}=\frac{13}{6}} \atop{{\frac{8}{a} -\frac{1}{b}=1}} \right. \\ \\ \left \{ {{6(5b+4a)=13ab} \atop {8b-a=ab}} \right. \\ \\ \left \{ {{6(5b+4a)=13ab} \atop {-104b+13a=-13ab}} \right.\\ --------\\ -74b+37a=0\\ a=2b$

$\frac{1}{x^2+xy} =\frac{2}{y^2+xy} \\ \\ 2x^2+2xy=y^2+xy\\ \\ y^2-xy-2y^2=0\\ \\ D=9y^2=(3y)^2\\ \\ 1)x=(y+3y)/2=2y\\ \\ \frac{8}{4y^2+2y^2} -\frac{1}{y^2+2y^2} =1\\ \\ \frac{3}{3y^2} =1;y^2=1\\ \\ y_{1} =-1;x_{1} =-2\\ \\ y_{2} =1;x_{2} =2\\ \\ 2)x=(y-3y)/2=-y$

не удовлетворяет одз

ОТВЕТ: (-2;-1);(2;1)