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

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

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

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

не удовлетворяет одз

ОТВЕТ: (-2;-1);(2;1)