Question

x^2+3xy+2y^2=3 2x^2-xy+y^2=8 система уравнений, помогите решить

Answer 1

$left { {{x^2+3xy+2y^2=3, |cdot 8} atop {2x^2-xy+y^2=8, |cdot (-3)}} right.; ; left { {{8x^2+24xy+16y^2=24} atop {-6x^2+3xy-3y^2=-24}} right. oplus ; left { {{x^2+3xy+2y^2=3} atop {2x^2+27xy+13y^2=0}} right.\\\2x^2+27xy+13y^2=0, |:y^2ne 0\\2cdot (frac{x}{y})^2+27cdot (frac{x}{y})+13=0; ; ,; ; t=frac{x}{y}\\2t^2+27t+13=0; ; ,; ; D=625; ,; t_1=-frac{1}{2}; ; ,; ; t_2=-13\\a); frac{x}{y}=-frac{1}{2}; ; to ; ; y=-2x; ; ,; ; x^2+3xy+2y^2=x^2-6x^2+8x^2=3x^2$

$3x^2=3; ; to ; ; x^2=1; ,; ; x_1=-1; ,; x_2=+1\\y_1=-2cdot (-1)=2; ,; ; y_2=-2cdot 1=-2\\b); ; frac{x}{y}=-13; ; to ; ; x=-13y; ,; ; x^2+3xy+2y^2=169y^2-39y^2+2y^2=132y^2\\132y^2=3; ,; ; y^2=frac{1}{44} ; ; to ; ; y_1=-frac{1}{2sqrt{11}}; ,; y_2=+frac{1}{2sqrt{11}}\\x_1=frac{13}{2sqrt{11}}; ; ,; ; x_2=-frac{13}{2sqrt{11}}\\Otvet:; ; (-1,2); ,; (1,-2); ,; (frac{13}{2sqrt{11}},- frac{1}{2sqrt{11}}); ,; (-frac{13}{2sqrt{11}},frac{1}{2sqrt{11}}); .$

$P.S.; ; frac{1}{2sqrt{11}}=frac{sqrt{11}}{22}; ; ,; ; frac{13}{2sqrt{11}}=frac{13sqrt{11}}{22}; .$