Question

решите систему уравнений под номером 11

Answer 1

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

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