Question

№16 Решить систему уравнений

Answer 1

$displaystyle left { {{x+y=14}atop {frac{x}{y}+frac{y}{x}=2frac{1}{12}}}right.\\left { {{x=14-y} atop {frac{x^{2}+y^{2}}{xy}=2frac{1}{12}}}right.\\\x^{2}+y^{2}=2 frac{1}{12}xy\\x^{2}-2xy+y^{2}=frac{xy}{12}\\12(x-y)^{2}=xy\12(14-2y)^{2}=y(14-y)\12(196-56y+4y^{2})-14y+y^{2}=0 \2352-672y+48y^{2}-14y+y^{2}=0\49y^{2}-686y+2352=0 \ y^{2}-14y+48=0\\D=b^{2}-4ac=196-192=4\\y_{1,2}= frac{-bб sqrt{D}}{2a} \ \y_{1}=8 \ y_{2}=6 \ \ x_{1}=14-y_{1}=6\x_{2}=14-y_{2}=8$

Ответ: {(6;8),(8;6)}