Question

 помогите решить систему уравнений
(х+у=41
(√х÷у +√у÷х =3

Answer 1

$left { {{x+y=41} atop {sqrt{frac{x}{y}}+sqrt{frac{y}{x}}=3}} right.\\t=sqrt{frac{x}{y}},; frac{1}{t}=sqrt{frac{y}{x}},; xne 0,; yne 0,; t>0\\t+frac{1}{t}=3,; t^2-3t+1=0,\\D=9-4=5,; t_1=frac{3-sqrt5}{2},; t_2=frac{3+sqrt5}{2}\\sqrt{frac{x}{y}}=frac{3-sqrt5}{2},; frac{x}{y}=(frac{3-sqrt5}{2})^2=frac{14-6sqrt5}{4}\\x+y=41; to ; y=41-x; to ; frac{x}{41-x}=frac{14-6sqrt5}{4}\\4x=(41-x)(14-6sqrt5)\\4x=574-246sqrrt5-14x+6sqrt5x$

$(18-6sqrt5)x=574-246sqrt5,; ; x_1=frac{574-246sqrt5}{18-6sqrt5}=frac{287-123sqrt5}{9-3sqrt5}\\y_1=41-frac{287-1123sqrt5}{9-3sqrt5}\\sqrt{frac{x}{y}}=frac{3+sqrt5}{2}\\frac{x}{y}=(frac{3+sqrt5}{2})^2\\frac{x}{41-x}=frac{14+6sqrt5}{4}\\4x=(41-x)(14+6sqrt5)\\4x=574+246sqrt5-14x-6sqrt5x\\(18+6sqrt5)x=574+246sqrt5\\x_2=frac{287+123sqrt5}{9+3sqrt5}$

$y_2=41-frac{287+123sqrt5}{9+3sqrt5}$