Question

Решите систему: {3xy+2=y^3/x {2xy+1=x^3/y

Answer 1

$left { {{3xy+2= frac{y^3}{x} , |cdot x} atop {2xy+1= frac{x^3}{y}, |cdot y}} right. ; ; ; ; ODZ:; ; ; xne 0; ,; ; yne 0\\left { {{3x^2y+2x=y^3} atop {2xy^2+y=x^3}} right. ; ; (1times 2) ; ; left { {{(3x^2y+2x)cdot (2xy^2+y)=x^3y^3} atop {2xy^2+y=x^3}} right. \\ left { {{6x^3y^3+3x^2y^2+4x^2y^2+2xy=x^3y^3} atop {2xy^2+y=x^3}} right. ; ; left { {{5(xy)^3+7(xy)^2+2(xy)=0} atop {2xy^2+y-x^3=0}} right. \\t=xyne 0; ,; ; ; 5t^3+7t^2+2t=0; ,; ; tcdot (5t^2+7t+2)=0$

$tne 0; ; ; ili; ; ; 5t^2+7t+2=0\\D=49-4cdot 5cdot 2=9\\t_1=frac{-7-3}{10}=-1; ; ,; ; t_2=frac{-7+3}{10}=frac{-4}{10}=- frac{2}{5} \\a); ; xy=-1; ,; ; y=-frac{1}{x}\\2xy^2+y-x^3=2xcdot (-frac{1}{x})^2-frac{1}{x}-x^3=frac{2x}{x^2}-frac{1}{x}-x^3=\\=frac{2}{x}-frac{1}{x}-x^3=frac{1}{x}-x^3=frac{1-x^4}{x}\\frac{1-x^4}{x}=0; ; ; Rightarrow quad left { {{1-x^4=0} atop {xne 0}} right. \\1-x^4=0; ,; ; (1-x^2)(1+x^2)=0; ,; ; (1-x)(1+x)(1+x^2)=0; to$

$1-x=0; ; ; ili; ; ; 1+x=0; ; ; (1+x^2 textgreater 0)\\x_1=1; ,; ; x_2=-1\\y_1=-frac{1}{x}=-1; ,; ; y_2=1\\b); ; xy=-frac{2}{5}; ,; ; ; y=-frac{2}{5x}\\2xy^2+y-x^3=2xcdot (-frac{2}{5x})^2-frac{2}{5x}-x^3=2xcdot frac{4}{25, x^2}-frac{2}{5x}-x^3=\\=frac{8}{25, x}-frac{2}{5x}-x^3=frac{8-10-25x^4}{25x}=frac{-2-25x^4}{25x} =-frac{25x^4+2}{25x} \\-frac{25x^4+2}{25x}=0; ; Rightarrow ; ; left { {{25x^4+2=0} atop {xne 0}} right. ; ; left { {{x^4=-frac{2}{25}} atop {xne 0}} right.$

$Tak; kak; ; x^4 geq 0; ,; to ; ; x^4ne - frac{2}{25}; .\\Resenij; sistema; ne; imeet; .\\Otvet:; ; (-1,1); ,; ; (1,-1); .$